T-CERT Math 4-8 Module Transcript

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T-CERT Math 4-8 Module Transcript

Introduction Video

Dr. Tom Faulkenberry – Hi, I’m Dr. Tom Faulkenberry, director of the math and science teacher preparation program at

Texas A&M University at Commerce. I will be spending the next hour with you going over the mathematics TExES exam.

The format will be simple. I will present problems that represent the types of knowledge you will be expected to

possess. Your knowledge will be broken up into 4 domains, each containing several competencies. I will present a

written description of each domain and competency followed by an example problem that I will work in detail. You will

need a calculator for some of these problems. I will be using a TI 84 Plus Silver edition but any comparable calculator

should do the trick. Now, let’s get started.

Slide 1 - .49 seconds

Domain I: Number Concepts

This domain includes some elementary content, but it is nonetheless important for the middle school teacher to know very well.

Understanding numbers goes way beyond simple arithmetic. Ideas from number theory can be used to solve hard problems in novel ways. Also, one needs to know some history behind how the numbers we use today came to be. Why do we not use Roman numerals still?

The practice problems that follow all represent some of the different number concepts that you should be familiar with.

Slide 2 - 1 min .10 seconds

Competency 001:

The teacher understands the structure of number systems, the development of a sense of quantity, and the relationship

between quantity and symbolic representation.


Slide 3 – 1 min .18 seconds

Dr. Faulkenberry – In the following problem we have a red trapezoid and a green triangle adding together to give us

4/5th. Given that the sum is 4/5th, what fraction would be represented by the yellow trapezoid?

Now the thing to realize about this problem that gives us something to go on is the fact that 4/5th is the same as 4, 1/5

pieces. That is, if we can visualize the top sum, the red trapezoid plus the green triangle, as 4, 1/5 pieces, then that

would tell us what 1/5th represents and then we can use that knowledge to deduce what the yellow trapezoid is worth.

Now if we play around with this just a little bit, we can see that there would be 3 green triangles that would fit into the

red trapezoid as follows. I’ll just make a sketch. Now that gives us a total of 1, 2, 3, 4 pieces summing together to give

us 4/5th. That is each of these is 1/5th in magnitude so we get the fundamental equality that the green triangle is worth

1/5th. Now also, we can break this yellow hexagon into 6, 1/5th pieces so this would give us 6/5th or simply 1 and 1/5th.

So the yellow trapezoid would represent a total of 1 and 1/5th.


Competency 002:

The teacher understands number operations and computational algorithms.


Dr. Faulkenberry – When asked to multiply 107 times 5, a student quickly responds 535. Explain how basic properties of

arithmetic could have been used to produce the answer so quickly.

Now this is a fairly simple problem to do mentally if one uses some basic properties of arithmetic. What I had in mind on

this problem, and this is similar to problems like it that you will encounter, is something like the following. First, one

thinks about 107 times 5 in terms of a decomposition of 107, that is 107 is the same as 100 and 7. Now writing it in this

manner one should be quickly led to think about this in terms of the distributive property, namely that 100 plus 7, times

5, lets one distribute the 5 to the 100 and to the 7 giving us 500 and a 35, resulting in 535. Now this is an easy

calculation to do mentally simply because 107 time 5 is easily represented mentally as 100 and 7. 100 times 5 or 500 is

an easy mental calculation as is 7 times 5 putting them together as 535.



Competency 003:

The teacher understands ideas of number theory and uses numbers to model and solve problems within and outside of

mathematics.


Dr. Faulkenberry – What is the smallest positive integer that does not evenly divide 43 factorial? Now realize that is a

factorial, I’m not just being emphatic about the statement 43!

43 factorial of course is simply the repeated product of 43 to 42 to 41 all the way down to 3 times 2 times 1. We’re

looking for the smallest positive integer that does not evenly divide 43 factorial. Now to get a handle on this problem,

first notice that 2 divides 43 factorial because 2 shows up as a factor in this factorization as does 3 and similarly 4, 5 and

up to 40, 41 42, 43. So all of the integers, 1 thru 43, do, in fact, evenly divide 43 factorial, simply by virtue of showing up

in the factorization. So that means we should probably start above 43, namely 44, so let’s write it down as a possible

candidate. Does 44 divide 43 factorial? Well, you don’t have to get a calculator out for this. All you have to do is realize

44 is the same as 4 times 11. Now, 4 does in fact show up in the factorization of 43 factorial as does 11 and hence 44

does in fact evenly divide 43 factorial. By similar reasoning, 45 will also evenly divide 43 factorial. One way to

demonstrate this would be to write 45 as 9 times 5 and note that each of these, 5 and 9 do show up in the factorization

of 43 factorial. Similar is 46. 46 is 2 times 23 and each of these factors shows up in the factorization of 43 factorial. We

finally run into a snag however at 47. 47 is prime. Since 47 is prime it has no smaller factors that are going to show up in

the factorization of 43 factorial. 47 itself does not show up in the factorization and hence, this is our smallest positive

integer that does not evenly divide 43 factorial.


Domain II: Patterns and Algebra

Much of algebra is developed very early by students who notice patterns. Inferring the next number in a sequence is a skill that is attainable by most young children, yet many teachers do not realize that this skill is necessary for the development of algebra.

Why? Inferring the next member of a pattern involves finding an unknown…this is the central problem in algebra. From here, children develop knowledge about functions and functional relationships.


Hence, it is important that the well-prepared teacher of 4-8 mathematics has a substantial knowledge of patterns and algebra. The problems that follow represent the types of problems that you will be expected to solve.


Competency 004:

The teacher understands and uses mathematical reasoning to identify, extend, and analyze patterns and understands the relationships among variables, expressions, equations, inequalities, relations, and functions.


Dr. Faulkenberry – Consider the following pattern. For the first entry we have 1 dot. For the second entry we have a

triangular pattern of 3 dots. For the third entry we have a triangular pattern of 6 dots and for the fourth entry we have a

triangular pattern of 10 dots. The question is how many dots would comprise the 100th figure?

Now you don’t have to go through and draw all of the figures from 5 up to a 100 to see the answer to this. You can tap

into this problem in 1 of 2 ways. The first thing that you could do is realize that these are simply the triangular numbers

and if you remember a formula for the Nth triangular number then great. Chances are you probably don’t so what I

want to look at is a classical bit of mathematics that would get us to the answer to this problem. First of all, realize that

each of the numbers of dots in these entries can be realized as a repeated sum. For example, the first dot is simply 1 but

the second pattern of dots could be looked at as 1 plus 2. The third could be looked at as 1 plus 2 plus 3 and the fourth

could be looked at as 1 plus 2 plus 3 plus 4. This is in fact the generating sequence of the triangular numbers. Using this

pattern, then the 100th figure would simply be the sum of 1 plus 2 plus 3 plus 4 plus all the way up to 98 and 99 and

finally adding 100. This is a famous sum in mathematics. You may remember a story about Gauss and how he

discovered a problem similar to this while being a very young student. The method to get this is very simple. First of all

notice you have 100 terms in this sequence. Now, the outer two sums added together, let me underline them for

emphasis, the outer two parts of the sum added together would be 101. Now move in 1 unit on both ends of the

sequence. So now we have 2 plus 99. 2 plus 99 is also 101 as is 3 plus 98. In fact, each time you move in 1 from the left

and move back 1 from the right, the sum is still going to be 101. So what we have is some number of groups of 101. Our

task now is to figure out how many groups is this? Well, now remember, we have 100 terms in this sequence. The 101

comes from a pairing of the outer most terms. So how many pairs are there in 100 terms. The answer is of course 50

pairs. So what we have here are 50 groups of 101 which we can easily figure out is 50 x 101 or 5050. So the 100th figure

would be a bit too big for us to draw but would have 5050 dots.



Competency 005:

The teacher understands and uses linear functions to model and solve problems.

Slide 12 – 11 min

Dr. Faulkenberry – A full water tank begins with 50 gallons of water. It loses 3.4 gallons each day due to a leak. Write an

equation that models the amount of water in the tank as a function of the number of days after it is filled.

Now this is a linear function. The main reason we know it is a linear function is that it loses a constant amount or rather

it changes by a constant amount each day. In this case it’s a loss and that’s going to correspond to a negative slope.

Now to write an equation for a linear function you need some information. You need the slope and you need the initial

value. Now on a graph, the initial value corresponds precisely to the Y intercept. So to get this function, we should

probably give it some reasonable names for the variables such as A for amount of water. You can use anything that you

wish. And since the change is by days we’ll probably use a small d to represent the days after it is filled. Now the way I

would write this function is as follows. I would say that the amount of water is going to start at 50 gallons and each day

it is going to lose 3.4 gallons so this would correspond to a minus 3.4 times the number of days (3.4d). Now you can of

course write this as the standard mx plus b form as follows (-3.4d+50). Being flexible with the form is one of the

hallmarks of being well prepared with this. So either way we get a function that models the amount of water in this tank

as a function of the number of days after it is filled.


Competency 006:

The teacher understands and uses nonlinear functions and relations to model and solve problems.


Dr. Faulkenberry - Suppose you deposit $1200 into an account that earns 6.3% interest (compounded once per year).

First we want to write an equation that models the amount A in dollars in the account after year y. Then we want to

answer how many years will it take for your money to double and figure out whether that answer depends upon how

big our initial deposit is and why.


So this is a multipart question but we got to have something to start with. Now this is a standard type of problem in

finance and in any mathematics course that you have probably encountered. For doing any kind of compound interest,

you realize that interest is computed upon previous interest and hence it’s not just a simple multiplication anymore but

a repeated multiplication that is represented by exponential function. So what we get is something like this. Our

amount depends on starting with $1200 and then each year we are going to multiply by something that is going to

increase that $1200 by 6.3%. Now the way to realize that is that’s got to be the original value, which is a multiplication

of 1, plus an additional 6.3%, or .063 Hence the growth factor for our exponential function is 1.063. We are going to

raise that to the y power because each year that multiplication, the effect of that repeated interest, is a repeated

multiplication. So this answers part 1. This is our compound interest formula and this models the amount in the

account after year y. Now, how many years will it take for our money to double? Well if we started with $1200, that

means we simply need to figure out what is the solution of the equation 2400 equals 1200 times 1.063 to the y. Now

there are various ways I could solve this and I would like to demonstrate a couple. What I’m going to do before I do any

of that is I’m going to realize on both sides of this equation I have a factor of 1200 that I can simply divide and simplify

this equation just a little bit. So I get 2 equals 1.063 to the y. Now the first way I’m going to solve this equation is

algebraically. Now if I want to get y by itself I’ve got to get it out of its role as an exponent and a standard way to do this

is with a logarithm. So I can take a log of both side I’m going to use the natural log. And when I do this I use the nice

little property of logarithms and how they play with exponents to find out that this y is going to come down to the front.

So what I get is something like log 2 equals y times log of 1.063 leaving y equal to log 2 divided by log of 1.063. Now I’m

going to cover up part of the paper with my calculator. I apologize for that. Then just enter in that calculation log 2

divided by log of 1.063 and I’m going to get something equal to 11.35 approximately. So it’s going to take approximately

11 to 12 years for this money to double. Now another way I want to show you to work this involves using a graph. I’m

going to scoot this over and look at my original equation. Namely, the 2 equals 1.063 to the y. I could enter both of

those as graphs as 2 and 1.063, raised to the x in this case, because I’m working on my calculator, and graph these

functions. Now what values are these going to take? Well, you know that you have to have a horizontal line at 2 so you

don’t want your y scale to be that large and we also know that it’s going to take several years for this to happen and

since the variable, in this case x from my graphing calculator, represents years I want to go out several years so I’m going

to set my window from 0 up to, let’s say up to 20 years, assuming I didn’t know how long it was going to take todouble.

The y mins, there is no reason to look at negative values, and I’m going to go up to about 3, again because I need to see

y equals 2 somewhere. And we will just look at a graph. So there’s my horizontal line and there’s my growth function

and I can simply use the functions of my calculator to find where they intersect. And as you can see, I get 11.34 as my

solution again so there is a couple of different ways you could solve that especially if you didn’t remember exactly how

the algebra of logs goes. Now for the last question, does the answer in part 2 depend on the initial deposit? The answer

is no. The reason is because of this step that we did to get from the original equation down to this reduced equation.

Notice no matter what you start with here doubling it will simply result in this number here being 2 times what you

started with. In every case you can divide by that starting value and you will be left with precisely this equation. So no

matter what you start with it’s only the interest rate that is going to determine how long it takes for your money to

double.


Slide 15- 18 min .46 seconds

Competency 007:

The teacher uses and understands the conceptual foundations of calculus related to topics in middle school mathematics.


Dr. Faulkenberry – Consider the infinite sequence of red squares shown below. If the area of the large square, that is,

this entire square, is 2, what is the area of the infinite sequence of red squares?

Now this is an idea of calculus known as infinite sequences or an infinite series because we are going to be adding

together the terms of this infinite sequence. So the thing that we want to do is realize how does each smaller square,

each smaller red square, depend upon the previous larger one? Now as we can see, the red square, the largest red

square, is 1/4th of the total square. Similarly, this next largest red square is 1/4th of the largest one, etc., etc., etc. So

what we really have is the following infinite series. We start out with a red square that is an area ½. Why is it ½? Well

because we know the original

square has area 2 and 1/4th of 2 is indeed ½. The area of the next square is going to be ½ times 1/4th because well it’s,

1/4th of the area of the square that’s larger than it. Similarly, this one is going to be 1/4th of this guy. Now you notice a

pattern here. If it’s 1/4th of this term then that’s simply ½ times 1/4th times 1/4th and what I’m going to get for each

square is a similar pattern. Add ½ times ¼ to the 3rd power plus ½ times 1/4th to the 4th power etc. etc. and it’s an

infinite sequence that is going to go on forever. So that’s how I’m going to construct the area of these red squares

added together. Now down here I want to clean this up a little bit. This area. First of all notice that every term has a ½

in it so I’m going to pull that ½ out and what I’m left with is 1 plus 1/4th plus 1/4th squared plus 1/4th cubed etc. without

end. So now I’ve got to figure out a way to get a handle on this guy right here. But if you remember some of your basic,

and this goes back to college algebra and possibly some other courses that you took, when you have this geometric

sequence, that is an infinite geometric sequence, its sum can be had very simply. Namely, you take the first term 1 and

divide by 1 minus this common ratio which is 1/4th. So if we simplify that just a little bit, I get ½ times, now 1 over 1

minus 1/4th is the same as 1 over 3/4th and 1 over 3/4th is of course 4/3rd so this is ½ times 4/3rd which simplified will give

us 2/3rd. So the area of infinite sequence of red squares has a sum of 2/3rd.



Domian III: Geometry and Measurement

Geometry and measurement is an important branch of mathematics for the 4-8 mathematics teacher to have mastered. For most children, geometry is the transition from concrete ideas to abstract reasoning. This is due to the fact that we can see the things that should be true about a geometric figure; so the only thing remaining is the ability to use deductive reasoning to show that it must be true, regardless of our ability to see it.

Measurement is essential to functioning in the physical world. Most physical phenomenon are studied because they can be measured. Hence, it is essential for the well-prepared teacher of 4-8 mathematics to have a deep knowledge of the concepts of geometry and measurement.

The problems that follow are indicative of the types of problems you will be expected to solve.


Competency 008:

The teacher understands measurement as a process.


Dr. Faulenberry – A forest manager uses a clinometer to measure the angle of elevation from a point on the ground to

the top of a tree 500 feet away. The angle is measured to be 25.7 degrees. The tree is known to be 82.03 yards tall.

What is the percent error of the forest manager’s measurement? Now first of all, we need to have a talk about what

percent error really is. Percent error, which I’m going to denote as PE, is simply the difference in an experimental or trial

measurement value with an accepted or known value. So one way to look at this is you can look at it as the

experimental minus the known and this quantity, this difference, needs to be scaled by the known quantity. You want

to compare it to the known quantity. So when we find out our experimental height, so to speak for this tree, then we

can compare it to this known value of 82.3 yards and figure out what the percent error of our instrument is. So to do

this we are going to use some basic trigonometry. We have a point on the ground that we are measuring the height of

this tree from and I’m going to denote this tree just by a single line. Now this triangle is not necessarily drawn to scale

but it will give us some reasonable mental things to help us with the calculation. Now we measure this angle to be 25.7

degrees. We know that the horizontal distance from our clinometer to the base of the tree is 500 feet. What we don’t

know is this height h. But we know we can relate those quantities via the tangent function. Namely tangent of 25.7

degrees is the same as the height divided by 500, the opposite over the adjacent and hence height is 500 times tangent

of 25.7 degrees. So I’m going to pull in the calculator just real quickly. First of all, make sure that you are in degrees


mode and if you’re not make sure you are there because if you are in radiance mode you won’t get a very sensible

answer here. So I’m going to find out what 500 times tangent of 25.7 is and I get 240.63 and the units of this

measurement are in feet. Ok, so that gives us our experimental value. Now the tree is known to be 82.3 yards tall. 82.3

yards is the same as, say were going to compare it to, 82.3 yards is the same as 246.9 feet. So we want to compare our

experimental value of 240.63 feet to 246.9 feet, our known value. So, to do that, we’ll simply apply our concept of

percent error. We’ll find the difference between the two values. I’ll do the known, 240.63 and subtract 246.9. Now

you’re going to get a negative answer here but ignore the negative sign. Were just simply wanting a magnitude

difference. We’re going to compare that to this know 246.9 and that gives us a percent error of 2.54%. Notice the

answer, I’ve got a negative sign here, ignore the negative and we’ve got .02539 so I just round that to 2.54%. So, our

clinometer is off in its measurement by 2.54%.


Competency 009:

The teacher understands the geometric relationships and axiomatic structure of Euclidean geometry.


Dr. Faulkenberry – Suppose triangle ABC is an isosceles triangle with segment AC congruent to segment AB; this one and

this one. A student constructs segment AD as shown: Is segment BD congruent to segment CD and why?

So this is a problem about basic geometric construction. As you can see from the construction marks, this student has

constructed an angle bisector, namely angle BAC has been bisected by this segment. So what we want to do is go

through the steps of a proof that one could use to establish the congruence of segments BD and CD. Now first of all,

when one constructs an angle bisector, the first construction is to start with a compass point on the vertex and swing an

arc and you can see these two arcs have been swung. The purpose of that is to establish that this segment and this

segment are congruent because they come from a circle of the same radius. Now similarly, the next step is to come

from this point and swing an arc and this point and swing an arc. And again, these two segments that have been

constructed are congruent. Now notice what you have. What you have is a side, side, side triangle congruence, namely

this point to this point to this point this triangle is congruent to this triangle via side, side, side and since corresponding

parts of congruent triangles are congruent, we have that this angle and this angle are congruent. Ok, namely what we

have is angle BAD is congruent to angle CAD. Now that’s simply a proof of the angle bisector construction and that it

actually works. Now our question is whether segment BD is congruent to segment CD. We’re going to use this piece of

information. Namely, that this angle and this angle are congruent. So what can we do? Well first of all, we know

segment AC is congruent to segment AB. We also know that AD is congruent to itself and finally we know that this angle


and this angle are congruent. So what do we have? We have a, side angle side, congruence. So by side angle side that

gives us that triangle BAD is congruent to triangle CAD. And once again, segment BD and segment CD are corresponding

parts of these two congruent triangles. Hence, they are in fact, congruent so this is a true statement and that’s an

explanation of why that statement is true.


Competency 010:

The teacher analyzes the properties of two- and three-dimensional figures.


Dr. Faulkenberry – We’re asked to find the surface area and volume of the following prism with a square base.

Now we’re not given much measurement on this prism and that is going to be a bit of an issue for us because we’re

going to have to use the Pythagorean theorem several times to find some relevant measurement. I’m going to work on

surface area first and then volume will follow quiet easily from our work on surface area. For surface area, recall that we

are interested in the area of the two dimensional skin around this entire prism. So one of the ways you can do this is

simply unfold what is called a net. So I’m going to sketch the net really quickly. There’s my base and if you were to

actually peel this like a banana you’d be peeling the 4 triangles off of the sides like so. Here’s a sketch, certainly not to

scale. Now we know some attributes of this. We know that each of these sides is of length 4 and we also know that the

slant height, that is, the height of that little slanted portion, is 5. Now unfortunately that slant height doesn’t help us

immediately when we try to find the area of this triangle but it will help us. We’ll use Pythagorean Theorem, to figure

out what the height of this triangle is and realize all of these are completely congruent to that triangle. So now to use

the Pythagorean Theorem, I’m just going to sketch this part of the triangle I’m interested in namely this upper half

triangle. Notice that the base, it’s kind of a strange thing but I have my base over here on the left hand side so you’re

going to have to do some mental rotations to follow here. This base has length 2 and we have a hypotenuse of 5. What

we are interested in is figuring out what this length is. Well now let’s just call that X since this is a right triangle by the

way we have pulled it out in construction we know that 2 squared plus X squared equals 5 squared. I will write that

down which of course is 4 plus X squared equals 25. If we solve this we see that X squared is 21 or X is the square root

of 21. So what that tells us, that’s a little hard to see right there but we will say it as well, is that this height is root 21.

And so the area, I’m going to come over here to keep this out, the area of the triangle is going to be ½ of the base times

the height. This is ½ times 4 times square root of 21 which simplified is simply 2 times root of 21. So, the total surface

area is going to be this square base which is 4X4 or 16, 16 square units, plus 4 of these triangles and each triangle has

area 2 time root 21 which gives us a total of 16 plus 8 root 21. So our total surface area is this. You could simplify this

on a calculator if you wanted to. Square root of 21 is about 4.5 but I’m going to leave that with an exact answer. Now


for volume, we’ve got 1 last thing. If you were to imagine rotating this prism, until you saw it directly from its side, then

what you would have is a schematic kind of like this, where the link of this total side down here was 4. Hence of

coursethis is 2. This link would be precisely this length that we found over here earlier that is namely square root of 21.

What we don’t know is this height. So if we were going to try to find out of course we could use the Pythagorean

Theorem again and relate the sides 2 squared plus h squared equals, now 21 squared of squared. That’s going to be an

easy one for us. So h squared, equals 21 minus 4 equals 17 which of course tells us that h is the square root of 17. Now

why is that important? That is important because to find the volume of a prism, we simply take, it occurs to me we are

using the absolute wrong term here, this is a pyramid, not a prism, sorry about that. It’s one of those things you catch

when you are doing something like that. To find the volume of a pyramid, we simply take the area of the base and

multiply by the height and take 1/3rd of that. So in the remaining space I have left over here, the volume is going to be

1/3rd of my base area by my height. Well, that’s 1/3rd. Now my base area is 4X4 or 16 again and then my height I found

to be root 17. So if I work all that out, I get 16 times root 17 over 3. Again, you could simplify that but I’m going to leave

the exact answer. So again, the surface area and volume of the pyramid, we had to use some Pythagorean Theorem,

but that gave us some fairly reasonable computations.


Competency 011:

The teacher understands transformational geometry and relates algebra to geometry and trigonometry using the

Cartesian coordinate system.


Dr. Faulkenberry – An overhead light source projects triangle ABC to triangle A prime B prime C prime. Given that the

projection is a dilation, what is the perimeter of triangle ABC in terms of the variable X?

Now a dilation is a multiplicative transformation. That is, its change is governed by multiplication. So if you will notice

segment AB is transformed to segment A prime B prime via a multiply by 3 transformation. Similarly, this segment BC,

which has length 4, which is transformed to segment B prime C prime which will also have length 12 because it’s a

4times 3 transformation. Finally, segment AC is transformed to segment A prime C prime, via the same transformation.

Namely, that this length is multiplied by 3 to give us X. Now that leaves us with the question of what has to be

multiplied by 3 to give us X and of course the answer is X divided by 3. If I multiply X divided by 3 times 3 I do indeed get

X. And so now that we have that established, that is sort of a key part of this problem, the perimeter of triangle ABC is as

simple as adding the 3 component of triangle ABC. So our perimeter is 4 plus 4 plus X/3 or simply 8 plus X/3.



Domain IV: Probability and Statistics

Probability and statistics have historically been neglected as topics of substantial coverage in the mathematics curriculum. However, probabilistic and statistical reasoning are used every day to make common decisions in our world.

The well-prepared teacher of 4-8 mathematics must have a good working knowledge of the basic principles of descriptive and inferential statistics. The problems that follow are representative of the types of problems you’ll be expected to know how to do.


Competency 012:

The teacher understands how to use graphical and numerical techniques to explore data, characterize patterns, and describe departures from patterns.


Dr. Faulkenberry – The scores on a survey of attitudes toward mathematics and science are distributed according to the

following box and whisker plot. Would either 7 or 19 be considered as an extreme score or an outlier?

Now a box and whisker plot is a very nice way to look at the distribution of a set of data. For example, it gives us 5

important pieces of information; the lowest value, the highest value, the median, which is 12 in this case, and the 2

quartiles. The quartiles are simply the medians of the lower half and the upper half of the data. Now one of the things

you will notice about a box and whisker plot is that sometimes the whiskers are very short. For example, this side has a

much shorter whisker than this side and that’s a visual cue that 19 does not follow the distribution of the data as well as

7 does. So we have this visual indication that 19 might be considered extreme or an outlier. Now the rigorous way to do

this is to take what’s called the inner quartile range, otherwise known as the IQR. That’s simply the distance between

the two quartiles; the third quartile minus the first quartile, which in this case is 13-10 or 3. Now this inner quartile

range is used then as a metric upon which to gauge your outliers. For example, notice that 7 is exactly 1 of these boxes

away from 10. That is its 1 inner quartile range. But now 13 to 19, you could fit this box, which again we will call the

IQR, you could fit that 1, 2, 3 times between, sorry, you could fit that twice between the 13 and the 19. Since 19 is 2

times the inner quartile range away from its closest quartile, we consider 19 as an outlier and we would reject it as such.


Now, you may be asking what is the cut-off? In other words, how far away is considered ok? The cutoff value for this

kind of judgment is 1.5 times the IQR. Anything beyond 1.5 times the IQR is considered as an outlier. Anything within

1.5 times the IQR is considered OK.


Competency 013:

The teacher understands the theory of probability.


Dr. Faulkenberry – A game is defined as follows: You roll one six-sided die. You win if you roll an odd prime number and

you lose otherwise. It costs $1 to play. If you win, you win $3 (in addition to getting your dollar back). If you lose, you

lose your dollar. If you play the game 10 times, what is the probability that you’re going to win a net total of $6?

Now this is a problem that seems simple enough but it’s got a lot of very core concepts from probabilities and statistics

there. Now to start, let’s size of the game in terms of what’s a win and what’s a loss. So up here notice that we will win

if we roll a 3 or a 5, namely, an odd prime number. We will lose if we roll otherwise. So that’s a 1, 2, 4, or 6. That gives

us that the probability of a win is 1/3rd and the probability of a loss is 2/3rd. Now that’s going to be important

information for us in just a little while. Now first of all, note that we are going to play 10 times. To find out the

probability of winning 6 dollars we have got to figure out how we are going to win 6 dollars at all. We can do this pretty

systematically. Let’s use a little bit of algebra. First notice that our wins plus our losses is going to add up to 10 games.

We also know we are going to get 3 dollars for every win for a net total of 3 times the number of wins and we are going

to lose a dollar for every loss and if we add up our wins and losses we want to end up with 6 dollars(3w-l=6). Now this

gives us a system of 2 equations with 2 unknowns which we can quickly solve using the elimination method. Namely, we

get 4w=16 or w=4. So that tells us that to win 6 dollars we need 4 wins and we need 6 losses. Of course we can check

that out really quickly. 4 wins would give us 12 dollars and 6 loses would give us a loss of 6 dollars. Hence we get a net

total of 6 dollars. Now, what is the probability that we win 6 dollars? Well, that’s the same as the probability of playing

the game 10 times and winning 4 times and that is going to be governed by what is called a binomial distribution. Now

the reason I am pointing this out is because somewhere in your preparation you have probably come across this term

binomial distribution. Binomial distribution simply helps us find the probability of getting some number of successes out

of what are called binary trials namely win/loss type trials. So the probability of getting 4 wins, which is we will call X

equals 4. Well, the way you do that is very simple. First of all, you know that out of 10 games you’ve got to win 4 of

them. So how many ways are there to do that? Well, you’ve got 10 choose 4. This is a combination of 10 things taken 4

at a time. For each of those 4 wins the probability of that win is 1/3rd. So that’s going to give us one third to the 4th


power and for the remaining 6 loses, the probability of that happening is 6. Now putting all of this together we get a

simplification. First of all, 10 choose 4, that’s 10 factorial over 4 factorial times 10 minus 4 factorial, or 6 factorial times

one third to the 4th times 2/3 to the 6th. Now, 10 factorial, over, 6 factorial is 10 X 9 X 8 X 7 all over 4 X 3 X 2 X 1. Now

I’m going to get rid of a little bit of this stuff. I’m going to divide the 4 and the 2 into 8 and I’m going to divide the 3 into

the 9 leaving a 3 there. And now I’m going to get my calculator out so I can get an idea of what the numerical answer

would be. I’m going to get a 10 x 3 x 7. So that’s 210. I’m going to multiply the 210 times (1/3) to the 4th then I’m going

to multiply that by 2/3rd to the 6th and I hopefully will get a decimal answer between 0 and 1 and in fact I do. I get that

that’s approximately equal to .228 or in other words a .228 % chance of winning 6 dollars. So there’s my probability

based on the binomial distribution.


Competency 014:

The teacher understands the relationship among probability theory, sampling, and statistical inference, and how

statistical inference is used in making and evaluating predictions.


Dr. Faulkenberry – If it were applied to the entire population, an IQ test would have a mean of 100 and a standard

deviation of 15. What is the probability that a random sample of size 25 would have an average IQ greater than 103?

Now to do this we need a basic concept of what the normal distribution looks like. I’m going to sketch an ideal normal

distribution. So this normal distribution has a mean of Mu and a standard deviation of sigma. Now where do you find

that on the curve? Well, you look for the inflection points, namely the points where the concavity changes. Now the

idea being the distance from the Mu to these inflection points is exactly the standard deviation. Now one of the things

we know about normal curves is that we have some benchmark ideas of the area underneath these curves or that is the

percentage of data contained within certain bounds. So let’s just sketch those out. Within 1 standard deviation of the

mean, we know we have 68% of the data contained. That is, this portion right in here that is one standard deviation on

either side of Mu contains 68% of the data. Now similarly, we know something about within 2 standard deviations.

That’s 95% of the data. That would be from about here to here leaving about 5% of the data in these tails split up equal.

Now, to our problem. If were talking about taking a random sample of size 25 then that gives us a whole new

distributions of means. Namely, if we picked 25 people from here and computed their mean, that would give us one

data point down here and if we did it again it would give us another data point and do it again and give us another data

point. And if we look at how those stack up, what we notice is that we have the same mean but we have a skinnier

distribution. By the way, this is called the Central Limit Theorem. Central Limit Theorem says no matter what you start

with your still going to get a normal distribution with the same mean. It’s just going to have a standard deviation that is


different. The standard deviation for the Central Limit Theorem is going to be found by taking the original standard

deviation and dividing by the square root of the sample size. Now the sample size in our problem is size 25 and our

original standard deviation was 15 so if we put all this together, this gives us 15/5 or 3. So this new distribution has a

standard deviation of 3. Now we can use all of that information to our advantage. We are interested in what is the

probability of having an average IQ greater than 103. That is, we’re interested in what is the probability of being in this

right hand tail? That is having an IQ 3 points away from the mean. Now let’s use our information we talked about

before. Within 1 standard deviation is 68% of the data. That leaves 34% of the data out here in the tails. Now if were

interested in just this tail, that’s going to be half of that 34% or 17%. And that is indeed our probability of having an

average IQ greater than 103 with this sampling distribution is .17 or 17%.

Closing Comments – 49 min .28 seconds

Dr. Faulkenberry – This concludes the problem session. If you need further assistance, such as some extra problems to

work that are similar to the ones you’ve dealt with today, you can always do a Google search for the Praxis exam which

is a certification exam that many other states use. I want to give you some tips for your testing day. First of all, this may

sound old but, it still holds. Sleep well and eat well before hand. Look, you’ve got a 5 hour exam waiting on you so you

better be prepared for it. Make sure you bring a picture ID and some pencils. Some sharpened #2 pencils. You’re not

allowed to have any electronic devices on you so don’t bring a cell phone or an iPod or anything like that. Now if you’re

taking the Math 4-8 exam, you will be provided a calculator you can use. You cannot bring your own. However, those of

you taking the Math 8-12 exam, can bring your own graphing calculator if you wish. There is a list of approved calculator

on the TExES website but if you have the standard TI-84 or similar calculator, that’s going to be an ok one. Well, that

concludes our review session. Good luck and have a great day.

Documents

T-CERT Math 4-8 Module Transcript