Mathematics Guided Practice - THIS CITE IS KEPT SIMPLE:)historywithmac.weebly.com/uploads/5/9/7/4/59744935/mathgpb.pdf · of multiplication in this way. Conceptually, multiplication

MathematicsGuided Practice

2015-16: IndiaAuthors

Nick Cates, Nick SynodisEditor

Stephen Bergauer

M a t h e m a t i c s C u r r i c u l u m G u i d e | 2

INTRODUCTION OFTEN the most despised, feared, and overlooked event in Academic Decathlon, the mathematics section

is extremely fast-paced, covering material far removed from the theme (though Russia is extremely well-known for its mathematicians). This year’s mathematics curriculum includes material regarding permutations and combinations (10%), algebra (40%), and statistics (50%). That means that around 17-18 questions will be about statistics, 3-4 will be about permutations and combinations, and the remaining 14 or so will be about algebra – plan accordingly.

Because the math subject is so different, this guide is structured differently than any of our other guides (and any other guides you may have used in the past). We take the subjects that need to be covered and present them in an easy-to-digest fashion, giving intuitive examples and relevant practice problems that the Resource Guide neglects. There won’t be proofs or heavy mathematical rigor in this guide, because that’s not what you need to learn. Instead, you will notice the inclusion of relevant real-life examples and supplementary graphics that will aid you on your journey toward mathematical expertise.

Keeping that in mind, here is an organized list of the unique features of this guide:

A Calculator Guide: Because this test goes by so fast, being able to use your calculator is absolutely

essential. We have included relevant tables of calculator commands as well as special guides throughout of calculator functions and problem-solving techniques. A “Calculator Help” section will come after appropriate subjects (a glossary for calculator commands is at the end of the guide)::

Timed Quizzes and Exams: Did we mention that this test goes by quickly? It really does. After every few sections, there are practice problems with wording similar to that of tests you will see in competitions. We try to keep the “35 questions in 30 minutes” ratio steady so you know how quickly you need to pace yourself on actual mathematics exams—you should take 6 minutes to take a 7-question quiz.

Video Lecture Links: In each section, you will find a clickable YouTube link to the related Decademy Video Lecture for the subject you’re covering. These videos were created by someone not involved with the creation of this guide, and it’s always nice to get two points of view!

Independence of Each Section: Also in each section is a “prerequisites” section. You can start pretty much anywhere in the guide, given that you understand the prerequisites. It is not necessary to go in order.

With all that being said, I wish you the best of luck!

Some last tips:

Start with Statistics: If you have already gone through a bit of high school math, you likely will have a basic understanding of Sections 1 and 2 of this guide. Section 3 will be foreign to most and is the most tested section, so if you’re in a crunch, start there! Also, keep in mind that many students don’t ever bother studying math – so if you can even get a few stats questions that others don’t, you already have an advantage of several hundred points.

3 | M a t h e m a t i c s C u r r i c u l u m G u i d e

Use your Calculator, but not too much! While it is important that you understand all the formulas, rules, and definitions in this guide1, the more adept you are with using your calculator, the more success you will have on the actual tests. However, if you become reliant on your calculator, there might be questions that require understanding and not calculation, so be wary.

Use each relevant section on the quizzes: This is a guided exercise book, not a test! While it will slow you down (you need to be moving quickly on all mathematics quizzes/exams), practicing with the formulas in front of you will help you internalize them, which is the point of this guide. So if you’re taking a quiz on sections 3.7-3.9, have those pages in front of you for reference, but don’t sit there and read them—you won’t be able to finish the quizzes.

Lastly, don’t panic! This is the first year in a while where it is completely possible to get a high score on the mathematics exam without much prior knowledge, and I’m hoping this guide will get you there! Nick Synodis, President, Decademy

About the Authors

NICK SYNODIS is a former Decathlete from Mountain View Mesa High School in Mesa, Arizona. As team captain, Nick wrote thousands of questions as well as outlines and workbooks in order to prepare his state-runner-up team for Decathlon. He was a two-year Honors, scoring over 8,300 points at the 2011 Arizona State competition. He is currently attending Arizona State University as a Math major with an emphasis in Mathematics Education. NICHOLAS CATES (Calculator Guides) is a former Decathlete from Red Mountain High School in Mesa, Arizona and is currently studying Biomedical Engineering at Arizona State University. During his two years of Decathlon, he earned a math medal at every competition in which he competed. He has remained involved with Decathlon and currently serves as the scoring chair for Arizona Region IV.

About the Editor

STEPHEN BERGAUER is a former three-year Decathlete from Hamilton High School in Chandler, Arizona. He was the overall individual champion at the 2012 Arizona State competition, recording a top score of 9,051 points. He is attending Arizona State University, double-majoring in Math and Finance.

1 Hey, two math majors have to try somewhat, right?


COMBINATORICS TABLE OF CONTENTS

Section 1.1: Multiplication Principle…………………………

5

Section 1.2: Permutations…………………………………….

6

Quiz 1.1-1.2: Multiplication Principle and Permutations….

10

Section 1.3: Combinations……………………………………

11

Section 1 Final Exam.………………………………………….

14


SECTION 1.1: THE MULTIPLICATION PRINCIPLE Multiplication seems like a very basic concept, but it is actually highly nuanced. In elementary school, it is often introduced as shorthand for repeated addition. That is, expressions like 5+5+5+5+5+5 can be rewritten as 5*6 = 30. Although some mathematicians disagree with is definition, for the sake of simplicity we will think of multiplication in this way.

Conceptually, multiplication is important because real-life situations often involve some form of repeated addition. For example, if a person has 7 shirts, 5 pairs of pants, and 3 hats, then that person could wear 7*5*3 = 105 unique outfits made up of 1 shirt, 1 pair of pants, and 1 hat. Without using multiplication, you could sum fifteen sevens, seven fifteens, or even thirty-five threes—all of which would be quite tedious.

This is where the Multiplication Principle comes in handy. The essential idea is that if you have items (e.g., shirts, pants, or hats), each of which has a certain number of possibilities, then the total number of possibilities for a certain number of items is the product of the numbers of possibilities for each item. More formally, this product is n1*n2*n3 … nk where nk is the number of possibilities for the kth item.

Remember that this is just a generalization. Problems involving the Multiplication Principle can often be solved intuitively – you just need to figure out the possibilities and then multiply.

Let’s look at some examples. Suppose you want to call a friend, but you have no way of looking up his phone number. If you remember the first three digits of your friend’s seven-digit phone number, what is the maximum number of phone numbers will you have to call before you find the correct number?

The probability of getting the first three digits right is 1 (or 100%) – you already know what they are. Instead, you’re only concerned with the last four. So, you have four “items” or “slots,” and there are 10 numbers (0-9) that can fill each of the last four slots. Using the Multiplication Principle, you have 10*10*10*10 = 10,000 total possibilities. Since your friend’s number is among these possibilities, at most you will call 9,999 phone numbers before you find the correct one.

Let’s look at one more. A pizza parlor offers three pizza sizes, two different cheeses, and seven different toppings. How many unique pizzas can you order if you want a pizza of any size with both cheeses and one topping?

Be sure to read the question carefully on this one. You want one size, both cheeses, and one topping. Therefore, your choice of cheese does not factor into the calculation of total possibilities. You can thus order 3*7 = 21 different pizzas.

Related video lecture:

http://youtu.be/b4rhCTp2Wm8

Suggested prior reading: none


SECTION 1.2: PERMUTATIONS In standard English, we tend to use the word “combinations” to mean any arrangement or collection of items, whether “ordered” or not. For example, when you go to a fast-food restaurant and ask for a “combo” meal of a hamburger, fries, and soda, you could receive those three items in any order – whether the burger comes first or last doesn’t matter, as long as you get all three. However, we also refer to a lock on a safe as a “combination dial” – even though the numbers “1234” will not unlock a safe coded for “4321.”

In English, this confusion is not very important; however, it makes a big deal in math and probability. We therefore divide the two types of “combinations” into two categories: ordered and unordered. The first category is known as permutations – in permutations, order matters.

If, for example, you have three marbles—one red, one blue, and one yellow— and you order them in every possible way, then you will have six distinct permutations:

red, blue, yellow

red, yellow, blue

blue, red, yellow

blue, yellow, red

yellow, red, blue

yellow, blue, red

The number of different orders can be determined mathematically, too. Think of the marbles as filling three “slots.” There are three options for the first slot: red, blue, or yellow. One marble is chosen for the first slot, so it cannot be chosen for the second slot. There are now two options for the second slot. By the same logic, there is just one option for the last slot. Using the Multiplication Principle gives us 3*2*1 = 6 total possibilities – the same number that we got by listing all the possibilities.

The expression 3*2*1 is an example of a special product called the factorial, which is denoted by the symbol “!”2. The factorial is defined as the product of all whole numbers starting with the number before the “!” down to 1. The product 3*2*1 is 3!, read as “3 factorial.”

Let’s take a moment to really work with factorials, since they will be foreign to most students. We need a general formula, so, let’s look at some factorial expressions and see how they work.

Consider 25!, again, read “25 factorial.” Taking a look at our example from earlier, we should be able to figure out that 25!= 25⋅24 ⋅23⋅22 ⋅21⋅20 ⋅19 ⋅18 ⋅…⋅3⋅2 ⋅1. What if we tried to generalize this, and called 25 “n”? 24 is just n – 1, then, right? And 23 is n – 2?

2 A favorite anecdote of my calculus teacher’s was of a student who, upon first seeing a factorial, decided that it meant to say the number loudly, like in English. So, instead of “four factorial,” he would say “FOUR!”


http://youtu.be/FNT1eKnio4E

Suggested prior reading: 1.1


Continue this product and go all the way down to n – 23 and then n – 24 (which is 1), and you have yourself a formalized definition/”formula for factorial.

n!= n ⋅(n−1)⋅(n−2)⋅(n−3)⋅…⋅1

Basically, though this definition of factorial is not necessarily rigorous3, it will help you get an intuition for

“cancelling” factorials. For example, suppose you have

(n−1)!(n−3)!

. By our definition of factorial, we know that

this means

(n−1)!(n−3)!

= (n−1)(n−2)(n−3)(n− 4)(n−5)…1(n−3)(n− 4)(n−5)(n− 6)…1

. Look! Stuff cancels! Everything “(n-3)” and on is

contained in both the numerator and the denominator, so you can cancel, which leaves (n-1)(n-2) as the simplified form of this expression.

If you’re missing the intuition for this, simply consider some dividing factorials of actual numbers; if we want

to simplify the expression

7!4!

, we simply apply the definition of factorial and cancel, so

7!4!

= 7 ⋅6 ⋅5⋅4 ⋅3⋅2 ⋅14 ⋅3⋅2 ⋅1

= 7 ⋅6 ⋅5 = 210.

Understanding factorial notation is crucial because it is used to express the formula for permutations:

n Pk =

n!(n− k)!

where k is the number of items chosen from n total items.4

Let’s try an example to see how this formula is derived. Imagine you want to create a password that is six characters long, and you can choose from all 26 letters and numbers 0-9, with no repeated letters or numbers. The total number of items is 26+10 = 36, and you are choosing 6 of those items. For the first character, there are 36 possibilities; for the second, 35; for the third, 34—and so on. Using the Multiplication Principle yields 36*35*34*33*32*31 because there are six slots available.

The trick is seeing this product as the quotient of two factorials, namely

36!30!

Why is this true? 36! can be rewritten as 36*35*34*33*32*31*30!, so dividing by 30! will give the original result. Notice that for n=36 total items and k=6 chosen items, 30! is indeed (n-k)!. So, we have the same permutations formula as above. Another important aspect of permutations is that no item in the group of items can be used more than once. If this weren’t the case—if repetition is allowed—then the answer to the previous problem would be 366, since possibilities would not decrease with each character.

3 Obviously, you can’t have “n-‐3” if you’re calculating 2! – Thus, use this formula to remind you that you must keep subtracting 1 until you get to 1. 4 To remember to use a permutation, I always read “nPk” as “n pick k.” This is really the only commonly-used notation for permutations; there are a few more for combinations, which are covered in the next section.


Let’s take a look at some problems relating to factorials and permutations.

1. Simplify

(n− 4)!(n−3)(n−2)!

.

This question requires knowledge of the definition of factorial. The expression above can be rearranged as

(n− 4)!(n−3)(n−2)!

= (n− 4)(n−3)!(n−3)!

, since (n-4)!=(n-4)(n-3)(n-2)…(n-1) and (n-3)(n-2)! is the same as (n-3)!;

cancelling (n-3)!’s, we are left with n-4 as our final answer.

2. Evaluate

5!6!3!4!

.

The easiest way to do this is to expand the factorials and cancel when possible. From the expression above, we

have

5!6!3!4!

= 5⋅4 ⋅3⋅2 ⋅1⋅6 ⋅5⋅4 ⋅3⋅2 ⋅13⋅2 ⋅1⋅4 ⋅3⋅2 ⋅1

= 5⋅4 ⋅6 ⋅5= 600 .

3. Simplify )!2()!2()!1()!3(

−++−nnnn

.

The expression )!2()!2()!1()!3(

−++−nnnn

can be rearranged as ⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

)!2()!1(

)!2()!3(

nn

nn

. Using the definition of the

factorial, we know that in general n! = n(n-1)(n-2)(n-3)…(3)(2)(1). The same pattern applies here. Since (n-2)! = (n-2)(n-3)! and (n+2)! = (n+2)(n+1)!, we can substitute:

(n−3)!(n−2)(n−3)!

⎛⎝⎜

⎞⎠⎟

(n+1)!(n+2)(n+1)!

⎛⎝⎜

⎞⎠⎟= 1

(n−2)(n+2)

This is the most simplified answer.

4. Prove that 0! = 1.

To prove that 0! = 1, we need to look at the relationship between 3! and 4!, 9! and 10!, or any n! and (n+1)!. Taking the first example, you know that 4! is the same as 4*3! (4! = 4*3!). Dividing by 4 on both sides yields 3! = 4!/4. Likewise, 9! = 10!/10. So, in general, n! = (n+1)!/(n+1). Plugging in 0 for n gives 0! = (0+1)!/(0+1) = 1!/1 = 1. Therefore, 0! = 1, as desired.5

5. How many four-digit integers can be formed from the digits 0-9 such that no digit is used more than once?

This problem is very commonly asked on permutation/combination exams because it tricks many students. Most students will just use the basic multiplication principles and take 10*9*8*7; however, this is incorrect because zero cannot be the first digit – if it were, we wouldn’t have a four-digit integer but a three-digit one. Thus, there are 9 options for the first digit. There are also 9 options for the second digit, since 0 can now be

5 If this doesn’t make complete sense, don’t worry – just remember that 0! is 1. Even in higher level math courses, this is simply left as “an exercise to the reader,” which math speak for “That’s the definition. Just let me teach.”


used, and one digit has already been used. There are 8 options for the third digit, and 7 options for the fourth digit. There are a total of 9*9*8*7 = 4,536 four-digit numbers. This is equivalent to 10P4 – 9*8*7; by subtracting that last term, re remove all numbers with zero as a first digit.

6. A certain number of people can be lined up 1,816,214,400 ways. If these people are chosen from a group of 15, how many people are lined up?

Notice that the problem gives the number of permutations and the number of people who can be lined up, and it asks for the number of people who are actually lined up. In other words, we have n = 15 and nPk = 1,816,214,400, and we want to find k. Therefore:

)!15(!151816214400k

Pkn −==

Some manipulation yields:

7201816214400

!15)!15( ==− k

Recall that 6! = 720 (it’s good to memorize factorials that come up often; if you don’t remember, just use your calculator). So, 15 – k = 6; k = 9. Nine people are lined up.

7. How many different unique ways can the letters in the word “POTATO” be rearranged?

At first glance, this seems like a simple multiplication principles problem – each letter can only be used once, so 6 letters gives an answer of 6!. However, there is a problem here: some of the letters are repeated. For example, there are two Ts in the word “POTATO”; the arrangements “TTPAOO,” where the third letter “T” is placed first, and “TTPAOO,” where the fifth letter “T” is placed first, are not actually unique. To eliminate the double counting, we need to divide the total number of permutations of letters by the number of permutations of each letter. For the word “POTATO,” that would be the same as dividing 6! by 1!1!2!2!, since there is 1 “P,” 1 “A,” 2 “Ts”, and 2 “Os”. This gives us 6!/4, or 180.

In summary, you should know three things about permutations:

(1) Permutations are arrangements of k items chosen from n total items in which the order of the chosen items matters.

(2) Permutations involve items that cannot be used more than once.

(3) The permutations formula:

n Pk =

n!(n− k)!

M a t h e m a t i c s C u r r i c u l u m G u i d e | 1 0

QUIZ 1.1-1.2: MULTIPLICATION PRINCIPLE AND PERMUTATIONS Suggested time for quiz: 6 minutes

Answers on page 123

1. A combination lock can be programmed to open to any 3-letter sequence. How many sequences are possible if no letter is repeated?

2. Morse code uses a sequence of dots and dashes to represent letters and words. How many sequences are possible using AT MOST 4 symbols?

3. Kathryn has 5 classic novels. She wants to arrange 3 of the 5 novels side by side. How many arrangements of 3 books are possible?

4. In a primary election, 8 candidates sought their party’s nomination for president. In how many ways could voters rank their first, second, and third choices?

5. In how many ways can the letters in the word MISSISSIPPI be arranged?

6. Bob buys 3 Cherry Colas, 2 Regular Colas, and 2 Vanilla Colas. He puts them in his refrigerator to eat one a day for the next week. Assuming that Colas of the same flavor are indistinguishable6, in how many ways can he select Colas to drink for the next week?

7. A printer has 5 A’s, 4 C’s, 2 D’s, and 2 E’s. How many different “words” (they don’t have to make sense) are possible that use all of these letters?

6 With generic soda, probably more likely than not.

1 1 | M a t h e m a t i c s C u r r i c u l u m G u i d e

SECTION 1.3: COMBINATIONS The other type of choosing, where order doesn’t matter, is a true combination.7 But combinations aren’t entirely distinct from permutations; in fact, they are related to each other.

To see what exactly this relationship is, let’s return to Section 1.2’s marble example. With three marbles (one red, one blue, and one yellow), you have 6 possible ways to order them when that order matters. But what if you don’t care what order they are arranged – instead, you just want to know which marbles are there? If you have three marbles and you choose three, there’s only one possible combination – one red, one blue, and one yellow. In fact, this pattern will always hold true: there will always be fewer combinations than permutations for a given set.

Let’s say that instead of 3 marbles in 3 slots, you have 4 marbles (one red, one blue, one yellow, and one green), and you want to pick three.. Then there are 4P3 = 24 permutations. But how many combinations are there? This time, you can have red, blue, yellow or even red, blue, green as a combination. We’ll list them all:

red, blue, yellow

red, blue, green

yellow, red, green

yellow, blue, green

Notice that for each, order is not taken into account. Also recognize that these four ways are four of the 24 permutations. What makes them combinations is the introduction of a new marble or marble configuration – basically, we change the marbles actually present, not the order of those there. So, we know that there are 24 permutations versus 4 combinations. The relating factor must be 6. But why?

Each of the four listed combinations is only one possible order of 3 items. We know that 3 items have 3! = 6 permutations. So multiplying the number of combinations by 3! should give the number of permutations. As it turns out, 4*3! does equal 24. From this, you can say 3!*4C3 = 4P3, where 4C3 is the number of combinations. Dividing gives:

4 C3 = 4 P3

3!= 4!

1!3!= 4

This allows us to generalize. If k items are chosen from n total items, the formula for combinations is as follows:

7 So, in essence, the fast food combo truly is a combination; a combination lock on a safe is really a permutation. Stupid English.


http://youtu.be/KmiSw7iTIvA

Suggested prior reading: 1.1, 1.2


n Ck =

n!(n− k)!k !

Combinations are also represented by

nk

⎛⎝⎜

⎞⎠⎟

, read as “n choose k”; it has the same formula as nCk.

Like permutations, combinations involve items that cannot be used more than once. This makes sense because the only difference between combinations and permutations is whether order matters.8

To further distinguish permutations from combinations, let’s look at a grouping many of you should be familiar with: swimming at the Summer Olympics.9 In all Olympic swimming events, there are two preliminary rounds (called heats) before the final medal round. In the first heat, the top 16 of the 32 swimmers advance; in the second, the top 8 of 16 make it to the finals. The heats are great examples of combinations: either a swimmer advances or doesn’t. Finishing first or eighth makes no difference – all that matters is whether a swimmer is in the advancing group, not the order in which she finished10. However, in the final, the top three finishers are awarded gold, silver, and bronze medals. Here, the selection is a permutation: the order of the top three finishers does matter, since gold is better than silver and silver better than bronze.

Let’s look at some sample problems:

1. A teacher divides his class into study groups, each of which consists of 3 sophomores, 1 junior, and 2 seniors. If there are 8 sophomores, 11 juniors, and 9 seniors in the class, how many different study groups are possible?

You are considering three separate combinations for this problem: 8 choose 3, 11 choose 1, and 9 choose 2. The trick is to know what do with them. You would multiply them because of the Multiplication Principle (each combination represents a number of possibilities). So, you have:

22176)36)(11)(56(29

111

38

==⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

There are 22,176 possible study groups.

2. If the class in Problem 1 decided to form a committee consisting of 5 sophomores, 10 juniors, and 7 seniors, how many different committees are possible? How does this number compare to the number of possible study groups?

The product of 8 choose 5, 11 choose 10, and 9 choose 7 is

8 This is actually not true, but we’re going to assume it is for our purposes. Combinations with repetitions are beyond the scope of this material. 9 This could also apply to many Olympic sports (like track), but I always preferred swimming. 10 This is not exactly true – the top qualifiers get the better lanes in the middle of the pool – but it’s close enough for our purposes.


22176)36)(11)(56(79

1011

58

==⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛,

There are 22,176 possible committees, which is precisely the same number as in Problem 1. Thus, there is no difference between the number of committees and the number of study groups.11

3. If a situation involves choosing 5 items from a group, by what factor would you have to multiply the number of permutations to yield the number of combinations?

Since)!5(!

5 −=nnPn

and!5)!5(

!5 −=nnCn , nP5 = 5!*nC5. Dividing gives (1/5!)nP5 = (1/120)nP5 = nC5. So, you would

multiply the number of permutations by 1/120 to yield the number of combinations.

To recap, you should know three things about combinations:

(1) Combinations are arrangements of k items chosen from n total items in which the order of the chosen items does not matter.

(2) Combinations involve items that cannot be used more than once.

(3) The combinations formula:

11 This should surprise you – after all, the number of people selected was different in each case! However, there’s actually a very good reason that the answer turned out to be the same: combinations and permutations are both symmetrical, meaning that they have “pairs” of numbers. To understand this, think about combinations this way: if we are choosing a group of 3 people from 7 choices, we are also “picking” another group of 4 – those four people not chosen are in their own group, in a sense. So, 7C3 and 7C4 are unsurprisingly the same value. The same applies for any two choices that add up the total – for instance, in the above problem, 8C3=8C5, 11C1=11C10, and 9C2=9C7.

n Ck =

n!(n− k)!k !


QUIZ 1.3: COMBINATIONS Suggested Time: 6 minutes

Answers on page 123

1. How many teams of 3 people can be formed from a group of 8 people?

2. A student has 10 math problems to work on. In how many ways can the student select 3 questions to work on?

3. Due to a new law that forces a manager to pay for health insurance for his employees, the manager must lay off 4 of his 12 employees. In how many ways can the 4 be chosen?

4. In a game of poker, five cards are dealt from a standard 52-card deck. How many of such hands have exactly 2 hearts?

5. In a game of poker, five cards are dealt from a standard 52-card deck. In how many ways can a full house—aces full of 9’s (3 aces and 2 9’s) occur?

6. In how many ways can a committee consisting of 2 faculty members and 3 students be formed if 6 faculty members and 10 students are eligible to serve on the committee?

7. How many different vertical arrangements are possible for 10 blocks if 2 are red, 3 are blue, and 5 are purple?


CALCULATOR HELP: PERMUTATIONS AND COMBINATIONS The first tool of enumeration that will come in handy is the factorial. To access the factorial: 1. First type in the number you wish to factorial.

Then, press “MATH.” The screen to the right

2. Use the arrow keys to scroll over to “PRB.” Then, use the arrow keys again to scroll down to “!” and press “ENTER.”

3. Once back at the homescreen, simply press “ENTER” again. The factorial will be evaluated.

The next tools are the permutation and combination. The syntax for these functions is as follows: n nPr k (Permutation) OR n nCr k (Combination) This evaluates the number of permutations (or combinations) of k objects taken out of a group of n objects. To access the permutation or combination function:


1. First enter n, where n is total number of objects in the group. Do not press “ENTER” here.

2. Follow the exact same procedure as you did for the factorial. This time, select “nPr” or “nCr” instead of “!.”

3. Now enter k, where k is the number of objects for which you are determining the permutations or combinations (in other words, the number of items you are choosing). Press “ENTER” to reveal the answer.


SECTION 1 FINAL EXAM Suggested Time: 9 Minutes

Answers on page 124

1. In a game of poker, you are dealt 5 cards. How many such hands have cards of a SINGLE suit (for example, ONLY hearts)?

2. How many “words” (they don’t have to make sense) can be formed using all the letters of the six-letter word “MAMMAL”?

3. Taking a road trip from New York to Phoenix, Rachel wants to stop in Chicago. If she has 5 different routes to choose from in driving from New York to Chicago and 3 routes to choose from in driving from Chicago to Phoenix, in how many ways can she travel from New York to Phoenix?

4. Shanan, Jacquie, and Patrick have different birthdays. If we listed all the possible ways this could occur, how many different birthday scenarios would there be? Assume that there are 365 days in a year.

5. A “bit” is a 0 or a 1. A six-bit string is a sequence of length 6 consisting of 0’s and 1’s. How many six-bit strings contain exactly 2 1’s?

6. A student has 12 questions on an exam and is allowed to answer the questions in any order. In how many different orders could the student answer these questions?

7. How many different 9-letter words (they don’t have to make sense) can be formed from the letters in the word ECONOMICS?

8. In how many different ways can the letters of ACDEC be arranged?

9. In how many ways can a seven-class schedule be constructed with the same classes?


10. From 1500 lottery tickets that are sold, 3 tickets are to be selected for 1st, 2nd, and 3rd prizes. How many possible outcomes are there?

11. A journalist makes a pre-season forecast of the top 15 university football teams (in order) from among 50 major university teams. How many different possibilities are there?


ALGEBRA TABLE OF CONTENTS

Section 2.1: Sequences and Series…………………………..

21

Section 2.2: Sigma Notation…………………………………

24

Quiz 2.1-2.2: Sequences, Series, and Sigma Notation……..

26

Section 2.3: Arithmetic Sequences and Series…………..…

27

Section 2.4: Geometric Sequences and Series...…………...

29

Quiz 2.3-2.4 (Section 2 Midpoint): Arithmetic and Geometric Sequences and Series………………………….……………..

37

Section 2.5: Adding and Subtracting Polynomials…………

39

Section 2.6: Multiplying Polynomials………………….……

41

Quiz 2.5-2.6: Adding, Subtracting, and Multiplying Polynomials……………………………………………………

43

Section 2.7: Binomial Expansions Section 2.7.1: “Tricks” with Multiplying Binomials……….. Section 2.7.2: Binomial Expansion Theorem………………. Quiz 2.7: Binomial Expansions………………………………

44 47 51

Section 2.8: Finance Math Section 2.8.1: Compound Interest………………………….. Section 2.8.2: Annuities and Loans………………………….

52 55

Quiz 2.8: Finance Math….………..………………………….. 61


Section 2 Final Exam…………………………………………..

62


SECTION 2.1: SEQUENCES AND SERIES Mathematics is full of repetition and patterns. You may or may not have encountered “pattern” problems in elementary school, where you would have something like “square, pentagon, hexagon, heptagon,…” – Your job was to (1) figure out the pattern and (2) say what the next item in the pattern was. For this, the pattern is that each polygon/shape has 1 more side than the last, and the next item in the sequence has 8 sides (4,5,6,7…?), so it’s an octagon. In mathematics, we call these sets with noticeable patterns sequences.

One common example is that multiplication can be defined as repeated addition; by the same logic, division can be defined as repeated subtraction. Let’s take a look at that second example to see if it makes sense.

Before the use of calculators, mathematicians would have to take extremely large numbers and divide them by other large numbers, and it would be quite tedious. To do this, they constructed the “Division Algorithm12,” which basically states that division is just repeated subtraction.

Let’s try an example: Divide 82 by 7.

82 = 82 – 7*0

82 – 7 = 82 – 7*1

75 – 7 = 82 – 7*2

68 – 7 = 82 – 7*3

61 – 7 = 82 – 7*4

54 – 7 = 82 – 7*5

47 – 7 = 82 – 7*6

40 – 7 = 82 – 7*7

33 – 7 = 82 – 7*8

. . .

19 – 7 = 82 – 7*10

12 – 7 = 82 – 7*11

so,

5 = 82 – 7*11

5 + 7*11 = 82

12 Which is taught in elementary schools as “long division” with a weird little bar “ ” that nobody uses once they

progress past it.


The Division Algorithm ends13 when you reach a nonnegative number less than 7 (in this case, 5). The number 5 is the remainder, and the number of multiples of 7 that were subtracted (11, as shown in this example), is the quotient. So, dividing 82 by 7 gives us 11 with a remainder of 5.

We can list the repeated subtractions of 7 from 82 in a data set, specifically called a sequence. Data sets and sequences are the same thing, but—typically—mathematicians use the word “sequence” to talk about numbers that are listed in a specific order. The sequence for the above case of the Division Algorithm is as follows:

{82, 82-7*1,82-7*2,82-7*3…82-7*11}

or14

{82,75,68,61,54,47,…5}

This sequence of numbers represents the numbers 82 down to 5, which each one having a common difference—meaning each successive number is different from the last by a certain amount—of 7.

Sequences need not be numbers but can also be other things, like shapes and days of the week. Shapes and months of the year can represent sequences too.

The list {triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, …, n-gon} represents the sequence of polygons each with 1 more side than the last.

The list {January, February, March, April, May, …, September, October} lists the months of the year from January to October, each “1 month more than” the last.

What is the 3rd term of the months-of-the-year sequence above? Clearly, it is March.

A general way of saying, “what term is this?” is to refer to the index of a sequence. We write the index of a number (or “thing”) in a sequence as a subscript.

For example, suppose that a = days of the week, and ai represents all the days of the week. The subscript “i” can take on numbers from 1 to 7. So instead of {Monday, Tuesday, Wednesday, . . . , Saturday, Sunday} we can write {a1, a2, a3, a4, . . ., a6, a7}, where ai = days of the week and i = the index of each day (or, rather, what numbered “day of the week” it is).

One way to find a sequence is to define it recursively. This just means that, given one number, you start with a number(s) and define the rest of the numbers in the sequence by their relation to that number. For example, take the set

{4,8,12,16,20,24,…}

which can be written as {x1, x2, x3, x4 , x5, x6, . . .} where x is a number in the set and each subscript number is one of the “xi’s” of the set.

13 Don’t commit this to memory or spend too much time trying to understand it—it’s purely motivation for what we’re about to talk about. This is just designed to give you an introduction to series. 14 We typically use “…” to show that there are terms that follow a pattern between the listed terms


We write a recursive formula for this sequence/set where each number is 4 more than the last as xi = xi-1 + 4. However, this formula is too broad; it could start on ANY number, so, for a recursive formula for sequences, we must also include a starting value/initial term, x1.15

A series is simply the sum of the numbers in a sequence. There are special formulas for calculating the sum of a sequence, much like there are formulas for calculating the numbers themselves, and these will be discussed in the following sections (2.2 – 2.4). For now, we will just add the terms that we can easily list.

Some final notes on the basics of sequences and series:

(1) A sequence is an ordered set of data/things/numbers/etc. A series is the sum of the numbers of a sequence.

(2) The index of a sequence simply describes “which term” of the sequence an item is. For example, in the sequence {1,2,3,4}, where a0=1, the first term, a0, is 1, the second term, a1, is 2, and so on.

(3) A recursive formula is a formula that includes previous numbers in a sequence to calculate the rest of the numbers, using their indices16. We write a sequence as “each number is one more than the last” as xi=xi+1+1, making sure to define an initial value (x1=? or x0=?) so that we have somewhere to start.

Let’s look at a few sample problems. For each problem, write out a few terms of the sequence unless otherwise specified. Also sum the terms of the sequence, if possible.

1. The even numbers from 4 to 18

2. Starting with 5, each number is 1.5 more than the last

3. x1=1, x2=1, xi=xi-1+xi-2, from i=1 to i=7

4. Starting with 7, each number is two times the last. End at the 5th term.

Answers:

1. {4,6,8,10,12,14,16,18}. The sum is 88.

2. {5,6.5,8,9.5,11,13.5,15,16.5,…}. The sum is infinity.

3. {1,1,2,3,5,8,13}. The sum is 33.

4. {7,14,28,56,112}. The sum is 217.

15 When writing recursive series, I prefer to use an “x0” term for the initial value; x0 is simply x1 minus the common difference. When we begin writing explicit sequences later on, x0 will be much more useful than x1. 16 That’s the plural of index, not some weird dice game.


SECTION 2.2: SIGMA NOTATION FOR SERIES We have already mentioned that a series is the sum of the terms of a sequence. However, we have also seen that sequences can be very long and tedious to write out.

Luckily, mathematicians often come up with convenient ways to write things out17. They are often complicated looking, involving lots of subscripts and Greek letters, but once you understand the meaning behind all the symbols, they become manageable. Series are no different—mathematicians have come up with something called “Sigma notation” to represent series in a short statement.

The most important thing to remember before delving into sigma notation is what the index means. The index essentially represents which number (in order) a value is in a sequence/data set. So, to review:

The set ai from i=1 to i=n is represented by {a1, a2, a3, a4, a5, . . ., an} where i goes from1 to n.

Sigma notation is simply shorthand for writing a sum of certain numbers in sequences. Take the sequence above…what if we want to sum all the numbers between the 2nd and 6th term of the above sequence (inclusive)? Then we write a2 + a3 + a4 + a5 + a6 . Obviously, if we wanted to find the sum of the numbers between, say, the 5th and 118th term of the sequence, we wouldn’t want to write that out. It is because of inconveniences like those that Sigma Notation exists. We represent the sum of a sequence from i=1 to i=n by

ai

i=1

n

∑

where ∑ , the capital Greek letter “sigma” means “the sum of,” ai represents the “rule” for the numbers in

the sequence, i=1 (the bottom of the sigma18) represents where the sum STARTS, and n (the top of the sigma) represents where the sum ENDS. Thus, the above formula in Sigma Notation represents the sum a1 + a2 + a3 + a4 + a5 + . . . + an.

Now, consider some actual sequences. Suppose we have the set of even numbers, from 2 to 10, and we want to sum them using sigma notation. First, let’s write out the sequence {2,4,6,8,10}. Now, we can arbitrarily call 2 whatever we want, as far as indices go, but 0 or 1 makes the most sense. Since 2 is the “1st” positive even number, we’ll say a1=2. Then, a2=4, a3=6, etc.

17 Mathematicians are nothing if not dependably lazy. 18 Note that this bottom number can be anything. You may start a sum at i=0, and thus the first term of the series would be the “ao” term in your indexed sequence; starting at “ao” would give a sum of ao + a1 + a2 + … + an


http://youtu.be/Pd4NLJ8CGs8



Though it is obvious that each even number is 2 more than the last, how do we write this recursively? Looking back to Section 2.1, we know that this can be written (since we know an initial value, a1=2), as ai=ai-1+2. What would we do if we wanted to know the 6th term of this sequence? We’d simply plug into the formula, using that a6=a6-1+2=a5+2=10+2=12.

However, we’ve digressed: We want to write the sum of the even numbers from 2 to 10 in sigma notation.

Thus, we can write

2kk=1

5

∑ or

2+2ii=0

4

∑ , depending on how you think about the sequence; the world is your

oyster19, so I won’t restrict your creativity and tell you that one method is “right” – there are many possible answers, depending on where we start.

Let’s look at some examples. If you find writing sequence rules difficult, you might want to revisit 2.1.

1. Find the sum from the 3rd to 6th term of {1,1,2,3,5,8,13,21,34}. Also, express this sequence recursively in sigma notation.

xi−2 + xi−1

i=3

6

∑ =18

This is the famous “Fibonacci sequence,” where each number in the sequence is the sum of the previous two numbers. There is no way for you to know this other than by examining each term and seeing how it relates to the terms around it. After that, the only trick is realizing that you have to start the sum at the third term (i=3), since the sequence (and thus the series) depends on the previous two numbers.

2. The sum of k3 from k=0 to k=4

k 3

k=0

4

∑ = 0+1+ 8+27+ 64 =100

3. The sum of {2,4,6,8,10,12}, x0=0, x1=2.

xi−1+2 =

k=1

6

∑ 2+ 4+ 6+ 8+10+12 = 42

The fact that an initial value is given might tip you off that this sequence is defined recursively. Given that x0=0

and x1=2, we can establish that each number is two more than the last, like our example from earlier.

19 This will be the only reference to literature you will see in this Math Guide, if not in any math book ever…


QUIZ 2.1-2.2: SEQUENCES, SERIES, AND SIGMA NOTATION Suggested Time: 6 minutes

Answers on page 125

1. Evaluate

k 2 −1k=3

6

∑ .

2. Given the sequence {10,7,4,1,-2,-5,-8,…} and that 10 = a0, write the sum of the terms in this sequence using sigma notation.

3. Evaluate

2k

k=1

5

∑⎛⎝⎜⎞⎠⎟− 1+ k

k=1

3

∑⎛⎝⎜⎞⎠⎟

.

4. Given that x0=5, x1=7, and xi=xi-1+2, what is x7?

5. Starting with c1=10, each number in a sequence is 0.75 times the last. What is the sum of the first 5 numbers of the sequence?

6. Write “the sum of the first 11 odd numbers” using sigma notation.

7. Find the sum of the series

(3n + (−1)n

n=1

8

∑ ) .


SECTION 2.3: ARITHMETIC

SEQUENCES AND SERIES This section is simply a formalization of parts of sections 2.1 and 2.2. From 2,1, you know that a sequence is an ordered set of numbers, and you know that some sequences have a common difference that is a certain constant number “between” them. We can use these two facts along with a basic understand of Sigma Notation to work with Arithmetic Sequences and Series.

All arithmetic sequences are sequences with a constant/common difference. The sequence {2,4,6,8,10} is arithmetic because each number is 2 more than the last; the sequence {3,-4,-11,-18,-25} is arithmetic because each number is -7 more (another way of saying 7 less) than the last, and {a1, a1 + d, a1+2d, a1+3d,a1+4d, …, a1 + nd} is arithmetic because each term is d more than the number preceding it.

“Each number is d more than the last” can be defined recursively as xi=xi-1+d. The direct formula, though20, for finding the terms of an arithmetic sequence, is

an=a1+(n-1)*d

where an is any term of the sequence, a1 is the initial value of the sequence, n is the number (or index) of the value you’re trying to find in the sequence, and d is the common difference.21

For example, given the sequence {5,12,19,26,33,…}, find the 10th term. a1=5, d=7, an=unknown, n=10 (“10th term”). So an=a1(n-1)*d, or an = 5(10-1)*7 = 5*9*7=315.

To find the sum of an arithmetic series, we simply use our knowledge of arithmetic sequences. The formula22 for the sum of an arithmetic sequence is:

Sn = (a1+an )⋅n

2

where S is the sum, a1 is the initial value, an is the nth value, and n is the term number.23

This formula can be written in sigma notation as

20 You should almost always use this one. It is also called the “explicit form.” 21 I prefer to use an alternate form of the direct formula that uses a0 instead of a1. Basically, we can distribute d by multiplying it by n and -1; this gives us a1+dn-d. If we separate the terms “a1-d,” we can see that this is equal to a0 – the first term minus the common difference must equal the previous “zero” term. Thus, another form of the direct equation is an=a0+nd, which I find easier to remember. Also, this form is identical to the slope-intersect form of a line: “n” is the x-value, “d” is the slope (which is the same mathematically as the common difference), “an” is y, and a0 is the y-intercept – the value of an when n=0. 22 Unless absolutely necessary, it is in your best interest to stick to the simple “S” sum form of this equation. It’s less complicated and thus easier to remember. 23 A derivation of this formula is in the 1st Decademy Mathematics Video Lecture linked above.

Related video lectures:

http://youtu.be/DHVgU0rbwk4

http://youtu.be/OBR84R35Orc



(a1+ (i −1)⋅d)

i=1

n

∑ = [2a+ (n−1)⋅d]⋅n2

To wrap things up:

(1) An arithmetic sequence is defined by the formula an=a + d(n-1), where an is a particular term, d is the

common difference, and a is the initial value of the sequence. Arithmetic sequences will always have a

common difference, which is what makes them unique.

(2) The sum of an arithmetic sequence can be expressed by the formula Sn= n/2 * (a+an), where Sn is the sum

from the initial value to the nth term, an is the nth term, and n is the number of numbers in the sequence.

Let’s try some examples:

1. Find the 15th term of the sequence {-4,-11,-18,…}

2. Find the sum from i=2 to i=8 of the sequence with a1=5 and d=5

3. Given a6=25 and a1=5, find the sum from the 1st to 6th term. Also, find d.

Answers:

1. an=a1+(n-1)*d. So a15=-4+(15-1)*(-7)=-4+14*-7=-102

2.

(5+ (i −1)⋅5)i=2

8

∑ = [2 ⋅5+ (8−1)⋅5]⋅ 82=180

3. S6=(5+25)*6/2=90. 25=5+(6-1)*d, so d=(25-5)/5, thus d = 4.


SECTION 2.4: GEOMETRIC SEQUENCES AND SERIES

This section, like section 2.3, is simply a formalization of the rules of sequences and series.

A geometric sequence is simply a sequence in which each term is a common multiple of each other. We call this the common ratio and label it “r”. For example, the data set {1,2,4,8,16,32,64,…2n} is the geometric sequence with a1=1 and r=2 – each term is double the previous one.

The data set {5, 5/2, 5/4, 5/8, 5/16, 5/32, . . .

52n−1 } is the geometric

sequence with a1=5 and r=1/2. The data set {a, ar, ar2, ar3, ar4, …, arn-1} is the geometric sequence with a1=a and r=r.

Thus, we define a geometric sequence recursively by xi=r*xi-1 and directly24 by an=a1*rn-1, where r is the common ratio, an is the nth term in a sequence, and a1 is the initial value.

With the direct formula, you are now equipped to find any term in a geometric sequence. Find the 6th term of the sequence {10,30,90,…}.

We know that a1=10 and that the common ratio between 10 and 30, 30 and 90 = 3 (r=3), so we can plug that in.

a6=10(3)6-1=10(3)5=2430.

All that’s left—much like for arithmetic sequences in section 2.3—is how to calculate the sum terms of geometric sequences (i.e., the sum of geometric series).

The general formula for the sum of a geometric series is

S = a1 ⋅r

k −a1

r −1= a1

r k −1r −1

where S is the sum, a1 is the initial value, r is the common ratio, and k represents the term number (Sk means

the sum from 1 to the kth term).

There is a special case for geometric series with an infinite amount of terms. With most common ratios, a geometric series from 1 to infinity would sum to infinity. That is, the series would diverge. However, some geometric sums converge, that is, they add to numbers OTHER than infinity. Infinite geometric series with common ratios 0<r<1 (|r|<1) converge25. Otherwise, such series diverge.26

24 Again, use the direct formula whenever possible.

25 This means that they sum to a specific number—in fact, they all converge to

a1

1− r


http://youtu.be/DHVgU0rbwk4

http://youtu.be/OBR84R35Orc



We end with Sigma Notation formulas for finite 27and infinite geometric series

Finite geometric series:

a1 ⋅r

i−1 = a1 ⋅rk −a1

r −1i=1

k

∑ = a1(r k −1)

r −1

Infinite geometric series (-1<r<1; |r|<1):

a1 ⋅r

i−1

i=1

∞

∑ = a1

1− r

26 Divergent series are of particular interest to Calc II students and are very important in math, but they are beyond the scope of this material. 27 That is, “countable” or “calculable” or simply not being infinite. 28 Why? The question asks for only “positive values for x,” so it’s a good bet that the positive values for x are in the positive half of the inequality. If that side only gives negative x, solve the other half. 29 Factoring is not covered in this guide or in the USAD® resource guide, so we suggest that you use your graphing calculator to solve for solutions.

A NOTE ON CONVERGENCE Something that you may be tested on is figuring out “when” a series will converge. For example, a test might

ask “For which of the following positive values of x does the series

2−5x

x 2 + 6⎛⎝⎜

⎞⎠⎟n=0

∞

∑n

converge?” Unfortunately,

this question presupposes that you understand (1) absolute values and (2) solving quadratics/inequalities involving quadratics, which, given the material in the Mathematics Resource Guide, is only somewhat relevant. To solve this problem, you must recall that, in order for an infinite geometric series to converge, |r|<1 (or, better, -1<r<1). Thus, for the above problem, we have -1 < -5x/(x2+6) < 1; we can multiply the entire inequality by (x^2+6) and then add 5x to the entire inequality to obtain -x2+5x -6 < 0 < x2+5x+6. For most of these problems, you should only have to worry about the right side, so we want to examine 0<x2+2x+1.28 You can either factor29 this or graph it to see when this statement is true. For this function “x2+2x+1,” we simply need to find what point it becomes “greater than 0”—looking at the graph, there are 2 candidates, x=-2 and x=-3. At this point in the multiple choice question, you can eliminate anything that does not involve {-3,-2,2,3}, keeping in mind that we are looking for positive values of x, so we’ll just be dealing with 2 and 3. At this point, we need to figure out if the answer will be “x>{2,3}” or “x<{2,3}.” We finish the problem by plugging in numbers less than or greater than your candidates, in this case {2,3}; we can see that “x>3” is the answer, since anything greater than 3 (try 4) will give r as a fraction, as required for convergence, and anything between 2 and 3 (try 2.5) will not, so the series would diverge.


If r is not -1<r<1, then the series will diverge to infinity.

Let’s try some sample problems:

1. Find the sum of the terms of sequence {100,50,25,25/2,25/4}.

2. Calculate

5⋅5k−1

k=2

5

∑

3. What is 3

125−14

in sigma notation?

4. To what value does

5(.75)i−1

i=2

∞

∑ converge?

All that being said, this problem is not easy, relevant, or even possible using the material strictly given in the Resource Guide. If a question is asked like this on an exam, it might be in your best interest to “challenge” the problem to the test proctor; however, you should be able to get the problem if you understand the process above.


Answers:

1. 100

( 12 )5 −112 −1

= 7754

2. Note that the initial value here is when k=2, not k=1, so the sum starts at i=2. Thus, 25

54 −14−1

= 3900

3. rk=125, r-1=4, so r=5, thus 5k=125, so k=3. Therefore, this expression is equal to

3⋅5i−1

i=1

3

∑

4. Like in problem 2, we have to note that “a1” = 5*(.75)1, so the sum is equal to

5(.75)1

1− .75= 5( 3

4 )14

=15

CALCULATOR HELP: SEQUENCES Given a direct or recursive formula, you can easily generate the corresponding sequence using your calculator. Unfortunately, there is no way to go from the sequence back to the formula—you will have to do that by hand.

Given the recursive formula: x1 = 3 xi = 2xi-1 - 2

1. Type in the first term. It’s just 3 in this case. Press

“ENTER.”


2. Now, type in the xi expression. However, in place of xi-1, we will use “Ans.” You can get “Ans” by pressing “2ND” then the “(—)” key. Press “ENTER” when you’re done.

3. Now, we can just keep pressing the “ENTER” key to generate more terms. 3 is our first term, 4 is the second term, 6 is our 3rd term, etc. Press “ENTER” to go as far out as you need.

Given the direct formula:

xk = 2k + 7

1. Press “2ND” then “STAT” (selecting “LIST”).


2. Scroll over to “OPS” with your arrow keys, then press “5” (selecting “seq”).

3. Now, enter the direct formula (using the variable key as necessary). Press the comma key, then press the variable key, then press the comma key again, then enter the lower bound, then press the comma key a third time, then enter the upper bound. If you want a full sequence, start it at 1 and go to some finite number, but if you want to find specific terms you can start it and end it at that term. In this particular example, I chose to find the first 10 terms of the sequence, so I chose 1 and 10 as my lower and upper bounds, respectively.

Now you can use your arrow keys to scroll left and right to see any part of the sequence that doesn’t fit on the screen.

CALCULATOR HELP: SUMMATIONS Given a recursive or direct formula, the corresponding sequence can be generated by your calculator. However, to sum a sequence with the calculator, you must have the direct formula. If you’re given the recursive formula, you will have to generate the sequence with the calculator and then add up the terms individually. There are two ways you’ll likely see summations on the test. The first is with just words. The question may ask “find the sum of the first 5 terms of this sequence” (or something similar to that). The second way is with sigma notation, which makes this even easier—you just need to know where to look for the information.

𝟐𝒌 − 𝟏𝟏𝟎

𝒌!𝟏

This is a typical sigma. The direct formula is the expression to the right of the sigma (2k-1 in this case). The lower bound is the number directly underneath the sigma (1), and the upper bound is the number above the sigma (10).


Example: Given 𝐱𝐤 = 𝟐 ∗ (𝟏

𝟑)𝐤!𝟏 , find the sum of the first 5 terms of this sequence.

1. Press “2ND” then “STAT” (selecting “LIST”).

2. Scroll over to “MATH” with your arrow keys, then press “5” (selecting “sum(”).

3. Repeat step #1.

4. Scroll over to “OPS” with your arrow keys, then press “5” (selecting “seq(”).

Your homescreen should now read: sum(seq(


5. Enter the sequence parameters. First, enter the direct formula (using the variable key as necessary). Press the comma key, then press the variable key, then press the comma key, then enter the lower bound, then press the comma key, then enter the upper bound. Press “ENTER” when you’re done. It will generate the sum.

Use the exact same process for sigma notation.


QUIZ 2.3-2.4 (SECTION 2 MIDPOINT): ARITHMETIC AND

GEOMETRIC SEQUENCES AND SERIES Suggested Time: 10 minutes

Answers on page 125

1. The sequence an ,n =1,2,3,…, is a geometric sequence. Assume a1 = 5 and a3 = 45. Find the value of a5 .

2. The sequence an ,n =1,2,3,…, is an arithmetic sequence. Assume a1 = 27 and a3 = 5. Find the value of a9 .

3. The sequence an ,n =1,2,3,…, is a geometric sequence. Assume a2 = 5 and a4 = 125. Evaluate

ann=2

6

∑ .

4. Evaluate 5(

12

)n−1

n=1

∞

∑.

5. Evaluate 6(

13

)n−1

n=4

∞

∑.

6. Evaluate 6(2)n

n=6

∞

∑.

7. The sequence an ,n =1,2,3,…, is an arithmetic sequence. Assume a1 = 5 and that the common difference is

8. The sequence an ,n =1,2,3,…, is a geometric sequence. Assume a1 = 5 and that the common ratio is 4. Find

the value of a5 .

9. Given the arithmetic sequence where a1 = 5 and the common difference is 3, find the sum from the 1st to 8th

term.


10. The sum of the geometric series

5(3)n

n=2

6

∑ is

11. The sum 3 + 6 + 12 + 24 + 48 + 96 can BEST be represented in Sigma Notation as

12. The sum of the infinite series

6(17

)n

n=3

∞

∑ is


SECTION 2.5: ADDING AND SUBTRACTING POLYNOMIALS Polynomials are things that have all sorts of mathematical definitions, depending on the level of math you’re at. A polynomial, simply put, is something where you multiply a number of constants by “x,” where x has to have a whole number power.

That definition is practically useless, though, as it doesn’t teach us how to do anything or display any mathematical rigor. Let’s look at some polynomials instead and see what we can gather from them…

A. x2+x+1

B. 1-2x+x3

C. 4x4+3x3-2x2+1

…

There are infinitely many possible polynomials that we can examine, but at least we can see what a polynomial “generally” looks like—there are constants in front of x’s, called coefficients, all added together. The coefficient of the “x2” term in Polynomial A is 1, the coefficient of the “x” term in Polynomial B is “-2,” and the coefficient of the “x4” term in Polynomial C is “4”.

Besides coefficients, there is also a degree element to polynomials. The degree of a polynomial is simply the highest power (exponent) in the polynomial expression. The degree of Polynomial A is 2 since x2 has the largest power, the degree of Polynomial B is 3, and the degree of Polynomial C is 4. We can think of a polynomial as a sequence. Say we always write our polynomials from lowest degree30 (0) to highest degree, so they are in the form

a0 + a1x + a2x2 + a3x3 + . . . + anxn

This means we can write a sequence as {a0, a1, a2, a3, . . ., an} for any polynomial in this form. Using Polynomial A from before as an example, a0, a1, and a2=1, but all greater powers (greater than a3) = 0. This polynomial would be represented as {1,1,1, 0,0,0,0,…} or simply {1,1,1}.

To make this sequence form of a polynomial make sense, we relate it to series and give it a Sigma Notation form:

ai x i

i=0

n

∑

Using knowledge from Section 2.1, we can write polynomials as sequences and series.

30 A term of degree 0 is called a constant term because it contains no variable “x.” Sometimes math definitions actually make sense.


http://youtu.be/VO_paonN68A



With that knowledge in mind, we can now do a number of things with polynomials that we can also do with sequences. For example, we can add and subtract.

To add and subtract sequences, you simply line up the terms of each (adding ai and bj sequences means you would add a1+b1, a2+b2, a3+b3, . . ., an+bm) and add them. For example,

{1,1,2,3,5,8}+{6,6,3,6,3,6}=

{1+6,1+6,2+3,3+6,5+8,8+6}=

{7,7,5,9,13,14}

If you really want, you could line up the terms

{1,1,2,3,5,8}

{6,6,3,6,3,6}

and add them up vertically, getting, again,

{7,7,5,9,8,14}

Pretend these two sequences were polynomials of the form we discussed before, where the first term is the 0-degree term, the next term is the 1st-degree term, etc. Then the polynomials being added are 1+x+2x2+3x3+5x4+8x5 and 6+6x+3x2+6x3+3x4+6x5, and the resulting polynomial after addition is 7+7x+5x2+9x3+8x4+14x5. If you’ve ever heard the expression “add like terms,” this is exactly what that means.31

Subtraction works the same way as it does with numbers—subtracting a polynomial is the same as adding the negative of that polynomial. So,

(1+x+x2) – (2-2x2+x4) =

(1+x+x2) + (-2+2x2-x4) =

-1+x+2x2-x4

1. Simplify the expression (1+ x + x 2 )− (−x 3 −2x 2 +2) .

2. What is the simplified form of ( x + 4 x 4 )+ (3x +3x 3 )− (2x − x 2 ) ?

3. Simplify the expression (−2x + x 2 )+ ( x + x 2 +3x 3 )− (3x 3 +2x 2 − x ) .

Answers: It is easiest to add one term at a time, starting with the highest degree/power.

1. x3+3x2+x-1; 2. 4x4+3x3+x2+2x; 3. 0

31 If you’ve ever taken a high school math class, I assume you’ve heard this term to death.


SECTION 2.6: MULTIPLYING POLYNOMIALS I would venture to say that every math problem has more than one possible way of being solved; some ways are just easier than others. One thing that we know for certain is this: some mathematicians are apparently very opinionated when it comes to the proper way to multiply polynomials. The USAD® math writer offers one way while denouncing the standard method (“FOIL32”) completely.

I will instead offer a different, yet quick and simple way to multiply polynomials. It always works, and it’s fast. All you need to remember is that, when multiplying powers of x, you add the powers (that is, x2 times x3 is x5).

We’ll get right to examples.

1. Evaluate (x+1)(x2-3).

We set up a “box” with each term in a different row/column. We then multiply the row by the column to get the number in each entry – so, for example, x2*x gives x3. Finally, we add together all of the terms and combine like terms to obtain our final answer.

x 1 x2 x3 x2

-3 -3x -3

So, (x+1)(x2-3)=x3+x2-3x-3.

When multiplying more than two polynomials, simply multiply two at a time.

2. Evaluate (x+1)(x2+x+1)(x3-5).

x2 x 1 x x3 x2 x 1 x2 x 1

So (x+1)(x2+x+1) = x3+x2+x2+x+x+1

= x3+2x2+2x+1

Then, to finish the problem, multiply (x3+2x2+2x+1)(x3-5).

32 . You probably learned this method in your Algebra classes. The box method is easier.


http://youtu.be/YZfhilEKYEU



x3 2x2 2x 1 x3 x6 2x5 2x4 x3

-5 -5x3 -10x2 -10x -5

So the product is x6+2x5-4x3-10x2-10x-5. Notice how I write from largest degree term to smallest—this is a convention, but it is not necessary. You can write starting with the smallest degree term, too.

Try using this “box method” to multiply two polynomials with 3 terms, x3-x+1 and x2-6x+9.

x3 -x +1

x2

-6x

9

You should end up with x5-6x4+8x3+7x2-15x+9.

Try a few more (make sure to make your own boxes!)

1. Evaluate (x3+2x+1)(x+2)(x-1).

2. Simplify the expression (x4+3x2+2x+1)(x+3x2+x3).

3. Evaluate (x+1)(x+1)(x+1)(x2+2).

Answers: (1) 5x4+4x3+6x-3; (2) 7x6+18x5+20x4+44x3+30x2+10x+1; (3) x5+3x4+5x3+7x2+6x+2


QUIZ 2.5-2.6: ADDING, SUBTRACTING, AND MULTIPLYING

POLYNOMIALS Suggested Time: 6 minutes

Answers on page 126

1. What is the simplified form of ( x 3 +2x +2x 2 −1)− (−3x −3x 2 +1) ?

2. Evaluate ( x 4 + x 3 +1)( x 2 − x −2) .

3. Evaluate ( x 4 −9)(3−2x + x 4 +2x + 6) .

4. What is the coefficient of the x 2 term of (2x +1)2 + (2x 4 +3) ?

5. The expansion of ( x +1)2 + (2x 2 +2)+ ( x +1)( x −1) has a term of the form ax . What is a ?

6. Evaluate ( x +1)( x 2 +2x +2)( x 3 +3x 2 +3x +3) .

7. Find the simplified form of

( x +1)( x 2 +2x +1)( x 3 +3x 2 +3x +1)−

( x 4 + 4 x 3 + 6x 2 + 4 x +1)( x +1)2


SECTION 2.7.1: “TRICKS” WITH MULTIPLYING

BINOMIALS We have already talked about how polynomials’ origins are complicated and difficult to explain. Though this might be a slight annoyance, it also implies that there is a great deal we can learn from polynomials without knowing exactly why things work the way they do. However, there is one term that we need to be very familiar with: binomials. “Bi” means two, and so a binomial is a polynomial with two terms. Usually, a binomial is a polynomial with the form (x+a)(x+b); many shortcuts exist for working with these types of polynomials. We will examine these in the next few sections.

Our first example is something not covered in the USAD® Mathematics Resource Guide, but something that you could very well be tested on—a “trick” of sorts. Let’s look at multiplying two polynomials together, reminiscing on the not-too-distant Section 2.6 of this guide:

Write ( x +9)( x −9) as a degree two polynomial.

We could use the “box” or “FOIL” method for polynomial multiplication and see that the above expression

equals x 2 − 81. However, this section is about finding shortcuts to things that would otherwise require brute force to figure out. Multiply out these polynomials and see if you can find a pattern:

( x +1)( x −1)=( x +2)( x −2)=( x +5)( x −5)=( x + 8)( x − 8)=( x +13)( x −13)=

It should be clear that the types of polynomials I’m picking are ones that have the same number, one added, one subtracted. Multiplying them together always gets x2 minus the square of that shared number; in symbols,

( x −a)( x +a)= ( x 2 −a2 )

We call this the “difference of two squares.” While it doesn’t relate to the rest of this section, it is still a nifty trick to know, and you may or may not come across it in competition, where it will definitely save you time.

As an aside, this difference of squares can be used to multiply “big” numbers in your head, assuming you know the squares of certain numbers. For example,

87*93 = (90-3)(90+3)= (902-92)=8100-81=8019.

95*105=(100-5)(100+5)=1002-52=10000-25=9975.

80*120=(100-20)(100+20)=1002-202=10000-400=9600.


Again, this is just something interesting. On the actual test, though, don’t mess around with this method; just use your calculator.

The next trick is fairly similar. Since we can multiply polynomials, we can also take powers of polynomials. For example,

(x+2)2 = (x+2)(x+2) = x2+4x+4

After a few examples, you should see a pattern, so try multiplying these out:

( x +3)2 =

( x +5)2 =

(2x + 4)2 =

(5− x )2 =

(6x − y )2 =

(5z 3 − x )2 =

The answers, in order are,

x 2 + 6x +9

x 2 +10x +25

4 x 2 +16x +16

25−10x + x 2

36x 2 −12xy + y 2

25z 6 −10xz 3 + x 2

The pattern is not as obvious with these, but once I tell you it, it will pop right out at you. The general formula for squaring a binomial is

( x +a)2 = x 2 +2ax +a2


This works for squaring any binomial. Let x equal the “first” term in the binomial and let “a” equal the second term, and the above formula will work. Keep in mind that there can be constants (you can let “x” equal 5x) or negative signs (you can let “a” equal -9) and you can use any power of x (you can let x = x3).

This formula is a special case of our next section, 2.7.2, The Binomial Expansion Theorem, which involves taking ANY power of binomials, not just squaring them. However, this “formula” is listed separately because it is more convenient and appears the most often. Now that you know the formula, revisit the problems from the previous page and see if you can write down the answer without using “brute force” to multiply the binomial squared.


SECTION 2.7.2: BINOMIAL EXPANSION THEOREM In Section 2.7.1 of this guide, we have already learned a few semi-useful “tricks” regarding how to multiply polynomials. However, those tricks are only useful with binomials of degree 2. In this section, we combine our knowledge from a few prior sections 33into one general theorem.

Let’s start with an example.

Expand (x+a)5.

Right off the bat, you know that expanding (x+a)5 is going to take a lot of work and a lot of time34. Let’s examine the expanded form:

a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5

To see if we can find a pattern, let’s look at a number of powers, not just 5. Let’s expand the following:

( x +a)2

( x +a)3

( x +a)4

( x +a)5

( x +a)6

To save you some time (a lot of time), I’ll expand (x+a)n from n=2 to n=6 for you…In order,

( x +a)2 = a2 +2ax + x 2

( x +a)3 = a3 +3a2 x +3ax 2 + x 3

( x +a)4 = a4 + 4a3 x + 6a2 x 2 + 4ax 3 + x 4

( x +a)5 = a5 +5a4 x +10a3 x 2 +10a2 x 3

+5ax 4 + x 5

( x +a)6 = a6 + 6a5 x +15a4 x 2 +20a3 x 3

+15a2 x 4 + 6ax 5 + x 6

There is definitely a pattern here. It may not be clear what the pattern is, but it is definitely clear that a pattern exists. Looking at the constant terms in from of each term of these expanded polynomials gives us a hint at what the pattern might be…let’s list them.

33 The Binomial Expansion Theorem is definitely over-tested. You will encounter at least one (if not two) questions related to it on each Mathematics test, so make sure to practice it a lot! Two questions on a Mathematics test is 57 points a person…so if every scorer on your team can get these questions right, that’s a guaranteed 342 points each competition! 34 Which you won’t have on the Mathematics test…


http://youtu.be/es9Ywv2c-mk

Suggested prior reading: 1.3,2.2,2.6,2.7.1


First, let’s get rid of the “a’s” and also add in (x+a)1 and (x+a)0 to the pattern so that it becomes more clear…

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

You might recognize this pattern, and it has a definite name. It’s called Pascal’s Triangle, and it comes up all over mathematics35.

The pattern arises from taking the terms in one row, adding the ones next to each other together, and writing their sum between them. That is, look at the second row—you add these two numbers together and write their sum one row down between the two numbers.

There’s no good way to explain that in words, so let’s look at another graphic.

Following the arrows, you can now tell which two numbers sum to the numbers in the next row.

Fortunately, there is nothing you need to memorize about Pascal’s Triangle except for the fact that the numbers in the next row are derived by summing numbers above them.

The Binomial Expansion Theorem takes Pascal’s Triangle and puts it into a formula. With this formula, you can find any term of any expanded binomial.

The formula brings us back to Section 1.3 of this guide, when we covered combinations36 and Sigma Notation. A binomial can be expanded in the following manner:

35 Check out http://youtu.be/Yhlv5Aeuo_k?t=2m45s for a look at some cool things about Pascal’s triangle 36 Which you should use a calculator to compute—never do them by hand on a test!


( x + y )n =

nk

⎛⎝⎜

⎞⎠⎟k=0

n

∑ ⋅ x k ⋅ y n−k

where, as mentioned in Section 2.7.1, x and y can be anything (positive or negative numbers…polynomials of higher order…etc.)

Each term37 of the resulting polynomial can be expressed by the formula above formula without the summation, or

nk

⎛⎝⎜

⎞⎠⎟⋅ x k ⋅ y n−k

The following graphic displays the connection of Pascal’s Triangle to the Binomial Expansion Theorem.

Using either Pascal’s triangle or the Binomial Expansion Theorem (or a combination of both), you can figure out the coefficients in front of each term of an expanded polynomial.

Looking at the graphics above, we can equate each row to a power of ten. The row with just “1” is the “(x+a)0” row, the next is the “(x+a)1” and so on. Then, each entry is a coefficient of each term in the polynomial, starting with the x0 term (the constant) and ending with the xn term (with n being whichever row you’re working in).

The most important thing to take from this is that the Binomial Expansion Theorem requires a lot of practice, since the formula is fairly involved. Don’t worry if you don’t get it yet—that’s what the practice problems are for:

1. What is the x4 term of (x+2)6?

2. Expand (3x+4)4.

3. The expansion of (7x+6)5 has a term of the form ax3. What is a?

4. Evaluate (x+1)(2x+2)3

37 USAD® will always ask questions like “what is the coefficient of the x6 term . . .,” so this is extremely useful


Answers:

1.

64

⎛⎝⎜

⎞⎠⎟

( x )4 (6)2 =540;

2. (3x+4)4=

44

⎛⎝⎜

⎞⎠⎟

(3x )4 (4)0 +43

⎛⎝⎜

⎞⎠⎟

(3x )3(4)1+42

⎛⎝⎜

⎞⎠⎟

(3x )2(4)2 +41⎛⎝⎜

⎞⎠⎟

(3x )(4)3 +40

⎛⎝⎜

⎞⎠⎟

(3x )0(4)4

= 81x4+432x3+864x2+768x+256

3.

53

⎛⎝⎜

⎞⎠⎟

(7x )3(6)2 =123480 – remember, 5 is the degree, you want the x3 term, hence “5 choose 3,” and the

exponents of the components must add to n (k and n-k, in this case 3 and 5-3=2).

4. First, expand (2x+2)3 and multiply this expansion by (x+1).

(2x +2)3 =

33

⎛⎝⎜

⎞⎠⎟

(2x )3(2)0 +32

⎛⎝⎜

⎞⎠⎟

(2x )2(2)1+31⎛⎝⎜

⎞⎠⎟

(2x )1(2)2 +30

⎛⎝⎜

⎞⎠⎟

(2x )0(2)3 = 8x3+24x2+24x+8. Multiplying this by

x+1 gives us 8x4+32x3+48x2+32x+8.


QUIZ 2.7: BINOMIAL EXPANSIONS Suggested Time: 6 minutes

Answers on page 127

1. Evaluate ( x 3 −2x 2 )( x 3 +2x 2 ) .

2. The expansion (2x +2)4 has a term of the form ax 3 . What is a ?

3. The expansion of ( x +3)3 + ( x +2)2 + x +1 has a term of the form ax . What is a ?

4. Expand (9x + 1

3)3 .

5. Evaluate [( x +1)2 −9][( x +1)2 +9] .

6. The expansion of (6x + x 2 )3 has a term of the form ax 4 . What is a ?

7. The expanded form of (

12

x + 13

)4 has two powers ( x n ) whose coefficients are equal. Which two powers are

they?


SECTION 2.8.1: FINANCIAL MATH 1 COMPOUND INTEREST AND CONTINUOUS GROWTH

And now for something completely different!

While financial math isn’t directly related to algebra, permutations, statistics, or anything else covered in this guide, it is one of the most immediately useful types of mathematics you can learn, and it requires some mathematical literacy (or a graphing calculator) to understand.

That being said, we will start with the idea of compound interest. Many of you might have savings or checking accounts; if you don’t, your parents almost certainly do. Why would people give up their money and put it in a bank, where they can’t immediately access it? There are a few reasons, but one chief cause is that banks pay interest in exchange for the use of deposits38. Interest is always calculated as a percent of the “lump sum” that is earned.

Compound interest is nothing more than interest that accrues over time – meaning that, after some time, you will be earning “interest on interest.” For example, say you have an interest rate of 10% APR39 compounded monthly. That means that every month, your account will earn 10%/12 (or about .83%) on its previous balance. If you start with $100, you will earn about 83 cents in interest after one month; however, during the second month, you will earn that .83% interest on the entire $100.83, earning you $.84 in interest that month. While this growth may start slow, it can pick up rather quickly: after one year, you will have $110.47; after 10 years, $270; after 30 years, just under $2000 – twenty times your original investment.

The formal equation for compound interest is

A= P(1+ rn )nt

where A is the accumulated (or total) value, P is the principal (or starting) amount, r is the annual interest rate, n is how often the interest is compounded in a year (monthly means n = 12, daily means n = 365, yearly means n = 1, biannually means n = 0.5, and quarterly means n = 4), and t is the amount of time that your money is in the account. It is essential that you keep t and n in the same units (months, years, etc.)

Let’s look at some examples of this type of compound interest. You will want to use your calculator to solve these.

1. Suppose you put $500 into an account with 2% interest compounded quarterly. How much will you have after 4 years?

38 Banks, in turn, make money by lending savers’ deposits to borrowers, who pay higher interest rates than the bank gives savers. For more, see the Econ section. 39 APR means “Annual Percentage Rate” and refers to the actual, stated percentage banks will pay you in interest in one year. “APY” is another commonly-used term and refers to the actual gain you see – meaning the interest earned due to your principal and the effects of compounding.


Interest:

http://youtu.be/thpq4x5Hsdw

Euler’s constant:

http://youtu.be/_qaXlB3HUY8



In this problem, P=500, A=to be solved, r=0.02 (don’t forget to convert percentages to decimals), n=4, and t=4, since we’re counting in quarters, and 4 years is 16 quarters of a year. So A=500*(1+0.02/4)^16 = $541.53.40

2. Arianne puts an unknown amount of money into a fund with an interest rate of 5% compounded monthly. After 36 months, the fund amounts to $2000. How much money did she start with?

Answer: P=to be solved, r=0.05, n=12, t=3, A=2000. 2000=P*(1+0.05/12)^(36). So P = $1721.95.

3. Danny puts $250 into an account that, after 24 months, grows to $300. Assuming interest is compounded daily, what is the APR on his account?

Answer: P=250, A=300, t=2, n=365, r=to be solved. 300=250*(1+r/365)^(730). Solve for r. r=.9117, so r=9.12%.

There is a special case for compound interest where, instead of being added at specific time intervals, the interest grows continuously. Continuous growth is modeled using the natural number, “e,” which is approximately 2.7182841. It should be noted that continuous growth is not a realistic “real life” example of how interest grows – there is no such interest account where interest is compounded this way, but USAD® (and many high school math textbooks) still treat continuously compounded interest as something that can happen. In reality, most banks compound daily; the difference between continuous and daily compounding is miniscule.42

The USAD® Mathematics Resource Guide goes into detail about where “e” comes from, so, for redundancy’s

sake, we will not. Instead, we will tell you that “e” is defined as the expression (1+ 1

n)n , where n is a huge

number (plug in something over 1 billion for n and see how closely this approximates 2.171828)43. You might

notice that our expression for defining “e,” (1+ 1

n)n , and the interest growth part of the formula for compound

40 A general rule of money—if you’re the one receiving the money, round down, since banks are stingy, but if they are getting money from you, round up. 41 This is why we never use “e” as a variable – it is always equal to this one number. e actually appears in many, many places in math, including differential equations, Taylor series, natural logs, and number theory. 42 Despite all the power we have from computing, imagine the processing speed it would take to constantly update and put infinitely small slivers of money into all depositors’ accounts – it’s simply not practical. 43 In calculus terms, e is the limit of (1+1/n)^n as n goes to infinity.


interest, P (1+ r

n)nt , look very similar. Through a bit of the magic of algebra, since “e” is the continuous rate of

growth for continuous growth, and since we’re still dealing with “interest” of sorts, the formula for continuous growth (of interest, cell colonies, or anything else) is known as the “Pert” equation,

A= Pert

Where r is the “rate” (interest rate, usually), t is time, P is the principal/initial value, and A is the value that has accumulated over time.

Before we move on to slightly more complex financial math, let’s practice with continuous interest. One quick note: To solve these equations without a calculator often requires basic knowledge of logarithms44. However, we strongly recommend that, if you can’t already proficiently use logarithms that you stick to using a graphing calculator to solve. If given two situations, use the graphing function to compare intersection points.

4. If Ajay puts $20,000 into an account with continuous growth at an interest rate of 2%, how long will it take for his fund to double?

Answer: Remember, A =Pert. So P=20,000 and A=20,000*2=40,000. This equation can be reduced to 2=1ert, as can any equation where you’re asked find “double” or “triple” an amount. r=0.02, so 2=1e0.02t. On your calculator, graph y1=2, y2=e0.02t and see that they intersect at t=34.66 years.

44 Check out http://www.khanacademy.org/math/algebra/logarithms/v/introduction-to-logarithms for an intro to logarithms.


SECTION 2.8.2: FINANCIAL MATH 2 ANNUITIES AND LOANS

With a good understanding of section 2.8.1, this section won’t serve as too much of a road block.

With the compound interest problems we examined earlier, the people involved typically had a “lump sum” or large chunk of money to invest – the principal. In the real world, this is not always the case; money is not always readily available, sitting around waiting to be invested.

To compensate for this, investors created products known as annuities. An annuity is a similar interest-bearing account where, instead of putting a lump sum in at the beginning, smaller chunks are deposited at regular time intervals. Suppose you have $1000 dollars, but you aren’t quite ready to dump $1000 into an account all at once. You could instead put $50 into an account every month for 20 months, giving you a safer (or at least more reasonable) method of deposit.

The formula for annuities is as follows:

F = A⋅[(1+ i)n⋅t −1]

i

where A = the amount deposited at each time interval, i=r/n (r=rate, n=how often deposits are and how often interest is compounded), t is time, and F is the final value (accumulated value) of the annuity. It should be noted that the n used for “i” and the n used in the formula should always be equal, since USAD® does not provide a method for solving questions where this is not true.

The formula looks like a mess, but I challenge you to figure out how it is derived45.!

Where annuities are ways to gain money, loans are similar products in which people pay back money. A person taking out a loan borrows a lump sum at once and pays it back in installments. Mathematically, this is basically taking Section 2.8.1 and setting it equal to Section 2.8.2. Since you take out a lump sum of a loan, the amount that loan is “worth” will increase over time, using compound interest from Section 2.8.1. You will pay this back using annuities, covered in this section. Written in symbolic form,

L(1+ i)n⋅t = A⋅[(1+ i)n⋅t −1]

i

where L is the original loan amount, i=r/n, and all symbols are the same as prior.

Let’s look at an example:

45 Hint: Think of the formula for the sum of a geometric series


http://youtu.be/aN1kqj4ktIg

Suggested prior reading: 2.8.1


Suppose you take out a loan of $10,000. You are given 10 years to pay off the loan, and the interest rate on both the loan and annuity is 3%, compounded yearly. How much will each individual payment be?

10000(1+0.03/1)10 =

A[(1+0.03/1)10-1]/(0.03), so A=1172.31.

This makes sense. With interest, you know that after 10 years, you’re going to be paying a bit more than the $10,000 loaned to you, since you were given a period of time to pay off your loan.

That’s all there is for financial math. To recap, there are 4 different main types of financial math problems:

(1) Compound interest (lump sum gains money with interest added at certain points in time)

(2) Compound interest with continuous growth (lump sum gains money with interest compounded continuously)

(3) Annuities (deposits are at specific time intervals, like interest)

(4) Loans (Loans gain interest while you pay them back, and so do annuities, so set them equal to each other)

Let’s try some sample problems:

1. Old Man Jenkins is putting money into his retirement fund. If he puts $260 every month for 20 years with an effective (compounded yearly) interest rate of 5%, how much money will he have accumulated?

2. Nick is saving up to buy a car. He makes $1500 a month, and is willing to divert 10% of his income monthly into an annuity with monthly compounded interest rate 4.5%. How much money will he have in his “car fund” after 5 years?

3. Nickie takes out a student loan of $9,500 with interest r% compounded quarterly. Assuming that she can pay the loan off in 12 years with payments of $250 every quarter, what is r and what is i?


Answers:

1. A=$260, i=r/n=0.05/1, t=20. So F = 260*[(1+0.05)^(20)-1]/(0.05) = $8597.15

2. A=$150, i=r/n=.045/12, t=5 years, so nt=60 months. So F = 150[(1+.045/12)^60-1]/(.045/12) = $10,071.80

3. This is a loan, so we set the loan equation equal to the annuity equation.

L=9500, A=250, i=r/n=r/4, t=12 years, so nt=48 quarters. Thus,

9500(1+r/4)^48=250[(1+r/4)^48-1]/(r/4) gives that r =3.99%. This means that i is about 1%, since i=r/n.

CALCULATOR HELP: COMPOUND INTEREST, LOANS, AND ANNUITIES Any question involving compound interest, loans, or annuities can be done with the TVM solver on your calculator. Its main advantages are that it completely eliminates the need to remember any of the long, complicated formulas while also saving you valuable time. Although the TVM solver may take some practice to acclimated, I highly recommend you use it on the test. To get to the TVM solver, follow these steps: 1) Press “APPS,” then scroll through your list of applications until you find

“Finance.” Finance may be your only application.


2) Press “ENTER” to select Finance. Press “ENTER” again on the next screen. You will see the following screen.

The TVM solver allows you to solve for 1 unknown variable from this list given that the other 6 have been inputted by the user. To input values, simply use your arrow keys to scroll, then type in numbers using your keypad, pressing “ENTER” after each entry. After you have inputted your 6 knowns, use the arrow keys to scroll to your unknown, press “ALPHA,” then “ENTER.” The unknown value will then appear.

Here is a quick overview of the variables before we jump into some examples.

N: N is the number of times the interest is compounded over the desired time period. In other words, it is simply the number of compounds per year, multiplied by the number of years.

I%: I% is the interest rate, in percent. Do not make the mistake of putting the decimal form here; 5% interest would be entered as 5, not 0.05.

PV: PV stands for the present or principal value. Usually it refers to the lump-sum with which you start the investment.

PMT: PMT stands for payment. It is the amount of money paid per unit time. FV: FV stands for future value. Usually it refers to the lump-sum that will hypothetically be there at the

end of the investment. P/Y and C/Y: This is simply the number of times interest is compounded per year. When you enter P/Y,

C/Y will automatically change to the same number; therefore, you don’t need to worry about C/Y. PMT: The PMT at the bottom should always be selected as “END.” Never change this.

Lastly, PV and FV need to be of opposite signs. The sign designation denotes where the money is going. If the money is leaving you, the sign is negative. If you are retrieving the money, it is positive. For example, for a bank deposit, PV would be negative, since you are giving away the money in the present. However, you will pull the money back out in the future, so the FV would be positive. The best way to teach you how to use TVM solver is to go through examples.


Example 1

Nick deposits $250,000 in his bank account, which earns 8% interest, compounded daily. When Nick withdraws his

money 4 years from now, how much will he have?

Solution This is a fairly basic compound interest question and, in fact, does not require the use of TVM; however, we will use it for practice. We know the interest rate is 8%, so enter ‘8’ next to I%, and that the money is compounded daily, so enter ‘365’ next to P/Y. You’ll notice that C/Y automatically changes to 365 as well. N is just the number of compounds, so we can enter ‘365*4’ next to N46. Nick deposited $250,000 at the beginning, so that will be our Principal Value. Since the money is leaving Nick, it is negative. Enter -250000 next to PV. Because Nick is not making a daily payment, PMT = 0. We have entered our 6 unknowns, and can now solve for the 7th, FV. Scroll to it, press “ALPHA,” then press “ENTER.” You should get $344,269.87 as your answer.

* * *

Example 2

Stephen needs money to buy some trophy polish, so he takes out a $1000 loan. The bank charges him 6% interest,

compounded monthly. If Stephen plans to pay back the loan in 3 years, how much will he owe at that time?

Solution Again, we’ll start with an easy problem. We know I% is 6, P/Y is 12, N is going to be 12*3, and PV is 1000. Note that PV is positive here since it is a loan—Stephen is retrieving money. Since Stephen is not making a monthly payment, PMT is still zero. Now that we have entered our 6 known variables, we just need to solve for the last unknown (FV). Scroll to it and solve to get -$1196.68.

46 You don’t even have to do this in your head; simply enter 365*4 and the calculator will do the arithmetic. Anything to save time.


* * *

Let’s try an annuity now.

Example 3

Katie, an aspiring theoretical physicist, decides to invest a portion of her paycheck into a retirement fund each

month. She decides to invest $150 per month for 15 years. The interest rate on her investment is 9%, compounded monthly. How much will she have in her account at the end of 15 years?

Solution

Again, we’ll start with a relatively basic annuity. I% is 9, P/Y is 12, and N is 12*15. Now let’s figure out PV. You may be tempted to say PV = -150, since Katie is investing $150 to begin the investment. However, remember that PV is only for a lump sum. Since Katie does not actually invest a chunk of money that she lets sit there, her PV is actually 0.47 Here, however, Katie is making a monthly payment, so PMT is not zero. Since she is depositing $150 every month, PMT = -150 (don’t forget the negative!). Now that we have entered our 6 knowns, you know the drill. Solve for FV to find that Katie will have $56,760.87 in her account 15 years from now.48

47 Remember, Katie does not begin her investing until after the first month. 48 Another related question could ask how much total interest Katie has earned. To find this, we subtract the total deposits from this number, or ($150)(12 months)(15 years) = $27000. So, over the course of those fifteen years, Katie earned nearly $30000 in interest—more than she actually invested.


QUIZ 2.8: FINANCE MATH Suggested Time: 8 minutes

Answers on page 128

1. Stephen, at age 25, decides to invest in an IRA. He will put aside $2000 per year for the next 40 years. How much will he have at age 65 if his rate of return is assumed to be 10% per year?49

2. How many quarters (rounded to the nearest quarter) will it take for $500 to grow to $700 if it is invested at APR 4% compounded quarterly?

3. Brian borrows $1600 at 6% interest compounded monthly. He will make monthly payments for 10 years to repay the loan. What is the size of the each payment (rounded to the nearest dollar)?

4. How many months will it take to save $500,000 if you place $500 per month in an account paying 6% compounded monthly?

5. Griffin invests $75 every week in a retirement fund that earns 4% annual interest compounded weekly for 35 years. How much money (rounded to the nearest dollar) will he have in his fund when he retires?

6. A plant grows at a rate of 4.5% for 3 years. Assuming continuous growth and that the plant starts at 12 inches, how tall will the plant be after 3 years?

7. Christine has $120. How much more money would she earn by putting her money as a lump sum in an account that yields 5% interest compounded monthly for a year than if she were to deposit $10 a month for 12 months into an annuity with the same interest rate?

49 If only this were actually possible…


SECTION 2 FINAL EXAM

Suggested Time: 10 minutes

Answers on page 129

1. The sum of the infinite series

12n=3

∞

∑ 14

⎛⎝⎜

⎞⎠⎟

12 n

is

2. The expansion of (7x +2)5 has a term of the form ax 5 . What is a ?

3. The sequence an ,n = 0,1,2,3,…, is an arithmetic sequence. Assume a0 = 3 and a2 = 1. Find the value of a4 .

4. You take out a loan of $500 at 2% annual interest, compounded semi-annually. How much money do you owe after 5 years?

5. Every week, you put $15 into an account bearing some 5.2% percent interest compounded weekly. After how many years (rounded to the nearest tenth of a year) will you have $3000?

6. Find the difference of ( x +2)4 − ( x +3)2 .

7. How many quarters (rounded to the nearest quarter) will it take for $1000 to double in amount if it is invested at APR 6% compounded quarterly?

8. Find the sum of the first 6 terms of a geometric series {a0 ,a1,a2 , ...,an } with a0 = 2 and common ratio 3.

9. The expansion of ( x +1)4 has how many terms with coefficient 4?

10. Amy invests $75 every month in a college fund for her daughter that earns 4% monthly interest for 18 years. How much money (rounded to the nearest dollar) will she have in his fund after she retires?


11. Evaluate

2+7nn=1

10

∑ .

12. The coefficient of the x 5 term of ( x +1)( x 3 +3x 2 +3x +1)( x 2 +2x +1) is equal to the coefficient to the x term of

a. (x+1)4

b. (x+1)3

c. (x+2)5

d. (x+2)6

e. (x+1)6


STATISTICS TABLE OF CONTENTS

Section 3.1: Mean, Median, and Mode……………………..

66

Section 3.2: Range……………………………………………

68

Quiz 3.1-3.2: Mean, Median, Mode, and Range……………

69

Section 3.3: The Five-Number Summary – IQR and Quartiles………………………………………………………

71

Section 3.4: Outliers…………………………………………

74

Quiz 3.3-3.4: Quartiles, IQR, and Outliers……………..…..

76

Section 3.5: Variance and Standard Deviation……………

77

Section 3.6: Z-scores…………………………………………

81

Quiz 3.5-3.6: Variance, Standard Deviation, and Z-scores………………………………..………………………..

84

Section 3 Midterm Exam: Statistics Part 1…………………

86

Section 3.7: Basic Probability……………………………….

89

Section 3.8: Independent Events………………..…………

91

Section 3.9: Dependent Events and Conditional Probability……………………..…………………………….

93

Quiz 3.7-3.9: Probability…………………………………………

95

Section 3.10: Expected Value………………………………... 97


Section 3.11: Probability Distributions………………………………………………

100

Quiz 3.10-3.11: Expected Value and Probability Distributions………………………………...……………

105

Section 3.12: Binomial Distributions………………..…

106

Quiz 3.12: Binomial Distributions………………..…….

113

Section 3.13: Normal Distribution………………..……

114

Quiz 3.13: Normal Distribution………………..………

120

Section 3 Final Exam……………………………..…….

121


SECTION 3.1: MEAN, MEDIAN, AND MODE Statistics is the mathematics of collecting, organizing, and interpreting numerical data. To a non-mathematician, however, a lot of statistical work can look like nothing more than a bunch of numbers on a page. Even to someone with a background in mathematics, the sheer breadth of information that can be displayed and analyzed with statistical methods can be overwhelming. In this section, we will only cover the foundational methods of elementary statistics – nothing here is overly difficult or challenging.50

When they first encounter data sets, statisticians use three different measurements to describe them: location, variability, and distribution. In this section, we will look at the first measurement, location, which tells us where the data is centered and gives us some frame of reference for the data.

One of the most used and most important measurements made by a statistician is the measure of central tendency, which gives an idea of a “typical” value of a data set. In other words, measures of central tendency attempt to determine the value at which data seem to “cluster around,” or a number that serves as the “central value” of a data set. The three most used51measures of central tendency are the mean, median, and mode.

Of these measures, the mean (noted as either µ or 𝑥) is the most common. The mean52 is simply the “average” of a set of numbers. To calculate the mean, you take the sum of the numbers in a data set and divide that by the number of numbers in that set.

For example, try to find the mean of the data set {5,10,25,10,3,7}. This should be a fairly simple calculation. First, we add up all the numbers—5+10+25+10+3+7=60. Then, we take the sum and divide that by number of the numbers in the data set. There are 6 numbers in the data set, so the mean is 60/6, or 10.

A “problem” with the use of the mean as a measure of central tendency arises when outliers come into play. Outliers are numbers in data sets that are distant from the rest of the numbers in the data53. The mean is not resistant to outliers, meaning that when an outlier is included in a data set, it greatly skews a mean in one direction (large outliers make the mean bigger than it would be otherwise, and small outliers make the mean smaller).

Imagine the data set used in the example above, but modified to accommodate an extra value, 1340. Then, the data set would be {5, 10 , 25,10 ,3 ,7 ,1340} and the mean would be 1400/7, or 200, much larger than the original mean without the outlier. If we added a very small number, like -1460, the new mean would be -

50 As a Math major with an emphasis in Stats, I strongly emphasize that you should explore stats more if you find this section interesting – unlike theoretical, relatively inapplicable formal math, stats is directly applicable to reality and is an incredibly valuable field. 51 These are the only 3 that you will be tested on as well as the only 3 that you will encounter in elementary statistics. 52 The official term for the mean we describe is “arithmetic mean,” which distinguishes the term from both the geometric and harmonic means. The reason for this name becomes clear from its method of calculation: addition and division are basic arithmetic operations. 53 Specific methods of determining outliers are outlined in Section 3.4 of this guide.


http://youtu.be/gTS_Xi_5w5A



1400/7, or -200. Obviously this mean does not describe the actual location of the data – most of the data is not centered around 200 or -200 but is instead clustered around 10, with one massive outlier skewing our mean.

The second measure of central tendency, the median, is literally the middle of the data set, separating 54the “upper half” of the data from the “lower half.” The median is found by ordering55 a set of numbers from lowest value to highest value and then finding the middle number in this ordered set. Of course, this only makes sense for sets of numbers with an odd number of numbers. The convention for calculating the median of a set of numbers with an even number of numbers makes sense, though: Take the two “middle” values and find their mean; that number is the median of the entire data set.

For example: find the median of the data set {5, 10, 25, 10, 3, 7}.

First, we have to order the numbers from least to greatest. The result is a data set, {3 ,5 ,7 ,10 ,10 ,25}. Note that this set has two “middle” values, 7 and 10, so we have to take the mean of these two values to find the median. Thus, the median is (10+7)/2 = 17/2 = 8.5.

Unlike the mean, however, the median is resistant to outliers. To see this, let’s add 1,000,000 to the data set we’ve been using; the median will barely change, from 8.5 to 10, whereas the mean would change to over 100000. Because each value is counted the same as all the rest, a very small or very large value won’t skew the data.

The last measure of central tendency, unlike the other two, isn’t something you “calculate”—it’s simply something you observe. The mode is the most common value of a group—the number that, in a data set, occurs the most often56. A data set typically has 1 mode, but there can be multiple modes57 or no modes58 for a data set.

For example: find the mode of the data set {5, 10, 25, 10, 3, 7}.

If we take a quick look at this data set, we can see that the number “10” occurs twice, so the mode of the data set is 10.

54 This will be further explored in Section 3.3 of this guide 55 This is the easiest step to forget. So don’t. 56 A mode is not useful for small data sets, since the odds of the same number occurring multiple times are fairly random and aren’t representative of the sample. However, with large data sets, the mode will generally be around the mean. 57 A data set with two modes is called bimodal 58 If all the numbers in the set occur only once each


SECTION 3.2: RANGE In Section 3.1, we talked about three main methods for describing a database: location, variability, and distribution. Mean, median, and mode are all excellent measures of location; however, they do not paint the whole picture of a database.

To see this, let’s look at an example:

Find the mean, median, and mode of the data set {1,1,2,3,5}. Compare these values to the mean, median, and mode of the data set {-15, -14, -8, 0, 1, 1, 2, 3, 4, 5, 16, 17, 19.2}

At first glance, these data sets can look very different – the second contains many more data points as well as more extreme values than the first. After a few calculations, we can see that the first data set has a mean of 2.4 (1+1+2+3+5=12 and 12/5 = 2.4), a median of 2 (2 is the middle number since the data set is in order), and a mode of 1 (since 1 occurs most frequently in the data set). Using the same methods on the second data set, it becomes obvious that the set has the same data centers as the other—a mean of 2.4, a median of 2, and a mode of 1.

How can two very different data sets have the same central tendency? Intuitively, we know that these two data sets are very different, and so we need to look past mean/median/mode and examine other qualities. First, look at how spread out the data in our second set is compared to the first. The lowest number is negative (and quite so), while the highest number is much larger than any number in the first data set. Although there are many ways to measure this difference in variability, we will look at a fairly basic one here known as the range – simply the difference between the smallest and largest values of a data set. In the above example, the difference between the largest and smallest value of the second data set is 34.2 because 19.2 minus (-15) is 34.2; the difference between the largest and smallest numbers in the first data set is only 4. This difference highlights how spread out the second data set is compared to the first.

Let’s look at an example: find the range of the data set {5, 10, 15, 20, 25, 30, 1000}.

The largest number of this set is 1000, and the smallest is 5. Thus, we take the difference of the largest and smallest, and 1000-5 is 995. The range of this data set is 995.

While the range is a useful “rough” guideline for the variability of a data set, it is not particularly useful. Much like the mean, the range is not resistant to outliers. Outliers will always be the smallest and/or largest values of a data set, and these numbers are the only numbers used in the calculation of range, so the presence of huge outliers might make the range a completely useless measurement. For example, let’s say we have the data set {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1000}. Here, the data set is very clustered around a mean of about 7.5; however, the range is 1000 because of the massive outlier 1000. We can compare this to another data set, like {0,100,200,300,400,500,600,700,800,900,1000}; here, the range is also 1000, but the data is evenly distributed around a mean of 500. Without reporting the mean, the range is completely useless; even with the mean, it only describes extreme values and doesn’t help us with the overall variability of the data. In Section 3.5, we will look at better techniques for describing variability.


http://youtu.be/8q0f0qn1hJk



QUIZ 3.1-3.2: MEAN, MEDIAN, MODE, AND RANGE Suggested Time: 6 minutes

Answers on page 130

1. The mean of the data set {10,15,20,30,90,x,x} is 45. What is ( x +1)2 − ( x −1)( x +1) ? a. 149 b. 74 c. 77 d. 152 e. 75 2. The range of the data set {-22,-7,0,11,y,3} is 33. Which of the following could be the value of y? a. 33 b. 22 c. 3 d. -23 e. -30 3. The median of the data set {10,16,9,22,11,z} is 10.5. Which of the following could be the value of z? a. 23 b. 22 c. 12 d. 11 e. 0 4. Which of the following data sets has 4 modes? a. {1,1,2,3,2,6,6,0,7,8,1,1,2} b. {1,2,3,4} c. {1,2,1,3,1,4,1} d. {2,1,1,8,7,0,6,2,6,3,2,1,1} e. {6,3,2,3,7,1,6,0,2,1} 5. For which of the following data sets is the mode twice the mean? a. {-1,1,0,1,-1,2,2,2} b. {0,1,2,4,8,8} c. {-27,9,18,9,9,-18} d. {8,8,-4,-8,0} e. {15,10,0,10,-15} 6. Find the mean, median, mode, and range for the following data set: {1,1,2,3,5,8,13,21,34}.


7. The mean of the data set {1,4,4,4,4,4,4,5,5,5} is equal to the data set’s a. median b. mode c. range d. all of the above e. none of the above


SECTION 3.3: THE FIVE-NUMBER SUMMARY – IQR AND

QUARTILES Since there are a lot of definitions and vocabulary words for Statistics, there is often a steep “learning curve59” associated with Statistics classes for the typical high school mathematics student; for this reason, Stats courses are often omitted60 from high school curricula.

But fret not! While Statistics is a pain with learning definitions, it is a gift from the mathematical heavens when it comes to useful diagrams and graphs. One such graphical tool is the Box Plot, which is a graphical display of the Five-Number Summary.

However, before we get to making Box Plots, we have to first start with the underlying concept behind them, and that’s quartiles.

Quartiles do just what their name implies61—they divide a data set into 4 equal parts. The lower quartile (called Q1) is the median of the data below the median of a data set and the upper quartile (called Q3) is the median of the data above the median of a data set. A key point to make here is that, like the median, Q1 and Q3 are numbers, not ranges: however, Q1, the median, and Q3 do divide the data into four parts, which are ranges.

The four equal parts, then, are as follows: 1) the data between the minimum value of the data set and Q1, 2) the data between Q1 and the median, 3) the data between the median and Q3 , and 4) the data between Q3 and the maximum value of the data set. Because these 4 parts are equal, we can better describe a data set by saying that “25% of the data lies between the first quartile and the mean,” or “25% of the data lies above Q3”.

Let’s look at an example:

Find the median, Q1, and Q3 of the data set {9,6,8,4,5,1,6,3,1}.

The median is simply the middle of the data set. However, first, we must put the numbers of the data set in increasing order. Thus, the data set is {1,1,3,4,5,6,6,8,9}. The middle number is 5, so the median is five.

Now we can split the data set in two parts, one part below the median (where Q1 lies) and one part above (where Q3 lies). The lower half of the data set is {1,1,3,4,5} and the upper half is {5,6,6,8,9}. Thus, Q1=3 and Q3=6.

The key trick with calculating the median, Q1, and Q3 is whether to include the median or not in our two separate halves. For example, in the above example, the median was 5 and was included in both the lower half and the upper half, influencing our Q1 and Q3. However, if we modify our data set slightly to {1,1,3,4,6,6,8,9}, the median remains the same (5), but we now divide the halves as {1,1,3,4} and {6,6,8,9} - both our Q1 and Q3 change as a result. How do we tell whether to include the median? There are a few ways, but the easiest one is

59 Or, really, a “getting used to it” curve. 60 Mathematical logic also gets this treatment; strangely, though, geometry—which has a similar nature—does not. 61 That is, if you recognize that “quart” usually means something involving the number 4.


http://youtu.be/8q0f0qn1hJk



to check whether the median is an actual data point. In the original data set, the median 5 was also a value in the data set, so we include it in both. In the modified set, 5 was not actually a value, and so we do not include it in our divisions. Another, equivalent way of wording this is to say that sets with an odd number of entries will include the median, while sets with an even number of entries will not include the median.

If, at this point, you can already easily visualize quartiles (“the middle of the lower half” …”the middle of the upper half”), then you should not have any problem answering the quartile-related problems you might see on tests. To take you a step further (and to review what we have discussed so far), we will discuss the Five-Number Summary.

As shown in the simple graphic above, the Five-Number Summary is a great way of organizing data. It consists of the minimum (the smallest member/lower extreme of a data set), Q1 (the middle of the lower half of the data set—between the minimum and Q2), Q2 (the median, or middle of the entire data set), Q3 (the middle of the upper half of the data set—between Q2 and the maximum), and the maximum (the largest member/upper extreme of a data set).

As promised earlier, the Five-Number Summary can be represented graphically. We call this a Box Plot62.

The Box Plot graphs our Five-Number Summary as five points, connected by a horizontal line. We draw vertical lines through the median, Q1, and Q3, and then connect these lines via a vertical line – hence the “Box” Plot.

As an exercise of your understanding, see if you can create a data set whose Five-Number summary can be represented by the box plot above (estimate the five numbers).

62 Box plots will not be tested, but they are standard in statistics classes because they help illustrate quartiles. We present them in this guide to help you intuitively understand quartiles. They are also known as “box-and-whisker plots,” which matches their shape.

Five Number

Summary

Minimum

Lower Quartile / Q1

Median / Q2

Upper Quartile/ Q3

Maximum


{ } The box plot above was extracted from the data set {18, 27, 34, 52, 54, 59, 61, 68, 78, 78, 82, 85, 87, 91, 93, 100}, but any data set with

median = 68 lower quartile = 52 upper quartile = 87 minimum = 18 and maximum =100 will do.

Finally, we can also say that 50% of the data lies between the quartiles, Q1 and Q3—this data, between the quartiles, is called the interquartile range (IQR). 63

Represented symbolically,

IQR = Q3 – Q1

Let’s try an example.

Using the data provided in the last example, calculate the IQR. What percent of the data in the set is between Q1 and Q3?

Our data set is {18, 27, 34, 52, 54, 59, 61, 68, 78, 78, 82, 85, 87, 91, 93, 100}. We can calculate or recall that Q3 = 87 and Q1 = 52. Then, the IQR is simply Q3 minus Q1, or 87 – 52, which is 35.

The answer to the second part of the question is simply a definition. “the data in the set…between Q1 and Q3” represents two “quartiles,” each representing 25% of the data, which means that between Q1 and Q3—the interquartile range, or the “box” part of the box plot—50% of the data is represented.

63 The word “range” makes this definition easier to understand—the range is the difference between the maximum and minimum of a set, so the interquartile range is the difference between the upper and lower quartiles.


SECTION 3.4: OUTLIERS So far, we have used the word “outlier” imprecisely to describe an entry in a data set that is numerically distant from the rest of the set. What precisely does “numerically distant” mean, though?

Now that we are familiar with Interquartile Range64, we are now equipped with the tools to figure out exactly what an outlier is.

Formally, an outlier is a number that lies 1.5*IQR above Q3 or 1.5*IQR below Q1. With that in mind, the easiest way to internalize this is to practice.65

Determine whether or not the following sets have outliers using the IQR test for outliers.

1. Median = 5, Q3=13, Q1=3 , minimum = -6, maximum = 19.

IQR = Q3 – Q1 = 13 – 3 = 10. So 1.5*IQR = 1.5*10 = 15. Thus, any number that is more than 15 above Q3=13 or 15 below Q1=3 is an outlier. In other words, anything above 28 (but not including 28) is an outlier, and anything below -12 (but not including -12) is an outlier. There are no outliers in this set.

2. {-11, 10, 10, 20, 40, 40, 86}

IQR = Q3 – Q1 = 40 – 10 = 30. So 1.5*IQR = 1.5*10 = 45. Thus, any number that is more than 45 above Q3=40 or 45 below Q1=10 is an outlier. In other words, anything above 85 (but not including 85) is an outlier, and anything below -35 (but not including -35) is an outlier. There is one outlier—80—in this set.

3. {51, 65, 39, 5}

64 IQR = Q3 – Q1, for reference. 65 Remember that we are only concerned about Q3, Q1, and 1.5*IQR. I always accidentally use the median in the definition of IQR, but this is wrong.

Suggested prior reading: 3.1-3.3


Unfortunately, in this set, there are 4 terms, which means that, in order to find the median (which you don’t need to find for this example…), Q1, and Q3, you have to take averages. First, we need to order the set: {5, 39, 51, 65}. Then, to find Q3, we find the median between the two numbers above the median, which is another mean—(51+65)/2=58=Q3. Q1 is the median between the two numbers below the median. So (39+5)/2=22=Q1. Then, IQR=Q3 – Q1 = 58 – 22 = 30. At this point, the problem becomes like the first two: 1.5*IQR = 1.5*30 = 45, and outliers will exist at values less than (Q1 – IQR = -23) and greater than (Q3

+ IQR = 103). So, there are no outliers.


QUIZ 3.3-3.4: QUARTILES, IQR, AND OUTLIERS Suggested Time: 5 minutes

Answers on page 131 1. Find Q1 for the data set {1,5,9,13,22}. 2. Find Q3 for the data set {-51,7,24,0,-1,51}. 3. Find the IQR (interquartile range) for the data set {-55,5,0,10,17}. 4. The IQR of the data set {4,10,20,25,x} is 18. Which of the following could be the value of x? a. 6 b. 10 c. 20 d. 25 e. 26 5. Using the IQR test for outliers, would 80 be an outlier for the data set {5,30,35,40,45}? 6. Find the IQR of the data set {1,1,2,3,5,8,13,21}.


SECTION 3.5: VARIANCE AND STANDARD DEVIATION Before we talk about anything new, let’s review the three main descriptive measurements for data: location, variability, and distribution. We have thoroughly examined location using our measures of central tendency – mean, median, and mode. We have also developed some measures of variability, like range and IQR; however, these numbers are fairly crude measurements of variability. In this section, we will look at more refined methods for analyzing variability.

Instead of only looking at the difference between the maximum and minimum values, a better measure of variability would involve comparing each data value to the mean – the farther points are from the mean, the more “spread out” the data must be. Statisticians use a measurement called variance to do exactly this. Formally, variance is the squared sum of the differences between all the values of the data set and the mean of the data set, all over the number of entries in the data set. Symbolically, we write,

σ 2 =

( xi − x )2

i=1

n

∑n

where σ2 represents variance, x represents the mean, xi represents each individual number66 in the data set,

and n is the number of numbers in the data set.

This formula can seem very unwieldy, but it’s actually not that hard to calculate for small data sets. Let’s look at an example:

Calculate σ2 for the data set {-101, -83,-20,60,42,114}.

When calculating variance, we first need to find the mean for a given set of data. In this case, the mean is (-101-83-20+60+42+114)/(6) = 2. Next, we need to find the distance from each point to the mean and square

each value67. So, we find that σ2 = (-101-2)2/6 + (-83-2)2/2 + (-20-2)2/6 + (60-2)2/6, + (42-2)2/6 + (114-2)2/6 =

25138/3, or about 8379.3.

Try one for yourself:

Calculate σ2 for the data set {-11,-5,-4,4,5,7,8,12}

66 Think of the ordered members of a data set as {x1, x2, x3, x4, . . ., xn}, so the index (i=1,2,3,…,n) of the sum will correspond to each number in order. 67 We square each value to eliminate meaningless negative values – distance is an absolute value, so variance is always positive.


Variance: http://youtu.be/tJo--2AcwEU

Standard Deviation:

http://youtu.be/qAJsThHybdU



The mean is (-11+(-5)+(-4)+4+5+7+8+12)/8 = 16/8 = 2. Using the same mean and formula as before but with

data set 2, we get that σ2 = 53.5.

However, the variance is not an immediately useful measurement because it doesn’t preserve units. Say you’re finding the variance of a data set of volumes, measured in liters (L). Since the variance is a squared difference, the variance would be reported in liters-squared (L2).

Fortunately, statisticians have a more useful measurement directly related to the variance. It’s called the

standard deviation, and it’s simply the square root of variance ( σ 2 ). Written symbolically:

σ =

( xi − x )2

i=1

n

∑n

where all symbols mean the same as they did before, and σ is the standard deviation.68

Let’s practice finding variances and standard deviations. Remember that typically, you will be given a data set without knowing the mean, so you will have to calculate that first.

1. Calculate the variance and standard deviation for the data {-5,3,14,8,97}.

2. Calculate the standard deviation for the data set {-11,-5,-4,4,5,7,8,12}.

Answers: 1. σ2 =1393.04, σ =37.32 2. 7.314 – use the tips in the calculator section (up next) to check your

answers.

We will be using standard deviations for the next few sections, so make absolutely sure that you understand them before you continue. Standard deviations are the “standard” way of finding how “dispersed” a data set is, which will become more evident as we explore z-scores and the normal distribution.

CALCULATOR HELP: DESCRIPTIVE STATISTICS Given a finite set of numbers, the mean, median, range, first quartile, third quartile, IQR (Q3-Q1), variance, and standard deviation can be found easily using your calculator by following these steps. Example Set: 1, 2, 3, 4, 5, 6

68 Because we will use standard deviation and almost never use the variance, standard deviation is given its own symbol,σ , while variance is written as standard deviation squared, σ

2.


1. Enter the set of numbers into L1. You can do this by pressing “STAT” then “ENTER” (choosing “Edit…”). It will bring up a screen containing 6 lists, labeled L1 through L6.

2. Scroll over to L1 using the arrow keys, and clear the list if it is not already clear by scrolling up until the cursor is highlighting “L1,” pressing the “CLEAR” button, then pressing “ENTER.”

3. Begin to enter the list by typing in the numbers one-by-one, pressing “ENTER” after each number. The list will grow vertically.

4. When you have entered the entire set of numbers into L1, quit back to the homescreen by pressing “2ND” then “MODE” (which selects “QUIT”).


5. Once back to the homescreen, press “STAT” again. A new screen will pop up.

6. Scroll to the right using the arrow key to highlight “CALC.” Then press “ENTER.” You will automatically be taken back to the homescreen.

7. The homescreen will now read “1-Var Stats” with the cursor flashing to the right. Now, you need to press “2ND” then the number “1” (selecting “L1”). Now press “ENTER.” A list of symbols and numbers will appear.

You can scroll down and back up using the arrow keys to see the rest of the information. Now, the only thing left to do is find the value you need.

Mean: The mean is the 𝐱 value. It would be 3.5 in this case. Median: The median is the Med value. It is also 3.5 in this case. Range: The range is not explicitly given by the calculator, but finding it is simple. Take the maxX value

and subtract the minX value from it. This gives you the range. It would be 5 in this case. Q1: The first (lower) quartile is the Q1 value. It would be 2 in this case. Q3: The third (upper) quartile is the Q3 value. It would be 5 in this case. IQR: Like the range, the IQR is not explicitly given by the calculator. To find it, simply take the Q3 value

and subtract the Q1 value. This gives you the IQR. It would be 3 in this case. Standard Deviation: The standard deviation is the σx value. It would be 1.708 in this case. Variance: The variance is not explicitly given by the calculator. To find it, take the σx value and square

it. It would be 1.7082 = 2.917 in this case.


SECTION 3.6: Z-SCORES Learning all this statistical jargon might not be everyone’s cup of tea, so, with this section, we’ll start with a real life example.

A sample of scores for an Academic Decathlon competition69 is as follows:

43845.7 33391.5 27968.7 42970.3 31500.0 27382.8 42754.1 31489.9 27068.7 42168.6 31164.9 26449.0 40570.2 31024.4 25913.0 38650.6 30864.6 25833.1 38466.4 30649.4 25021.2 37068.2 30540.0 24959.7 35463.0 30191.3 22501.5 34814.6 28478.7 18853.9

with mean x = 31585.6 and standard deviation σ = 6535.8.

Academic Decathlon is all about the competition, and often teams wonder not only how well they did (their personal score) but also how they fared in comparison to others. For example, the top team score above, 43,845.7, is about 850 points above the second-place team. Is this difference meaningful? What about an individual top-scoring student who wins the competition by the same 850 points? Is this number “more” meaningful? How can we put the number “850” in context? In essence, a central question of statistics is not only the absolute difference between values but also their relative positions to the data set as a whole. Using single data entries, the mean of all scores, and the standard deviation of all scores, we can make these comparisons.

The standard way of doing this—finding where a score lies in relation to others—is called the z-score (or standard score70). The formula for the z-score of a data point is:

z = x − x

σ

where z is the z-score, x is an individual data point, x is the mean, and σ is the standard deviation. In plain language, the z-score is how many standard deviations a “score” is above a mean—if a z-score is negative, then it is z standard deviations below the mean.

69If you recognize that these scores are from a competition you recently participated in, you’re spending too much time on the Internet and not enough time studying. 70 Called this because it is heavily involved in the “standardization” of data


http://youtu.be/HECBPvz-Ca8



Historically71, this is what “grading on a curve” has always meant. Scores that were 2 standard deviations (or “2 sigmas72”) above the mean were “A’s,” scores that were between 1 and 2 standard deviations above the mean were “B’s,” scores that were within 1 standard deviation of the mean (up or down) were “C’s,” scores that were between 1 and 2 standard deviations below the mean were “D’s,” and anything lower was an “F”.

Looking back at our scores on the left from the Decathlon competition, we are now equipped to answer questions about z-scores. For example, how many standard deviations above the mean did the top team score? We can quickly bust out our calculators and see that z = (43845.7 – 31585.6)/6535.8=1.876, so the top scorers were 1.876 standard deviations above the mean.

Then, we could ask what a team would have to score to break the “2 sigma mark,” looking at our analogy to “grades” earlier and thus implying that this team should receive an “A”. To do this, we simply plug in “2” for z and solve for the rest. We get 2 = (x - 31585.6)/6535.8, so73 2*6535.8 + 31585.6 = x. This yields that x = 44657.2, so a team with a score of 44,657.2 would break the “2 sigma barrier” and score 2 standard deviations above the mean.

Now, you’re equipped74 to answer any z-score questions the test may throw at you. Before you move on to solving z-score problems, consider this: The sum of all z-scores in a data set is 0. Why?

Consider scores on a Calculus exam for two different teachers.

Teacher 1:

67 48 90 92 86 60 70 32 100 48

Teacher 2:

75 60 75 80 90 75 85 80 78 69

1. For Teacher 1’s exam, what is the z-score of the student that received a 100? What is the z-score for the student that received a 90?

2. For Teacher 2’s exam, what is the z-score for the student that received an 80? What is the z-score for the student that received a 90?

3. Compare the z-score of the students that received a 90 in each class. Which z-score is higher, and what does this say about the student’s relative standing 75in his/her class?

71 We say “historically” because many teachers/professors have switched to the arbitrary “add a few, take a few” method to “curving” a test. 72 That’s what the Greek letter “σ ” is called. 73 This is simply algebraic manipulation to show intermediate steps. 74 Somewhat equipped, that is. Now you just have to mentally prepare for what happens if you’re given nothing more than a data set. You have to manually calculate the mean and standard deviation (and, technically, the variance before that!) before you even consider the z-score. 75 If a student’s z-‐score is higher than another’s, this means he/she did relatively better. That is the whole point of z-‐scores—to be able to compare people in a “normal”/standardized manner. See our Normal Distribution section for more.


Answers:

1. First, calculate the mean and standard deviation of the set to be used in the z-score formula. The mean, µ, is 69.3, and the standard deviation, σ , is 21.345. It would be in your best interest to use a calculator (as outlined in the previous section) to calculate these two values, if you can.

z-score (score of 100) = (100 – 69.3)/21.345 = 1.438

z-score (score of 90) = (90 – 69.3)/21.345 = .969

2. Following the methods above, the mean, µ, is 76.7, and the standard deviation, σ , is 7.849. It would be in your best interest to use a calculator (as outlined in the previous section) to calculate these two values, if you can.

z-score (score of 80) = (80 – 76.7)/7.849 = 0.420

z-score (score of 90) = (90 – 76.7)/7.849 = 1.694

3. The student in Teacher 2’s class with the score of 90 did relatively better, since 1.694 is greater than 1.438.


QUIZ 3.5-3.6: VARIANCE, STANDARD DEVIATION, AND Z-SCORE Suggested time: 6 minutes

Answers on page 131

1. Calculate the variance of the data set {1,5,6}.

2. The variance of the data set {-1,0,1,x} is 2.1875. What is the value of x? 3. Given the data set {10,20,36,44}, calculate the variance and standard deviation (approximate to 3 decimal places). 4. Given a data set with µ = 5 and σ = 1, what is the z-score for 7? Use the following data set for questions 5-6 90 92 97 92 93 100 87 5. Calculate the variance and standard deviation of the data set. 6. Find the z-score for the score of 97. How many standard deviations above the mean is 97?


Use the following data set for questions 7-8 x x -10 -15 x -20

7. The variance of the above data set is

1003

. Assuming that x is NOT -15, what is the value of x?

a. -10 b. -5 c. 0 d. 5 e. 10 8. Assuming that x is -15, what is the z-score of -20?


SECTION 3 MIDPOINT EXAM: STATISTICS PART 1 Suggested time: 12 minutes

Answers on page 132 1. Which of the following data sets is bimodal? a. {0,1,1,2} b. {1,1,1,0} c. {2,2,1,1,5} d. {2,1,0,3} e. {2,1}

2. The standard deviation of the set {1,1,x,y} is 2 . Which of the following could be the value of x+y? a. -1 b. 0 c. 1 d. 2 e. 3 3. What is the IQR (interquartile range) of the data set {45, 50, 66, 75, 42}? a. 25 b. 25.5 c. 26 d. 27 e. 27.5 4. The product of the mean and median of the data set {-4,-2,1,0,4, 5} is a.

13

b. 12

c. 32

d. 1 e.

23

5. Given Q3=15, Q1=5, and Median=10, using the IQR test for outliers, we know that outliers exist BELOW a. 15 b. 10 c. 5 d. -5 e. -10


6. The median for the data set {152, 160, -5, 44, m} is 50. Which of the following could be the value of m? a. 46 b. 56 c. 160 d. -5 e. none of the above 7. Calculate the variance for the data set {50,51,50,50,50,100}. a. 18.569 b. 50 c. 344.583 d. 2500 e. 287.435 8. The range of the data set {-22,-7,0,11,y,3} is 44. Which of the following could be the value of y? a. 33 b. 22 c. 3 d. -23 e. -30 Use the following table of scores on a twelve-question statistics exam for questions 9-12. 3 12 4 8 9 7 4 5 7 6 9. Which of the following is the LARGEST for the data set of scores? The a. mean b. range c. mode d. median e. standard deviation 10. Calculate the standard deviation of the data set. a. 6.650 b. 6.5 c. 2.653 d. 2.579 e. 2.549 11. What is the z-score for the score of 12? a. 2.391 b. 2.204 c. 2.157 d. 2.133 e. 2.075


12. Calculate 1.5*IQR for the exam. Of the given scores, how many outliers are there for the exam? a. 4; there is 1 outlier b. 4, there are no outliers c. 6; there are 2 outliers d. 4; there are 2 outliers e. 6; there are no outliers


SECTION 3.7: BASIC PROBABILITY Probability is a fundamental building block of statistics76 and combines a number of mathematical concepts. By definition, probability measures the expectation that an event will occur or that a statement is true. To calculate a probability, you simply divide the number of specific occurrences of an event by the total number of outcomes.

That is,

P(event) = (number of ways event can occur)/(total number of outcomes)

This statement is not incredibly meaningful in words, so let’s try a specific event. Suppose you want to find the probability of rolling an even number with a fair 6-sided die. To do this, follow these 3 easy steps:

1. List the ways the event (in this case, rolling an even number) can occur

2. Count the number of possible occurrences of the event

3. Find the total number of outcomes (this is usually obvious) and divide the number in step 2 by this number

So, for the probability of rolling an even number, we have the following two sets:

Rolling an even number = {2,4,6}

Possible outcomes on a 6-sided die = {1,2,3,4,5,6}

There are 3 numbers in the first set and 6 numbers in the second, so for the probability of rolling an even number, we have

P(even)= 3

6= 1

2

Though some probability problems can get quite complicated, they all follow this basic probability model. Before we move on, here are some basic properties of probability:

1. Probabilities are written as decimals or fractions and must always be between 0 and 1 (inclusive) – written symbolically, it is always the case that 0 ≤ P(event )≤1.77

2. The “complement” of an event E, denoted E’, is the probability that E will NOT occur. That is, if E is “rolls an even number,” then E’ is “does NOT78 roll an even number”—so P(E) + P(E’) = 1, that is, P(E’) = 1 – P(E).

76 Especially the statistics that you are about to learn 3-4 sections from now 77 This makes sense: having a probability greater than 1 means that there are more desired outcomes than possible outcomes, which is by definition impossible. Similarly, a negative probability means that one of the two values must be negative – which also doesn’t make sense. 78 Which, due to the properties of integers, is equivalent to “rolls an ODD number”



3. The sum of all probabilities for a particular situation79 is 1. Since the complement of a probability involves all events that are NOT a certain event, this follows80 from definition (2) above.

Let’s try some sample problems.

Calculate the probability of the following events.

1. Rolling a number that is a multiple of 3 on a 6-sided die

2. Getting 3 “Heads” after flipping a fair coin 3 times

3. NOT getting the sequence “Heads, Tails, Heads, Tails” after flipping a coin 4 times

1. The possible multiples of 3 rolled with a 6-sided die are {3,6}, so there are two possibilities. We already know that there are 6 total outcomes for a 6-sided die—{1,2,3,4,5,6}, so P(multiple of 3)=2/6=1/3.

2. The possible outcomes for a fair coin are “Heads” or “Tails”. As far as the number of ways the event can occur, there is only 1: {H,H,H}. For the total number of outcomes, we have {H or T, H or T, H or T}, which gives a variety of outcomes: {H,H,T}, {H,T,T},{T,T,T},{T,H,H},{T,T,H},{T,H,T}, {H,T,H},{H,H,H}. There are 9 different outcomes (23, since there are 3 flips with two outcomes—H or T—each), so we have P(H,H,H)=1/8.

3. Since we know “NOT” just means “find the complement,” which is 1 – P(E), we should find P(E) first. We know—using the methods from the last problem—that not only is there only ONE unique occurrence of {H,T,H,T}, but also that there are 24=16 different possible outcomes (you can list them to the right if it helps). So P(H,T,H,T)=1/16, but since we need to find P(NOT H,T,H,T), we simply calculate 1 – P(H,T,H,T)=1-1/16=15/16.

79 Rolling a die, spinning a spinner, flipping a coin, etc. 80 Using a 6-sided die as an example, P(roll a 1)=1/6, P(2)=1/6, P(3)=1/6, P(4)=1/6, P(5)=1/6 and P(6)=1/6, so P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1.


SECTION 3.8: INDEPENDENT EVENTS As mentioned in the previous section, probability can get a bit complicated at times—what happens if you’re trying to find the probability of a number of events at once? What do you do if one of your events depends on the other? Luckily81, probability—like the rest of mathematics—has a clear and logical structure. By that, I simply mean that there are rules.

When there are multiple events occurring, there are two possibilities:

1) The occurrences of one event depend on the other

2) The occurrences of one event do NOT depend on the other

The first case involves events that are called dependent events. This section focuses on the second case, called independent events.

Some examples of independent events: (1) You pull different colored socks from a drawer and roll a pair of dice, (2) You flip a coin and spin a spinner, (3) You flip four different coins, (4) You find the blood-type of two different people, and basically any set of events where one thing does not depend on the other. After you read about dependent events, “independent event problems” will become glaringly obvious.

To calculate the probability of multiple events that are independent, you simply multiply each event’s probabilities together. That is,

P(A and B) = P(A)P(B)

Or, more generally,

P(E1 and E2 and E3 and . . . En) = P(E1)P(E2)P(E3)…P(En)

Let’s look at some examples:

Calculate the probability of the following independent events. If an event is not independent, write “DEPENDENT”.

1. Rolling a prime number on a 6-sided die and flipping “Heads” with a fair coin

2. Drawing a yellow and a blue marble from a bag of 3 red marbles, 3 blue marbles, 6 yellow marbles, and 8 green marbles. Assume you put the yellow marble back in the bag before randomly drawing another marble.

81 There might be a really lame probability joke there somewhere…


http://youtu.be/DQ1vQEtr0yA



3. Drawing a yellow marble and then another yellow marble from the same bag as above without putting the first marble back.

Answers:

1. Simply multiply the two probabilities together—the prime numbers on a 6-sided die are 2 and 5 (1 is NOT prime), so there are 2 outcomes and P(prime)=1/3. P(Heads) is clearly 1/2. Thus, P(Prime and Heads)=1/3*1/2=1/6.

2. Multiply the probabilities. P(Yellow)=6/20, P(Blue)=3/20, so P(Yellow and Blue)=18/400=9/200.

3. DEPENDENT—drawing the second marble depends on you drawing the first marble.


SECTION 3.9: DEPENDENT EVENTS AND CONDITIONAL

PROBABILITY We have already explored the two different “types” of probability—independent and dependent. In this section, we will explore dependent events and their probabilities.

The probability of two dependent events is the probability of the first event times the probability of the second event given that the first event occurred. The second part of that definition involves what is called conditional probability. The probability of “B given that A occurred” is written P(B|A).

As stated before (but now in symbols), the probability of dependent events A and B is

P(A and B)= P(A)P(B | A)

For once, the material makes your life easier—you will never82 have to calculate P(B|A) on a test. It will almost always be given on a table or easily discerned. What is the probability of drawing two Queens in a row—without replacement83—from a deck of cards? Using the above formula,

P(Q and Q)=P(Q)P(Q|Q), with P(Q)=4/52=1/13 and P(Q|Q)=3/5184=1/17, so P(Q and Q) = 1/13*1/17=1/221.

Let’s try some problems.

Calculate the probability of the following dependent events given P(A)= 0.5, P(B)=0.25, P(C)=.2, P(D)=0.05, P(B|A)=1/3, P(A|C)=.75, P(A|D)=.1, P(B|C)=.5, P(C|D)=2/3.

Assume that, with dependent events, “A and B” means A and then B, given A.

1. P(A and B)

2. P(B and C)

3. P(D and C)

4. P(C and B)

82 A possible formula, through quick algebraic manipulation, is P(B|A)=P(A and B)/P(B). 83 These two words are often your key toward deciding whether or not two events are dependent. 84 There is 1 less Queen and thus 1 less card in the deck.


http://youtu.be/E8LpcqoLMyM



Answers:

1. P(A and B) = P(A)P(B|A)=(0.5)(1/3)=1/6

2. P(B and C)=P(B)P(C|B) cannot be determined from the above data, since P(C|B) is not given

3. P(D and C) = P(D)P(C|D)=0.05*2/3=1/30

4. P(C and B)=P(C)P(B|C)=.2*.5=.1


QUIZ 3.7-3.9: PROBABILITY Suggested time: 8 minutes

Answers on page 133

The following is a table of the eight possible blood types along with the approximate percent of the United States population having that type. Use the table for questions 1-2.

O positive 38% A positive 34% B positive 9% O negative 7% A negative 6% AB positive 3% B negative 2% AB negative 1%

1. What is the probability that a randomly selected person has a blood type that is Rh-negative (meaning any blood type with the word “negative” next to it)?

2. What is the probability that a randomly selected person has a blood type that is not A, AB, or Rh-negative?

3. A box contains 12 flashlights, of which 5 are defective. All flashlights look the same and thus have equal probability of being chosen. What is the probability that, if 3 flashlights are removed from the box, at least 2 are defective?

4. What is the probability of flipping two fair coins and getting two tails given that, no matter what, at least one of the flips will be tails?

5. Suppose a school of 1000 people includes 70 Decathletes85 and 520 females. There are 40 females who are Decathletes. A person is chosen at random, and we know that said person is a female. What is the probability that the person is a Decathlete, given that the person is a female?

6. Two cards are drawn at random (without replacement) from a regular deck of 52 cards. What is the probability that the first card is a diamond and the second is red (meaning it has a suit of hearts or diamond)?

85 There are obviously a few weeks left until the district competition…


7. In a game of poker, a hand of 5 cards is dealt to each player. Find the probability of getting a hand containing only diamonds.

8. If E and F are events with P(E) = .2 and P(F) = .4, and P(E and F) = .1, find P(F|E).

9. You toss two fair dice. What is the probability of obtaining a sum of either 7 or 11?


SECTION 3.10: EXPECTED VALUE We’re about to jump into some more applied probability—the kind that actual statisticians use and think about.

First, though, let’s get some definitions out of the way:

1. A probability distribution is a set of outcomes in which the probability of all outcomes is between 0 and 1— 0 ≤ P(Ei )≤1 and the sum of all probabilities is 1.

2. The expected value of a probability distribution is the mean “payoff” of the distribution over a long time86.

The formula for the expected value is

E = P(Ei )⋅Ei

i=1

m

∑

where E is the expected value (you can remember the definition from this—the last time you saw a “bar” over

a letter was with the mean, x , so it makes sense that the expected value is the “mean” payoff) and Ei is each event in a distribution. For this formula to work, Ei has to have a “value” associated with it.

To help you wrap your mind around these formulas, let’s do an example.

Suppose you’re playing a lottery87 with $2 tickets, where the following holds true:

Event Win the lottery—$174,000,000

Smaller prize—$10000

Winning nothing— $0

Probability of Event

0.000000005707 0.000001541 0.9999984534

Remember that, since you’re paying $2 to play the lottery, all the winnings will be $2 less than what is written88.

86 The idea of “over a long time” cannot be stressed enough in probability theory—think of flipping a coin. If you flip a coin 6 times, you won’t always get 3 heads and 3 tails, but if you flip a fair coin 1 billion times, the probability of getting heads will be something like 0.49999999…much closer to the ideal 1/2. 87 This is a modified distribution of an actual lottery in Arizona. It’s also the same distribution used in Decademy Video Lecture 20.


http://youtu.be/xduuiaG9na8

Suggested prior reading: 3.7,3.9,2.2


To calculate the expected value of the game, you simply multiply the value of the each event times its probability and sum these products together. For the distribution above, this looks like $174,000,000*0. 000000005707+ $10000*0.000001541+0*0.9999984534 (keeping in mind the cost of the ticket, so subtract $2 from whatever you get). Another similar example is on our Facebook page (www.facebook.com/decademy/), specifically: https://www.facebook.com/Decademy/posts/501409423213761

It is important to note that the expected value of a single event is simply the probability of that event times the event’s value, i.e. P(winning the lottery)*value of lottery.

Typically we will want to find the expected value of an entire game. For this lottery, we multiply P(win the lottery)*(value of lottery89) + P(win $10,000)*(10000-2) + P(win nothing)*(0-2).

For this lottery, E turns out to be -0.99. That means, after playing the lottery a large number of times, you’re bound to lose $0.99 on average.

That means that the lottery is not a fair game, since “the house90” gets more money out of the game than you do on average.

We define a fair game to be a probability distribution with an expected value of 0. In other words, a game is fair when, over time, you neither win nor lose money.91

Let’s try some examples:

You are playing a number of “betting” games with a friend. Find the expected value of each game—also, decide whether or not each game is fair and, if not, decide who will more often have a net gain from the game.

1. You roll a die. If the result is even, you get $5; if not, you give your friend $5.

2. You are drawing cards from a standard deck of 52. If you pull an even number (2,4,6,8,10), you get $10. If not, your friend gets $8.

3. You buy a ticket from your friend for $2. There is a 5% chance that ticket will net you $100, and a 55% chance it will net you $2. The expected value of the game is $7, in your favor. What is the probability your ticket will net you $5? Besides these 3 outcomes, your ticket may also win nothing and thus net -$2.

88 That means if you win nothing, you actually lose $2. 89 This number is so big it’s not going to make a difference whether or not you subtract $2 90 The people in charge of the lottery; this term is more commonly used in gambling, where the dealer is the “house.” 91 Games in Vegas are by definition never fair – the casinos need to make money. Some games, however, have better odds than others; poker and blackjack are your best bet, while roulette and Keno have terrible odds. Over a lengthy period of time, though, you will lose – there’s no skirting around the laws of probability.


Answers:

1. P(even)=1/2, P(odd)=1/2. $5*1/2 + (-$5)*1/2 = 0. This is a fair game.

2. P(even card)=20/52=5/13. P(any other card)=32/52=8/13. $10*5/13 + (-$8)*8/13=-14/13=-1.0769. This game is not fair and is your friend’s favor.

3. P(net $100)=.05, P(net $2)=.55, P(net $5)=X, P(net -$2)=1-P(net $100 or $2 or $5)=1-(.6+X)=.4-X. Thus, .05*$100+.55*$2+X*$5+(.4-X)*(-$2)=$7. By algebraic manipulation, X=P(net $5)=.2429=24%.

M a t h e m a t i c s C u r r i c u l u m G u i d e | 1 0 0

SECTION 3.11: PROBABILITY DISTRIBUTIONS

Now that you’re an expert on expected values92, you probably understand a little bit more about how chance games work. With the expected value, you can figure out the mean pay-off of a game. However, by now you should know that the mean (“central tendency”) is not always the most important thing to know about a distribution or data set. What if you—a gambler—want to figure out how to capitalize on variations in game’s prize distribution? In other words, how do measures of variability, like variance, standard deviation, and z-score, factor into probability?93

In this section, we will examine probability distributions through the lens of expected value; this will give us some interesting formulas for our standard analysis tools. First, we’ll list the formulas for mean, variance, standard deviation, and z-score for “regular” data sets. Then, we will list their analogues for probability distributions.

Mean: x =

xii=1

n

∑n

Variance: σ 2 =

( xi − x )2

i=1

n

∑n

Standard Deviation: σ =

( xi − x )2

i=1

n

∑n

Z-score: z = x − x

σ

Remember that a probability distribution is a set of outcomes in which the probability of all outcomes is between 0 and 1— 0 ≤ P(Ei )≤1 and the sum of all probabilities is 1.

92Right? 93 Since none of you are gamblers (right?), perhaps another example will be a little more understandable. A few years back, Harvard mandated that its professors give grades with a mean GPA of 3.2. However, the variability was still up to the professors, which led to some interesting results. The best students wanted professors who “spread” the grades – that is, gave out many As as well as many Fs. The borderline students, however, wanted “clumpers,” professors who assigned most of their grades around a 3.2. In other words, the best students wanted high standard deviations because they were more likely to get a high grade; poorer students, in contrast, wanted low standard deviations because they were likely to be near the mean.


http://youtu.be/vaniXXkHBkU

Suggested prior reading: 3.7,3.10,3.5,2.2

1 0 1 | M a t h e m a t i c s C u r r i c u l u m G u i d e

That being said,

Mean (Expected Value):

E = P(Ei )⋅Ei

i=1

m

∑

Variance:

σ 2 = (Ei − E )

i=1

m

∑2

⋅P(Ei )

OR (easier to calculate with a calculator, though either is fine)

σ 2 = [Ei

2 ⋅P(Ei )]− E2

i=1

m

∑

Standard deviation: σ = (Ei − E )

i=1

m

∑2

⋅P(Ei )

OR

the square root of the second variance formula above

Z-score: z = E − E

σ (where E is a specific outcome)

Any example from the previous section will hold in this section. Check out the question here94: https://www.facebook.com/decathlonmemes/posts/293623727423956 to see a great example of what you might catch on exam regarding probability distributions and variance. The most important thing to take from this section is that is the foundation for the final two sections on Binomial and Normal Distributions.

94 The question reads “Suppose a game involves rolling a die. If a player rolls an even number; he must pay $1; if a player rolls a 3; he receives $2; and if a player rolls a 1 or a 5, he does not gain or lose anything. What is the variance of this probability distribution?” for which the answer is 41/36, explained here: http://imgur.com/opP63


CALCULATOR HELP: PROBABILITY OR FREQUENCY DISTRIBUTIONS Given a table of values and their corresponding probabilities (or frequencies), the expected value and variance can easily be determined with your calculator. Even if a table is not given, it is possible that you will be expected to create the table yourself from the problem using basic probability skills. Once the table is generated, however, the expected value and variance calculations are trivial. 1. The first step is to enter the table into L1 and L2. It

is critical that you put the ‘values’ in L1, and the probabilities/frequencies in L2. To enter the lists, press “STAT” then “ENTER” (choosing “Edit…”). It will bring up a screen containing 6 lists, labeled L1 through L6.

2. Scroll over to L1 using the arrow keys, and clear the list if it is not already clear by scrolling up until the cursor is highlighting “L1,” pressing the “CLEAR” button, then pressing “ENTER.” Repeat this for L2.

3. After both lists are cleared, scroll back over to L1 and begin entering the values by typing in the numbers one-by-one, pressing “ENTER” after each number. The list will grow vertically. Now repeat this by entering the probabilities/frequencies into L2. Both lists should be the same length when you are finished.


4. When you have entered both lists, quit back to the homescreen by pressing “2ND” then “MODE” (selects “QUIT”).

5. Once back to the homescreen, press “STAT” again. A new screen will appear.

6. Scroll to the right using the arrow key to highlight “CALC,” then press “ENTER.” You will automatically be taken back to the homescreen.

7. The homescreen will now read “1-Var Stats” with the cursor flashing to the right. Now, you need to press “2ND” then the number “1” (selecting “L1”).Then, press “2ND” then the number “2” (selecting “L2”). Press “ENTER.” A list of symbols and numbers will appear.


Although you can scroll down to see the rest of the list, the only two values you’re ever going to need are on the very first screen.

Expected Value: This will be the 𝐱 value. In this case, it is 2.05. Variance: Although variance is not explicitly given, it can easily be found by taking the σx value and

squaring it. In this case, it would be 0.9732 = 0.947


QUIZ 3.10-3.11: EXPECTED VALUE AND PROBABILITY

DISTRIBUTIONS Suggested time: 6 minutes

Answers on page 133

1. What is the expected number of tails in tossing a fair coin 3 times?

2. A special dice game is made where players can earn (and lose) money. The player recovers an amount of dollars equal to the number of dots on the face that turns up. However, when a 5 or 6 is rolled, the player will lose $5 or $6, respectively. What is the expected value of the game?

3. Suppose that 1000 tickets are sold for a raffle that has the following prizes: one $300 prize, two $100 prizes, and one hundred $1 prizes. The remaining tickets are worth $0. In order for the raffle to be nonprofit (to make the game fair), how much should the house charge for a raffle ticket?

4. Suppose that you receive $3 each time you obtain two heads when flipping a coin two times and $0 otherwise. What is the expected value of the game? In order to make the game fair, how much should you pay each time you flip a coin?

Use the following table for questions 5-7

Outcome E1 E2 E3 E4 Probability .4 .2 .1 .3 Payoff 2 3 -2 0

5. What is the expected value of the table/distribution?

6. Calculate the variance and standard deviation of the distribution given in the table.

7. What is the z-score of E3?


SECTION 3.12: THE BINOMIAL DISTRIBUTION

As we draw closer to knowing all the statistics you need to know to excel in Decathlon, ideas begin to build on each other and become more relevant to “real world” events. The Binomial probability Distribution comes up in quite a few places in “real life” probability, which makes sense since it’s fairly rigorous.

The Binomial Distribution occurs when the following conditions are met:

1) The probability distribution being discussed is composed of independent events

2) Events must either be calculable as SUCCESS or FAILURE (TRUE or FALSE…YES or NO).95

3) The probability of “SUCCESS” is constant across events.

With probability of success “p” and probability of failure “q,” the probability of “k” successes out of “n” events

is given by the formula P(k successes out of n events)=

nk

⎛⎝⎜

⎞⎠⎟

pkqn−k .

Easily THE MOST important thing about Binomial Distributions is that the probability of success and failure must add up to one (that is, p + q = 1). This follows from our earlier definitions of probability distributions, and also from the idea that if the probability of something is p, then the compliment of that probability is 1 – p. That is, 1 – p + p = 1 = p + q. Also note that with n events, n + 1 different possible outcomes exist, since you must always account for the fact that a SUCCESS/FAILURE may occur “0” out of n times. Knowing that the Binomial Distribution is simply a special case of a probability distribution, we can apply our knowledge of probability distributions and thus solve every question thrown at us. Let’s look at a table that covers binomial probabilities.

Suppose you have designed a computer program that makes an 8-question True-or-False exam. The following table depicts the various features of the scenario. Fill in the blanks when needed96.

# of “TRUE” answers Probability by Binomial Distribution

Decimal approx. P(Ei )⋅Ei

Ei

2 ⋅P(Ei )

0

80( )(.5)0 (.5)8

0.0039063 0 0

1

81( )(.5)1(.5)7

0.03125 0.03125 0.03125

2 0.109375 0.21875 0.4375

3 0.21875 0.65625 1.96875

95 As my father was fond of saying, “It’s a digital world – 0 or 1. Either you’re right or you’re wrong.” At the very least, he was right for binomial distributions. 96All you’re doing is changing the number you “Choose” and the exponents of the probability of p = TRUE and q = FALSE


http://youtu.be/8_HoGd1ZX2g

Suggested prior reading: 3.11,2.2


4

84( )(.5)4 (.5)4

0.273438 1.09375 4.375

5

85( )(.5)5 (.5)3

0.21875 1.09375 5.46875

6 0.109375 0.65625 3.9375

7 0.03125 0.21875 1.53125

8 0.0039063 0.03125 0.25

You should recognize “ P(Ei )⋅Ei ” as the formula for expected value. For this distribution, summing this column

in the table gets you 4, which makes sense. You expect to get 4 “TRUE” answers, knowing that TRUE has a 50% chance of occurring (and so does FALSE), and there are 8 occurrences. For a binomial distribution, we can skip any intermediate calculations and calculate the expected value with a simple formula:

E = np

Thus, for our distribution, we simply multiply n, the number of occurrences, by p, the probability of each occurrence (so 8*.5) and get the expected value, 4.

Summing the final column, which you should recognize as the component pieces of “variance97,” you get a similar result. By subtracting the square of the expected value from the sum of this column (see below), you get that the variance is 2.

There is a convenient formula for variance of a binomial distribution, too, and it is

σ2 = npq

and thus standard deviation is

σ = npq

With all that in mind98, here are some take-home points about binomial distributions:

1. You need to know 3 formulas: The formula for each term of the binomial distribution, the one for expected value, and the one for variance99

97 Using the “convenient” formula, σ 2 = [Ei

2 ⋅P(Ei )]− E2

i=1

m

∑

98 And that’s a lot. 99 Remember, standard deviation is always the square root of variance.


2. p + q = 1, and q must be the opposite of p. Everything you deal with in a binomial distribution is split into successes and failures.

3. Calculating the

nk

⎛⎝⎜

⎞⎠⎟

factor of the distribution is tedious, and should be done with a calculator whenever

possible. This is outlined in the Calculator Guide portion of this Practice Book.

4. You don’t always need to calculate the whole distribution. If a question about the prior table asked about cases “with more than 4 TRUE answers” then you would only need to deal with (and sum up) the probabilities when there are 5,6,7, and 8 TRUE answers.

CALCULATOR HELP: BINOMIAL DISTRIBUTIONS Binomial distributions are probability distributions with a very strict set of criteria. The criteria are as follows:

There must be a finite number of trials There must be only two possible outcomes—a “success” and a “failure” The trials are independent The probability of success or failure remains constant

If the above criteria are met, we have a binomial distribution. The best example of a binomial distribution is flipping a coin. Binomial distributions use the following variables:

n: Number of trials p: Probability of a success q: Probability of a failure

To find the expected value and variance of a binomial distribution, there’s no trick. Just remember the formulas.

E(X) = n*p V(X) = n*p*q

σ(X) = 𝐧 ∗ 𝐩 ∗ 𝐪 A faster way to find these probabilities is to use the following nifty calculator functions. This eliminates the need for messy formula and will save you a considerable amount of time. First, some clarifications. When asked to find the probability that an exact number of trials be successes, you will use the binompdf function. That’s pdf, with a p. When asked to find the probability that “at least” or “fewer than” a certain number of trials be successes (i.e., a range), you will use the binomcdf function. That’s cdf, with a c.


To get to these functions: 1) Press “2ND” then “VARS” (selecting “DISTR”). This

screen will appear.

2) To select binompdf (with a p), you have it easy! Just press “0.” You will automatically be taken back to the homescreen.

3) To select binomcdf (with a c), you have it a little bit harder. Instead of just pressing “0,” you need to press “ALPHA” then “MATH” (selecting A). If you feel uncomfortable doing that, you can just scroll up until you see binomcdf( on the list, and press “ENTER.”


4) After doing one of these, your homescreen will read “binompdf(“ with a flashing cursor (or binomcdf if you selected that one). Now, we need to enter the parameters. Enter “n” (the total number of trials), then press the “comma” key, then enter “p” (the probability of a success), then press the “comma” key again, then enter the desired number of successes. Now press “ENTER.” The probability will be generated right there on the screen.

The hard part of this calculator trick is determining whether binompdf or binomcdf needs to be used. To try to nail this home, let’s do a few examples.

Example 1

Nick flips a fair coin 50 times. Find the probability that he will flip exactly 30 heads.

Solution Since this is an “exactly” question, we know we’re going to be using binompdf. So we’re going to press “2ND,” “VARS,” and “0”. Our homescreen should look like this:

Now we enter our parameters. The total number of trials is 50. Press comma, then enter the probability of a success. If we designate a success as flipping a head, the probability of a success is 1/2. Press comma again, then enter the desired number of successes (30 in this case). Our homescreen should now look like this:

Press “ENTER.” The probability of Nick getting exactly 30 heads is 0.042.

* * *

Example 2


Stephen rolls a fair die 100 times. Find the probability that he rolls fewer than 12 threes.

Solution Since this is a “less than” problem, we’re going to use binomcdf. So we’re going to press “2ND,” “VARS,” “ALPHA,” and “MATH.” Our homescreen should look like this:

Now we enter our parameters. The total number of trials is 100. Press comma, then enter the probability of a success. If we designate a success as rolling a three, the probability of a success is 1/6. Press comma again, then enter the desired number of successes (12 in this case). Our homescreen should now look like this:

Press “ENTER.” The probability of Stephen getting less than 12 threes is 0.1297.

* * *

Example 3

Let’s say Katie, like Stephen, rolls a die 100 times. Find the probability that Katie rolls at least 20 fours.

Solution Since this is an “at least” problem, we’re going to use binomcdf. However, let’s pause here for a second. The binomcdf function is made for finding the probability that the number of successes is less than or equal to a number, not greater. So how will we find the probability that the number of successes is greater than a certain number? To do this, we utilize the fact that the total probability has to equal 1. Rather than finding the probability that we have at least 20 fours, we can find the probability that we have less than or equal to 19 fours, and subtract this from 1100. So we’re going to enter “1”, then “-“, then press “2ND,” “VARS,” “ALPHA,” and “MATH.” Our homescreen should look like this:

100 Note, we have to use 19 here because binomcdf finds the probability that the number of success is less than or equal to a certain number. Since we want find the probability that the number of successes is greater than or equal to 20, we need to subtract the probability that the number of successes is less than 20, which is the same as less than or equal to 19.


Now we enter our parameters. The total number of trials is 100. Press comma, then enter the probability of a success. If we designate a success as rolling a four, the probability of a success is 1/6. Press comma again, then enter the desired number of successes (19 in this case). Our homescreen should now look like this:

Press “ENTER.” The probability of Katie getting at least 20 fours is 0.2197.


QUIZ 3.12: BINOMIAL DISTRIBUTION

Suggested time: 6 minutes

Answers on page 135

1. A baseball pitcher gives up a hit one in every five pitches. If nine pitches are thrown, what is the probability that eight or more pitches result in hits?

2. A machine produces duck figurines—in order for a duck to be “sellable,” it must have its wings intact. 80% of the duck figurines produced meet these specifications. A sample of 6 ducks is taken from machine’s production line and placed in a box. What is the probability that at least three of them fail to meet the specifications (i.e., there is something wrong with their wings)?

3. Find the probability of obtaining exactly one tail in six tosses of a fair coin.

4. A multiple-choice test contains 100 questions, each with four choices. If a person guesses, what is the expected number of correct answers?

5. In flipping a fair coin five times, what is the expected number of tails?

6. What is the probability that a family with exactly 6 children will have 3 boys and 3 girls?

7. Find the probability of obtaining exactly 6 successes in 10 trials when the probability of success is 0.3


SECTION 3.13: THE NORMAL DISTRIBUTION

Ladies and gentlemen, above is the Normal Distribution, better known as the “bell curve” and the culmination of our study of statistics. The Normal Distribution occurs just about everywhere—often naturally in large sets of data or artificially in standardized tests—and uses a culmination of statistical knowledge (standard deviations and z-scores).

If you’ve taken a calculus course, you know how to find the area under a curve. However, that beautiful curve above is not simple, represented by the

equation (please do not remember this) P( X )= 1

2πσ 2 e−( X−µ )2

2σ 2. Calculating

the area of such a curve is not an easy feat101.

Luckily, you’ll never have to.

Here are some facts that you do need to know about the Normal Distribution:

1. Its mean, median, and mode are all equal.

2. It is a probability distribution, so the sum of all its probabilities must equal 1.

3. It is a continuous probability distribution, which means that—since it is a curve—there are infinitely many probabilities between 0 and 1 (.5, .635235235, .232, .0000001, etc.).

4. It is the connection between z-scores and probabilities

z = x − µ

σ

101 If, however, you are up to the challenge, try to integrate that probability function (or find someone on your team that can…you can use any fancy approximation methods that you like), and I’ll give you a prize. Send a solution to [email protected].


http://youtu.be/hVeq1kdwPFk

Suggested prior reading: 3.12, 3.5, 3.6


which is nothing new, except that we are using µ instead of x-bar (which is conventional for Normal Curves).

The connection, though, between z-scores and probabilities is huge and is illustrated through the Empirical Rule. This states that, in a set of data that is normally distributed, approximately 68% of the data falls within one sigma (standard deviation) of the mean, 95% falls within 2 sigmas of the mean, and 99.7% falls within 3 sigmas of the mean. This is also called the 68-95-99.7 rule, which is what you should commit to memory. This is illustrated in the graphic on the previous page.

How does this relate to z-scores? Well, the z-score by definition is simply how many standard deviations (sigmas) a score is away from the mean. So, by the Empirical Rule, 68% of the data has z-scores between -1 and 1, 95% has z-scores between -2 and 2, and 99.7% has z-scores between -3 and 3. Thus, now it makes even more sense why extremely large/small z-scores (like 5 or -5) are hard to come by. Also keep in mind the idea of “relative standing” from section 3.6, which basically just means that when comparing z-scores from different tests/data, the higher z-score has a higher relative standing.

Calculations with the z-score will pretty much always be using these 3 numbers (68, 95, 99.7) in one way or another, because otherwise you would have to use special technology102 or a “z-score table,” which is often used in an introductory statistics textbook.

Looking at the graphic at the top of the previous page, you can get quite a bit of intuitive information about the normal curve. For one, if you want to calculate the probability of something that is above/below one of these “empirical” values, all you have to do is subtract from 1 and find what’s left. However, this does not always do the trick, as if something is at the top (read: far right) of the curve, it is not on the far left of the curve. Often, you will have to divide probabilities by two to account for “only the top” or “only the bottom” of the Normal Curve.

For example, suppose the average American woman is 5 feet, 5 inches tall, and the standard deviation is 2 inches. Assuming height is normally distributed, a person that is 5’1” is taller than what percent of American women?

First off, you need to see how many standard deviations (sigmas) 5’1” is away from the mean (5’5”). Since it is 2 sigmas away, you know you’re going to be using 95% somehow. You know that 95% of women are within two sigmas (so between 5’1” and 5’9”) of the mean. The question asks who this person is TALLER than. That means you’re looking at the far left corner of the distribution, so you take the remaining 5% and cut it in half. So, this person is taller than 2.5% of Americans.

Now, you are equipped to answer the practice problems on Normal Distributions and…at this point in the guide (assuming you know your calculator tricks), everything on the Mathematics test. I wish you the best of luck, and I congratulate you for getting this far.

102 Google “R statistical software” if you’re interested. It’s free and great.


Consider the above graphic when answering any questions about the Normal Distribution. The following questions will ask a question and provide a related graphic.

1. What percent of the data lies above the red bar?

2. Given 10,000 people in a sample with mean=10 and standard deviation 8, according to the graphic above, how many people would lie under two standard deviations from the mean?

3. What percent of the data lies between -3 and +2 standard deviations of the mean?

4. What percent of the data lies outside the two red bars (outside two standard deviations of the mean)?

Answers: 1. We know that 50% of the data (cutting the data into an upper half of 50% and a lower half of 50%) is definitely included, so only worry about the sliver between -2 and 0 sigmas. We know that 95% of the data lies within two sigmas, and since we’re only considering the lower half, we cut this in half, which gets 47.5%. So, we have 50%+47.5%=97.5%

2. Using our answer from question 1, we know that “what’s leftover” is simply 1 – 0.975, since the probabilities in the Normal Distribution must add up to 1. This means 2.5% of the data fall in this range, so 0.025*10,000 gives us 250 people. The rest of the given data is irrelevant, though if the problem said “how many people…have a value of less than 6,” with the given information, the answer would be the same.

3. The area given is simply the “percent of the data within 2 standard deviations of the mean” plus an extra sliver on the left hand side. Thus, we know that the answer will be at least 95%. To figure out what that last little sliver is worth, consider how much of the data lies within 3 sigmas—99.7%. Thus, we find the difference between this and 95% (2 sigmas) and cut that in half—that is the value we’re looking for, which turns out to be 4.7/2=2.35%. Thus, we add 2.35%+95% and get 97.35% of the data between the two bars.

4. We know that 95% of the data lies between 2 sigmas, and to find what is “outside” this part of the distribution, we subtract that value from 1 due to the rules of probability distributions. Thus, 1 – 0.95 = 0.05 = 5% of the data.

CALCULATOR HELP: NORMAL DISTRIBUTIONS Normal distributions are much easier to deal with than binomial distributions. When questions ask you about normal distributions, they can either explicitly say “the sample is normally distributed,” or they may simply give you a mean (expected value) and standard deviation, which implies that the distribution is normal. In either case, you will be given the mean and standard deviation. With normal distributions, you will be using the normalcdf (with a c) function all the time. Unlike with binomial distributions, we will never use normalpdf (with a p). To access the normalcdf function: 1. Press “2ND” then “VARS” (selecting “DISTR”). The

following screen will appear.


2. To select normalcdf, just press “2.” You will automatically be taken back to the homescreen.

3. Now, we need to enter our parameters. The normalcdf function has the following syntax:

normalcdf(Lower bound, Upper bound, Mean, Standard Deviation)

To enter these, we will simply use the number pad to enter the numbers, and use the comma key to separate the entries. Press “ENTER” to reveal the probability.

Let’s look at some examples to hammer this home.

Example 1

Nick has a box of remote controls. The battery lives of the remote controls are normally distributed, with a mean of 50 hours and a standard deviation of 5 hours. Find the probability that a remote control selected at random will

require new batteries between 40 and 60 hours from now.

Solution Follow the steps above to get to the normalcdf function. Now, we need to enter our parameters. The lower bound is 40, the upper bound is 60, the mean is 50 hours, and the standard deviation is 5 hours. Thus, we should have:

normalcdf(40,60,50,5) Pressing “ENTER” yields that the probability of Nick needing to replace the batteries within 40 to 60 hours is 0.954.103

Example 2

103 Put another way, this question simply asks the odds of the batteries’ lifespans falling within two standard deviations of the mean; you can memorize this value as about 95%.


Stephen decides to go jogging every morning. Stephen’s average jog is 4.2 miles, with a standard deviation of 0.6

miles. Assuming this distribution is normal, find the probability that tomorrow morning, Stephen will jog more than 5 miles.

Solution

In this case, we’re trying to find the probability that Stephen jogs at least 5 miles. Thus, our lower bound is 5 miles. Our upper bound is technically infinity, but since our calculators can’t do that, let’s just choose a really big number (like 99999). The mean is 4.2, and the standard deviation is 0.6. Thus, we should have:

normalcdf(5,99999,4.2,0.6) Pressing “ENTER” yields that the probability of Stephen jogging more than 5 miles tomorrow is 0.0912.

Example 3

Katie sits in a lab all day to time how long it takes for boson particles to travel around her homemade particle accelerator. She finds that the mean time required for the particles to complete the circuit is 9 nanoseconds, with a

standard deviation of 2 nanoseconds. Find the probability that her next particle will take less than 8 nanoseconds to complete the circuit.

Solution

Since we’re trying to find the probability that it takes less than 8 nanoseconds, our upper bound is 8. Our lower bound would technically be negative infinity, but since our calculators can’t do that, we’ll make it something really negative (like -99999)104. The mean is 9, and the standard deviation is 2. Thus, we’ll have:

normalcdf(-99999,8,9,2) Pressing “ENTER” yields that the probability of Katie’s particle taking less than 8 nanoseconds is 0.309.

104 Be careful—we want a very small number in the negative direction, meaning that we want a “big” negative number, like -99999. In more math-y terms, we’re looking for a negative number with a very large magnitude. DO NOT pick zero or a value close to zero.


QUIZ 3.13: NORMAL DISTRIBUTION Suggested time: 6 minutes

Answers on page 135

1. At a high school, IQ scores are normally distributed, and the average IQ score of the 1200 students is 100, with a standard deviation of 15. How many students have an IQ between 85 and 115?

2. The scores on an Academic Decathlon® Art exam were normally distributed with a mean of 780 and a standard deviation of 70. What is the probability that a test chosen at random has a score between 850 and 920?

3. A student receives a score of 820 on his Academic Decathlon® Music exam at a regional competition, for which the mean is 730 and the standard deviation is 90. The same student receives a score of 890 on the Music exam at the state competition, for which the mean score is 810 and the standard deviation is 150. In which exam is his relative standing higher?

4. A normal distribution of scores has mean 100 and standard deviation 10. What percent of the distribution has scores less than 80?

5. Suppose height in a population is normally distributed around a mean of 1.7 meters. If the standard deviation is 0.1 meters, what percent of the population will be taller than 2 meters?

6. Suppose there are 5,000,000 people in a city where weight is normally distributed around a mean of 150 pounds with a standard deviation of 25 pounds. According to the distribution, how many people should weigh more than 125 pounds?

7. Corn flakes come in a box with a printed guarantee that the box holds a mean weight of 16 ounces of cereal. The standard deviation is 0.1 ounces. Supposing a normal distribution and that the manufacturer packages 600,000 boxes, how many boxes weigh between 15.7 and 16.3 ounces?


SECTION 3 FINAL EXAM Suggested time: 11 minutes105

Answers on page 135

1. Suppose a normal distribution with mean 50 and standard deviation 3.5. What percent of the data should lie between 43 and 60.5?

2. Find the probability of obtaining more than 8 successes in 12 trials when the probability of failure is 0.4. Give the answer as a whole number percent.

3. The mean of the data set {-40,-40,-20,0,50,y,y,y} is -6.25. What is the value of e y 2

?

Use the following probability distribution of a chance game for questions 4-7. Approximate to 3 decimal places when necessary.

Outcome Gain $1 Lose $5 Gain $2 Lose $5 Lose $100 Gain $138 Nothing Probability 1/12 1/4 1/12 1/4 1/24 1/24 1/4

4. Calculate the expected value for the game. Is the game fair?

5. Calculate the variance of the distribution.

6. Calculate the standard deviation of the distribution.

7. Which individual outcome has the greatest expected value?

8. What is the probability of rolling two dice and getting a sum (of the numbers on the faces) that is greater than 8? (Write your answer as a fraction)

105 You’re going to need it


9. A three-person jury must be selected at random from a pool of 12 people with 6 men and 6 women. What is the probability of selecting an all-male jury? page 469

10. Given that a flood has a 1% chance of occurring in any given year, what is the probability that a region wil experience at LEAST one 100-year flood during the next 100 years? Assume that 100-year floods in consecutive years are independent events. [1-.99^100]=0.634

Given the data set {8, 18, 36, 4, 14, 28, 40}, answer questions 11-14.

11. What is the median of the data set?

12. Find Q1 and Q3 for the data set.

13. Calculate IQR for the data set.

14. Suppose the number 75 is added to the data set and that the mean, lower quartile, upper quartile, and IQR stay the same. According to the IQR test for outliers, is 75 an outlier?

15. A survey found that supermarket scanners are overcharging customers at 30% of stores. If you shop at 3 supermarkets that use scanners, what is the probability that you will be overcharged in at least one store? [p470]


Quiz/Exam Answers and Explanations Section 1

Quiz 1.1-1.2: The Multiplication Principle and Permutations

Question # Answer Explanation 1 15,600 There are 26 options for the first letter, and since you may not use

the same letter twice, there are only 25 options for the second letter, leaving 24 for the third. Thus, there are 26*25*24=15,600 possible sequences.

2 30 “At MOST” means 1,2,3, or 4. Since each symbol may be a dot or dash, there are two options for each symbol. If there is one symbol, it can be a dot or dash, if there are 2 symbols, both can be a dot or dash, if there are 3, each of the three can be a dot or dash. Mathematically, this is represented as 2+2*2+2*2*2+2*2*2*2=30.

3 60 Kathryn has 5 ways to fill the first space (5 books), 4 ways to fill the second, and 3 ways to fill the third, so 5*4*3=60.

4 336 We have 3 spaces to fill and 8, then 7, then 6, to fill each space respectively, giving 8*7*6=336. Alternatively, you may use the permutation formula 8!/(8-3)! = 8!/5! = (8*7*6*5*4*3*2*1)/(5*4*3*2*1)=8*7*6=336.

5 34,650 For these types of problems, you simply take the total number of letters/things and divide it by each distinguishable “thing”. That is, divide 11!, the number of letters, by 1! (M) 4! (S) 4! (P) 2! (I), so 11!/(1!4!4!2!)=34,560.

6 210 Using the same method as the problem above with distinguishable permutations, we take 7!, the total number, and divide by 3!2!2!, so 7!/(3!2!2!)=210.

7 540,540 13!/(5!4!2!2!)=540,540, using the same method as the above two problems.

Quiz 1.3: Combinations

Question # Answer Explanation 1 56

You have 8 people, choose 3, so

83

⎛⎝⎜

⎞⎠⎟

=56. By definition of

combination,

83

⎛⎝⎜

⎞⎠⎟

=8!/(5!3!). Alternatively, you can use the

calculator function “nCr” with (8,3) as your input. 2 120

103

⎛⎝⎜

⎞⎠⎟

=120 by the methods above.

103

⎛⎝⎜

⎞⎠⎟

=10!/(3!7!)

3 495

124

⎛⎝⎜

⎞⎠⎟

=495 by the methods above.

4 712842 There are 13 hearts in a deck, so the 2 hearts will be selected from


those 13 cards. The other 3 cards must come from the remaining 39 cards that are NOT hearts. Using combinations and the multiplication principle,

132

⎛⎝⎜

⎞⎠⎟

393

⎛⎝⎜

⎞⎠⎟= 78 ⋅9139 = 712,842

5 24 The arrangements of the 3 aces or the 2 9’s does not matter. We again use combinations and the multiplication principle, since

there are

43

⎛⎝⎜

⎞⎠⎟

ways to get 3 aces from the 4 in a deck and

42

⎛⎝⎜

⎞⎠⎟

ways to get 2 9’s. Multiplying these together, we get 24.

6 1800 Multiply

103

⎛⎝⎜

⎞⎠⎟

62

⎛⎝⎜

⎞⎠⎟

we get 1800.

7 2520 This is much like the problems from section 1.2 where we have distinguishable permutations—there are 10! block arrangements, and 2! ways to order white blocks, 3! to order the red, 5! to order the blue. Thus, there are 10!/(2!3!5!) arrangements total.

Section 1 Final Exam

Question # Answer Explanation 1 5148 There are 13 choose 5 ways to pick 5 cards from a single suit,

and there are 4 suits, so 4 13

5⎛⎝⎜

⎞⎠⎟

=5148.

2 60 6!/(3!2!1!)=60. There are 6 letters, 3 M’s, 2 A’s, 1 L. By distinguishable permutations, we divide the total number of orders by the permutations of each letter.

3 15 5*3=15 by the multiplication principle. 4 48,228,110 In this problem, 3 birthdays are selected from a possible 365

days, and no birthday may repeat itself, so 365!/(365-3)!=365*364*363=48,228,180.

5 15 6 choose 2 = 15. 6 12! There are 12! ways to order 12 questions, by the definition of

permutation 7 90,720 Using the methods used in question 2,

9!/(2!2!)=90,720. 8 60 Using the methods used in questions 2 and 7, 5!/(2!)=5*4*3=60. 9 5040 Ordering 7 classes gives 7!=5040. 10 3,368,283,000 First, you must find the number of ways you can pick 3 tickets

from 1500, which is 1500 choose 3. Then, you must order these 3 tickets, which can be done in 3! ways;

15003

⎛⎝⎜

⎞⎠⎟⋅3! =3,368,283,000.


11 2.943*1024 Using the same methods as question 10,

5015

⎛⎝⎜

⎞⎠⎟⋅15! =2.943*1024

Section 2

Quiz 2.1-2.2: Sequences, Series, and Sigma Notation

Question # Answer Explanation 1 82 (32-1)+(42-1)+(52-1)+(62-1)=82 2

ai −3

i=1

7

∑ We know that a0=1, a1=-2, a2=-5, … -- Thus, each term is 3 less than the last. We start with the 2nd (a1, NOT a0) and end with the 8th, which correspond to a1 and a7 respectively.

3 53 Treat each sigma separately. The first is equal to 2+22+23+24+25 – (2+3+4) = 53.

4 19 This is a recursive sequence. We know that x2=x1+2,=7+2=9, and then each number is 2 more than the last, which gives that x7=19.

5 30.5078 The sequence is {10, 10(.75), 10(.75)2, 10(.75)3, 10(.75)4} Summing these gives 30.5078.

6

(1+2n)

n=0

10

∑ The odd numbers start with 1, and each number is 2 more than the last. We need to start with 1+0*n, or 1.

7 9840 The alternating series (-1)n will cancel in the end. Thus, we only need to sum 3n from n=1 to n=8. Thus, 3+32+33+34+…+3=9840. Alternatively, you can enter this into your calculator, using the sum(seq(formula,x,beginning,end,1)) command, so sum(seq(3^x+(-1)^x,x,1,8,1)).

Quiz 2.3-2.4 (Section 2 Midpoint):

Arithmetic and Geometric Sequences and Series

Question # Answer Explanation 1 405 We need to first find the common ratio, so, knowing that a3 and

a1 are 2 indices apart, we have that 45=5(r)2, 9=r2, and r=3. Thus, the sequence will continue, {5,15,45,5(3)3,5(3)4}; 5(3)4=405=a5.

2 -61 a3=a1+(n-1)*d so 5=27+(2)*d, and d=-11. Thus, continuing the sequence, we have {27,16,5,-6,-17,-28,-39,-50,-61} so a9=-61.

3 3905 Much like question 1, this question asks us to find unknown terms of a geometric sequence. The last line simply asks us to sum all the terms from index 2 to index 6.

4 10 We use the formula for the sum of an infinite series, a/(1-r). The first term is 5, so 5/(1-1/2)=10.

5 1/3 The first term is 6(1/3)4-1=6/27=2/9. Thus, we have (2/9)/(1-1/3)=1/3.


6 The series diverges

In order for us to be able to use the formula a/(1-r), |r| must be less than 1, which, in this case, is not true. It is obvious that the series diverges if you start to sum the terms and realize that it will never end.

7 61 With a common difference of 7, the sequence will be {5,12,19,26,33,40,47,54,61}, and a9=61.

8 1280 a5=5(4)5-1=1280

9 124 The formula for the sum of an arithmetic sequence is n/2(an+a1), i.e. 8/2(5+3*7+5)=4*31. Alternatively, you can just write down the terms {5,8,11,…,26,…} and sum the first through eighth terms.

10 5445 5(3)2+5(3)3+5(3)4+5(3)5+5(3)6=5445 11

3(2)n−1

n=1

6

∑ This is a geometric sequence with initial term 1 and common ratio 2.

12 1/49 The initial term is 6(1/7)3, and 6(1/7)3/(1-1/7)=1/49.

Quiz 2.5-2.6: Adding, Subtracting, and Multiplying Polynomials

Question # Answer Explanation 1 x 3 +5x 2 +5x −2 When adding and subtracting polynomials, we

need to make sure that coefficients stay the same. For this expression, we have x3+(2+3)x2+(2+3)x+(-1-1)

2 x 6 −3x 4 −2x 3 + x 2 − x −2 x2 -x -2 x4 x6 -x5 -2x4 x3 x5 -x4 -2x3

1 x2 -x -2 Adding up all the boxes, we get the answer to the left.

3 x 8 − 81 This expression is equal to (x4-9)(x4+9), which is a difference of two squares. If you don’t know that yet, you can use the table method above and get x8-81.

4 4 First, we multiply (2x+1)2 and get 4x2+4x+1, to which we add 2x4+3

5 a = 2 First, we multiply (x+1)2 which is equal to x2+2x+1, to which we add (x+1)(x-1), which is x2-1 as well as 2x2+2, which has an “x” term of 2.

6 x 6 + 6x 5 +16x 4 +26x 3 +27x 4 +18x + 6 There is no easy way to do this problem; it requires brute force. First, multiply x+1 and x2+2x+2; then, multiply that product by x3+3x2+3x+3.

7 0 You could go through quite a bit of multiplication to get this problem finished, or you could recognize that the first product of polynomials is equal to (x+1)6, and so is the product being subtracted from it; thus, the difference is 0.


Quiz 2.7: Binomial Expansions

Question # Answer Explanation 1 x 6 − 4 x 2 This is a difference of two squares. 2 64 Using the binomial expansion theorem, we use

43

⎛⎝⎜

⎞⎠⎟

(2x )3(2) , since we have a 4th degree

polynomial where we want to know the 3rd degree term’s coefficient.

3 32 Using a combination of the binomial expansion theorem and some basic “tricks” with polynomials, we get the polynomial x3+10x2+32x+32, which has “x” term with coefficient 32.

4

729x 3 + 81x 2 +3x + 1

27

Since this polynomial only has 3 terms, it might be easier to “brute force” to get the answer. However, to show how the binomial expansion

works, this expansion gives

32

⎛⎝⎜

⎞⎠⎟

(9x )2(13

)1 +

31⎛⎝⎜

⎞⎠⎟

(9x )1(13

)2 +

31⎛⎝⎜

⎞⎠⎟

(9x )1(13

)2 +

30

⎛⎝⎜

⎞⎠⎟

(9x )0(13

)3 .

5 x 4 + 4 x 3 + 6x 2 + 4 x − 80 This is a difference of two squares, equation to

(x+1)4-81. We can expand the first using brute force or the expansion theorem, and then we simply subtract 81.

6 108 The Binomial Expansion Theorem gets a bit confusing when there are 2 x’s, so it is best to brute force through this expansion, giving that

7 The x3 and x2 terms both have coefficients of 1/6

Use the Binomial Expansion Theorem to compare the coefficients of the expansions of this fourth degree polynomial.

Quiz 2.8: Financial Math

Question # Answer Explanation 1 $885,185.11

This is an annuity. F = 2000[

(1+ 0.10)40 −10.10

] . There are 40

deposits of $2000 per year, and i=10%. 2 34 quarters

700 = 500(1+ 0.044

)n

700 = 500(1.01)n

7 / 5= (1.01)n


Thus, graph 7/5 and 1.01 and see that they intersect at 33.81, which is approximately 34.

3 $18 With i=0.06/12, the loan amount = 1600 (keep in mind that the loan will have to add interest added to it so that you can set the loan equal to the annuity/equal payments), we have that

1600(1+ 0.0612

)120 = A[(1+ 0.06

12)120 −1

0.0612

]

So A=17.76, which is close to $18. 4 359.25

months (360 months)

F = A(1+ i)n −1

i

500,000 = 500(1+ 0.06

12)n −1

0.0612

6 = (1.005)n

Then, you can use logarithms or graph 1.005n and 6 and see that they intersect at 359.25.

5 $297,669

This is an annuity.

75[(1+ 0.0452

)35*52 −1]

0.0452

Which gives that he will have $297,669 after 35 years, when he retires.

6 13.73 ft A=12e0.045*3=13.73 feet. Continuous growth uses the formula A=Pert.

7 $3.35 For this question, we need to subtract an annuity from a lump sum gaining compound interest. It should make sense that the lump sum will have more money, since it starts with more money, and that money will gain more interest (due to having a higher balance). The difference can be represented as

120(1+ 0.0512

)12 −10[(1+ 0.05

12)12 −1

0.0512

]

which equals $3.35, only a small difference.


Question # Answer Explanation 1 1/8 To make things simpler, we can change


12n=3

∞

∑ 14

⎛⎝⎜

⎞⎠⎟

12 n

to

12n=3

∞

∑ 12

⎛⎝⎜

⎞⎠⎟

n

, since

12

⎛⎝⎜

⎞⎠⎟

2

= 14 . Now we have a

standard infinite series, and we use the formula a/(1-r) to calculate the sum, where a is the starting element, which in this case is (1/2)(1/2)3, or 1/16. Then, (1/16)/(1-1/2)=1/8.

2 16807

Using the Binomial Expansion Theorem,

nk

⎛⎝⎜

⎞⎠⎟⋅ x k ⋅ y n−k

. We have

a 5th degree polynomial, and we’re chooseing the x5 term, so we

have

55

⎛⎝⎜

⎞⎠⎟

(7x )5(2)5−5

=16807.

3 -1 Using that an=a0+(n-0)*d, we need to find the common difference. The reason that we use 0 instead of 1 as standard is because we are starting with 0 instead of 1 in the “n-1” (n-0) expression is because a0 is our starting element—think of the “n-1” element as a way of keeping track of the difference between the indices being compared. Thus, 1=3+(2)*d, so -2=2d and d=-1. Now we know that we’re dealing with an arithmetic sequence with common difference of -1, which will look like {3,2,1,0,-1,…}, where a4 =-1.

4 $552.31 Don’t be fooled by the word “loan,” this is just a compound interest loan, but instead of gaining money you owe money. Thus, 500(1+0.02/2)^(5*2)=552.31

5 3.5 years For this question, it would be best to put in each expression (y1=3000 and y2=the annuity) and graph them to see where they intersect, as that will be “t”.

Thus,

3000 =15[(1+ 0.052

52)52⋅t −1

0.05252

] and t=3.5.

6 x 4 + 8x 3 +23x 2 +26x +7 How you expand these two binomials does not matter (it might be easier to brute-force multiple out each polynomial), but they expand to x 4 + 8x 3 +24 x 2 +32x +16 and

x 2 + 6x +9 respectively, and their difference is

x 4 + 8x 3 +23x 2 +26x +7 . 7 47 2000=1000(1+0.06/4)^t. Graphing this, we see that the two

graphs (y1=2000, y2=(1+0.06/4)^x) intersect at 46.55, which approximates to 47.

8 728 While it might be tempting (and perfectly valid) to use the formula for summing geometric series—which, for this problem, would be 2*(1-36)/(1-3)—it will be just as easy to type into your calculator each term and sum them; hence, enter 2+2(3)+2(3)^2+2(3)^3+2(3)^4+2(3)^5=728

9 2 If you know Pascal’s triangle, this problem should go quickly


1 11

121 1331

14641

The 5th row (the 4th “non-zero” row) has terms 1,4,6,4,1, as does the polynomial (x+1)4, so there are 2 terms with 4 as a coefficient.

10 $23,669.40 This is an annuity (as are most “saving up” problems), with F=

75[(1+ 0.04

12)18⋅12 −1

0.0412

] .

11 405 Using that the sum of an arithmetic series is n/2*(an+a1), we have that 10/2*((70+2)+7+2) equals 405.

12 E The long string of polynomials given is equal to

( x +1)( x +1)3( x +1)2 , which is equal to ( x +1)6 . The polynomial

( x +1)6 has an x5 term with coefficient equal to its x term, both being 6.

Section 3

Quiz 3.1-3.2: Mean, Median, Mode, and Range

Question # Answer Explanation 1 D For problems like these, you’ll need to solve equations, either by

hand or by graphing each side of the equation and seeing where they intersect. By the definition of mean, we have that (10+15+20+30+90+x+x)/7=45, so x=75. The expression given, (x+1)2-(x-1)(x+1) is equal to (x+1)2-(x2-1)=(x2+2x+1)-(x2-1)=2x+2. Thus, the answer is D, 152.

2 C This question is multiple-choice for a reason. You’re looking for an answer that doesn’t change the highest or lowest number of the data set, due to the definition of range. With A, the range is 55, with B, the range is 44, with C, the range is 33 (nothing changes), with D, the range is 34, and with E, the range is 41, so the answer is C.

3 E This question is also necessarily multiple-choice. In its current state, the data set has no mean, as we don’t know whether or not z is in the middle, the top, or the bottom. Thus, we first must order the set and try each value. With A, {9,10,11,16,22,23}, the median is the mean of the middle two values, 13.5, with B, the median is still 13.5, with C, the data set becomes {9,10,11,12,22,23}, the median is 11.5, with D, the median is 11, and with E, the set becomes {0,9,10,11,16,22}, and the median becomes the mean of the middle two values, 10.5.

4 E The sets are messy, but after inspection, we see that E has 4


modes—6,2,3, and 1 occur twice each. 5 D {8,8,-4,8} has a mean of (8+8+8+(-4)+0)/5=20/5=4, and the mode

is 8, which is twice 4. 6 88/9, 5,

1, 33 Mean=(1+1+2+3+5+8+13+21+34)/9=88/9 Median=5, the middle of the data set Mode=1, which occurs the most often Range=34-1=33

7 D The mean is equal to (1+4+4+4+4+4+4+5+5+5)/10=4, which is equal to the median, the mode, and the range of the data set, which are all 4.

Quiz 3.3-3.4: Quartiles, IQR, and Outliers

Question # Answer Explanation 1 3 Q1 is the median of the lower half of the data, all of the numbers

less than 9, the median. Since this is an even number of numbers, we have to average 1 and 5, which gives us 3.

2 24 First, we must put the numbers in order: {-51,-1,0,7,24,51}. Q3 is the median of the upper half of the data, all of the numbers more than 3.5, the median. Thus, 24 is Q3.

3 41 First, we must put the numbers in order: {-55,0,5,10,17}. IQR=Q3-Q1=13.5-(-27.5)=41.

4 D With a question like this, it’s best to just try the possible answers. With A, the set is {4,6,10,20,25}, which has Q3=22.5, Q1=5, so IQR=Q3-Q1=17.5. Trying this for each letter gives numbers that are not 18, however, for D, we have the set {4,10,20,25,25}, and Q3-Q1=25-7=18=IQR.

5 No The IQR test for outliers requires that a number be distant by 1.5*IQR from both Q3 and Q1. IQR for this set is 42.5-17.5=25; thus, 1.5*IQR=37.5, so numbers greater than 80 or less than -20 are outliers. The question asks if 80 is, for which the answer is no.

6 9 The median is (3+5)/2=4. Q3=21/2, Q1=3/2. IQR=21/2-3/2=18/2=9.

Quiz 3.5-3.6: Variance, Standard Deviation, and Z-score

Question # Answer Explanation 1 14/3

The formula for variance is σ 2 =

( xi − x )2

i=1

n

∑n

. Thus, we must first

find the mean, which is (1+5+6)/3=12/3=4. Then the variance is [(1-4)2+(5-4)2+(6-4)2]/3=14/3.

2 3 This equation gets a bit messy, so it would be best to—if you are not comfortable solving equations by hand—graph 2.1875 and the equation and see where they intersect. The mean is (-1+0+1+x)/4, or x/4. Then, the variance is [(-1-x/4)2+(x/4)2+(1-x/4)2+ (x-x/4)2]/4=2.1875, so x=3.

3 176.75, The mean is 55/2. Using the variance formula, we get 176.75, and


13.295 since standard deviation is the square root of variance, (176.75)^(.5)=13.295 is the standard deviation.

4 2 z=(7-5)/1=2 5 16,4 See 1,2, and 3 above. The mean is 93. 6 1,1 Using the standard deviation and mean from #5, we have that

z=(97-93)/4=1, so the z-score is one, and by definition, this means that 97 is 1 standard deviation above the mean.

7 B This question is similar to 2 above. The mean is (x+x-10-15+x-20)/6=(3x-45)/6, and applying this to the variance formula (you should graph the equation if possible), we have that x=-5. Alternatively, you can plug in each number, and realize that none of them works but 5.

8 -1.73 First, we must find the mean when x=-15, which is -15. We also have to calculate the standard deviation for the new set, which is 2.887. With that, we have z=(-20+15)/2.887, so z=-1.732

Section 3 Midpoint Exam: Statistics Part 1

Question # Answer Explanation 1 C C has two modes, 2 and 1. 2 D The mean of this data set is (2+x+y)/4. Using this and the standard

deviation formula, we can substitute another variable “u” for x+y and graph the square root of 2 versus the standard deviation equation. This gets u=x+y=2.

3 D First, we must order the set {42,45,50,66,75}. Then, we have that Q3=141/2, Q1=87/2, so Q3-Q1=IQR=54/2, or 27.

4 A The median of the data set is the average of the middle two values, 0 and 1, which is 1/2. Multiplying this by the mean, 2/3, gives 1/3.

5 E Q3-Q1=10, and 1.5*IQR=15. Thus, outliers exist above 30 and below -10.

6 E After exhausting answers A through D, it becomes clear that no number will give that the median is 50.

7 C

Use the formula for variance, σ 2 =

( xi − x )2

i=1

n

∑n

, gives 344.583.

8 B The lowest number in the data set is -22, and the current highest is 11. The inclusion of 22 makes 22 the highest number, so the range becomes 44.

9 B Mean = 6.5, Median = 5.5, Mode = 7, Range = 9, and standard deviation is clearly less than 9.

10 D See question 9. Standard deviation is the square root of variance, for reference.

11 D (12-6.5)/2.579=2.133 12 E IQR=4, so 1.5*IQR=6, and there are no outliers, since no number is

greater than Q3+6 or Q1-6.


Quiz 3.7-3.9: Probability

Question # Answer Explanation 1 .16 P(O negative) + P(A negative) + P(AB negative) + P(B negative) =

.07+.07+.01+.02 = .16 2 .47 P(O positive) + P(B positive) = .47 3 .36363 “At least 2” means either exactly 2 or exactly 3 defective

flashlights. Since these events are independent, the sum of their probabilities will give the total probability. Thus, .31818+.04545 = .36363

4 1/3 The sample space for this is {TT, TH,HT}, since {HH} is not included in “given…”, so the probability of {TT} is 1/3.

5 1/13 P(E|F) means “the probability of E given F;” so, if we let E = “A person chosen at random is a Decathlete” and F = “A person chosen at random is female,” so P(E|F)=40/520=1/13.

6 25/204 P(The first card is diamond AND the second card is red) = P(The first card is diamond)*(P the second card is red given that the first card is diamond) = ¼*25/51, using 51 as the denominator since there is 1 less card (1 less red card, at that) than when we started.

7 .0004952 Multiplying “4 choose 1” and “15 choose 5,” we get the probability of getting a flush. To find the probability of a specific flush, we divide this by 4. As a reminder, the nCr(n,k) calculator function is incredibly useful for problems like these.

8 1/2 P(E and F) = P(F)*P(F|E), so .1=.2*P(F|E), giving that P(F|E) is ½. 9 2/9 The probability of the sum being 7 is 6/36, and the probability of

the sum being 11 is 2/36, so we have 8/36 (the probabilities added together), or 2/9 as the probability of either.

Quiz 3.10-3.11: Expected Value and Probability Distributions

Question # Answer Explanation 1 1.5 The sample space is S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}. Now we need

to split the sample space into specific events involving “number of tails”—that is, there could be 0 tails, {HHH}, 1 tails {HHT,HTH,THH}, 2 tails, {HTT,THT,TTH}, or 3 tails {TTT}. We assign “payoff” values for the number of heads, 0 when 0 tails, 1 when 1 tails, etc. – The expected number of heads can now be found by multiplying each payoff by its corresponding probability and finding the sum of these values. E = 0*1/8+1*3/8+2*3/8+3*1/8=3/2=1.5.

2 -$0.167 Event 1 2 3 4 5 6 Payoff $1 $2 $3 $4 -$5 -$6

By multiplying each event by the payoff value and summing them, we get the E=-$0.167.

3 $0.60 Multiplying each outcome by its payoff, we get $300*1/1000+$100*2/1000+$1*100/1000+$0*879/1000 =$0.60, so $0.60 tickets would make the game “fair”.

4 $0.75 The sample space is S={HT,TH,HH,TT}, with every outcome having a value of


0 except for {HH}, so we have that E=3*1/4 + 0*3/4 = ¾ = $0.75 for the game. Both questions are the same—the expected value of the game is how much you should pay each time you flip the coin to make it fair, so that everything equals out to $0.

5 1.2 Edistribution=.4*2+.2*3+.1(-2)+.3*0=1.2 6 2.216,1.489

The formula for variance is σ 2 = (Ei − E )

i=1

m

∑2

⋅P(Ei ), so we have that variance

is equal to (using table values) (2-1.2)2(.4)+(3-1.2)2(.2)+(-2-1.2)2(.1)+(0-1.2)2(.2) =2.216, and standard deviation is the square rot of variance, 1.489.

7 -2.15 E3 has a payoff of -2, and z-score is the mean (expected value) subtracted from the individual value E3 all over the standard deviation (from problem 6), thus z=(-2-1.2)/1.489=-2.15.

Quiz 3.12: Binomial Distribution

Question # Answer Explanation 1 .0000189 For almost all of the explanations below, we will (and you should)

use the calculator command binomcdf(trials,probability,value)—obviously, different calculators have different commands, but this general pattern usually follows. For reference, the binomial

probability can be calculated using

nk

⎛⎝⎜

⎞⎠⎟

pkqn−k

For “eight or more,” we simply add the probability of 8 pitches and the probability of 9 pictches, so binomcdf(9,.2,8)+ binomcdf(9,.2,9).

2 .0989 “At least 3” of the 6 means that we must add the probabilities of 3,4,5, and 6 failing respectively, so binomcdf(6,.2,3)+ binomcdf(6,.2,4)+ binomcdf(6,.2,5)+binomcdf(6,.2,6).

3 .0938 binomcdf(6,.5,1) 4 25 For binomial distributions, E=np, so E=100(.25), given n=100

questions and p=.25 (A, B, C, or D). 5 5/2 E=np, with p=1/2 and n=5. 6 0.3125 binomcdf(6,.5,3) 7 0.0368 binomcdf(10,.3,6)


Quiz 3.13: Normal Distribution

Question # Answer Explanation 1 68% By definition of normal distribution, 68% of the data should be

within 1 standard deviation of the mean, 85 and 115 are within 1 standard deviation of 100.

2 0.135 The score of 850 is 1 standard deviation above the mean, and the score of 920 is 2 standard deviations above. Thus, we are trying to find the probability of being between 1 and 2 standard deviations. Since we are in the top half of the distribution, we only look at 50% of the data. Of that 50%, 34% (half of 68%) should be between 0 and 1 standard deviations above the mean, and 47.5% (half of 95%) should be between 2 and 3 standard deviations. 47.5-34=13.5%, so the probability is 0.135.

3 The first test

For relative standing, we compare z-scores. If a z-score for one test is higher than the other, the student did comparably better on that test. Thus, (820-730)/90 = 1 = z-score for the first test, and (890-810)/150= 0.5333 = z-score for the second test. The z-score is higher for the first, so the he did comparably better on the first.

4 2.5% 80 is two standard deviations below the mean, so we are looking for the percent below this. We know that 95% of the data is within 2 standard deviations of the mean, and since we are dealing with the lower half of the data, we only care about half of this, so 2.5% of the data is less than 80.

5 .15% Anything taller than 2 meters is more than 3 standard deviations of the mean. Using the same methodology as the problem above, we have 99.7% of the data within 3 sigmas of the mean, leaving .3% not included. Since we are only considering the top half of the data, we take half of that, .15%, that is in the upper half and greater than 3 standard deviations from the mean.

6 4,200,000 125 pounds is 1 standard deviation below the mean, and since 68% of the data is within 1 standard deviation of the mean, we must first look at the lower half of the data, since all of the upper half of the data (50% of the total data) lies in the criterion “more than 125 pounds”. There is 1 standard deviation between the mean (150) and 125 lbs, meaning that there is 34% of the data that is greater than 125 lbs. Adding together 50%+34% we have that 84% of the data is greater than 125 lbs. Since the question asks for how many people, we take 0.84*5,000,000 and get 4,200,000 people weighing more than 125 lbs.

7 598,200 99.7% of the data is within 3 sigmas of the mean, so we take .997*600,000 and get that 598,200 people within this range.


Question # Answer Explanation 1 97.35% First, consider that 95% of the data is within two standard

deviations of the mean. Then, we only need consider the part “between 2 and 3 standard deviations”. Thus, take into account that 99.7% of the data is between -3 and 3 standard deviations. To


account for the probability between 2 and 3, we subtract 95% from 99.7% and divide our result by 2 (since we are only in the upper tail). Thus, 95 + (99.7-95)/2 = 97.35%.

2 44% Notice the question says that the probability of failure is .6, so the probability of success is .4. Then, binomcdf(12,.6,8)+ binomcdf(12,.6,9)+ binomcdf(12,.6,10)+ binomcdf(12,.6,11)+ binomcdf(12,.6,12)=.438=43.8%, which approximates to 44%

3 1 Given that the mean of {-40,-40,-20,0,50,y,y,y} is -6.25, we solve the equation (-40-40-20+0+50+3y)/8=-6.25, which gives that y=0. Then, we take e0^2, which is e0, which simplifies to 1.

4 -2/3 The expected value of the game is the sum of the products of each outcome and its probability. So, 1*1/12+(-5)*1/4+2*1/12+(-5)*1/4+(-100)*1/24+138*1/24+0*1/4=-2/3.

5 1222.640 The variance of a probability distribution is given by

σ 2 = (Ei − E )2

i=1

m

∑ ⋅P(Ei ) . Thus, using our answer from 4, we have

(1+2/3)^2*1/12+(-5+2/3)^2*1/4+(2+2/3)^2*1/12+(-5+2/3)^2*1/4+(-100+2/3)^2*1/24+(138+2/3)^2*1/24+(0+2/3)^2*1/4=1222.64

6 34.966 Standard deviation is the square root of variance, so (1222.64)^(.5)=34.966.

7 “Gain $138” has EV = 5.75

138*1/12=5.75, which is by far the largest expected value of an outcome.

8 5/18 P(sum of 9)+P(sum of 10)+P(sum of 11)+P(sum of 12)=4/36+3/36+2/36+1/36=10/36=5/18.

9 0.091 Because there are 6 men in the sample of 12 people, the probability of selecting the first male is 6/12. Then, when we pick the next person, there will be 1 less male and thus 1 less person, giving the probability of 5/11. By the same logic, the probability of picking a third male is 4/10. Thus, 6/12*5/11*4/10=0.091.

10 0.634 This problem requires you to take two things into consideration. First, because we have 100 consecutive events, we are going to multiply the probability P(E) by itself 100 times, which gives P(E)100. Second, because we are thinking of “at least one year,” we can consider the complement of “at least one year” as “no flood in one year,” which gives 1 – P(no flood in one year)100. Then, we take the complement of our original probability, 0.01, which is 1-0.01=0.99. Finally, we have 1 – (0.99)100=0.634 as our probability.

11 18 We order the set {4,8,14,18,28,36,40} and find that the middle value is 18.

12 Q1=8, Q3=36

Using the above ordered set, the middle of the lower half of the set {4,8,14} is 8, and the middle of the upper half of the set {28,36,40} is 36.

13 IQR=28 Q3-Q1=36-8=28 14 75 is not

an outlier

Using the IQR test for outliers, 1.5*IQR=1.5*28=42, so any number above 42+36=78 will be an outlier; 75 is not greater than 78.

15 0.657 The probability of 0 overcharges is P(0) = binomcdf(3,.3,0) = 0.343.


Then, we take the complement of this probability (since the complement of “0 overcharges” is “some [at least one] overcharge[s]), 1-P(0)=0.657.


GLOSSARY OF COMMON CALCULATOR FUNCTION LOCATIONS Function Syntax Keystrokes Notes

Factorial (!) n! "MATH", scroll to "PRB", scroll down to !

Where n is any integer ≥ 0

Permutation (nPr)

n nPr k "MATH", scroll to "PRB", scroll down to nPr

Total number of permutations of k objects from a group containing

n objects Combination

(nCr) n nCr k "MATH", scroll to "PRB", scroll

down to nCr Total number of combinations of k objects from a group containing

n objects seq seq(DF, X, Lower

Bound, Upper Bound

"2ND", "STAT", scroll to "OPS", scroll down to seq(

Where DF is the direct formula

sum sum(seq(DF, X, Lower Bound, Upper Bound

"2ND", "STAT", scroll to "MATH", scroll down to sum(

Where DF is the direct formula

TVM Solver N.A. "APPS", scroll to "Finance", scroll to TVM Solver

Use "ALPHA", "ENTER" to calculate unknown

1-Var Stats 1-Var Stats L1, L2 i) "STAT", scroll to "CALC", Scroll to 1-Var Stats

ii) "2ND", "1" for L1, "2ND", "2" for L2

L2 is only present when given probability or frequency

distribution

binompdf binompdf(n, p, x "2ND", "VARS", scroll to binompdf

Where n is the total number of trials, p is the probability of a success, and x is the desired

number of successes binomcdf binomcdf(n ,p, x "2ND", "VARS", scroll to

binomcdf Where n is the total number of trials, p is the probability of a success, and x is the desired

number of successes normalcdf normalcdf(Lower

Bound, Upper Bound, Mean, σ

"2ND", "VARS", scroll to normalcdf

Where σ is the standard deviation

Documents

Mathematics Guided Practice - THIS CITE IS KEPT SIMPLE:)historywithmac.weebly.com/uploads/5/9/7/4/59744935/mathgpb.pdf · of multiplication in this way. Conceptually, multiplication