Chapter 7 Random Variables and Probability Distributionsbtravers.weebly.com/uploads/6/7/2/9/6729909/...Chapter 7 Random Variables and Probability Distributions. x1 What Is A Random

Chapter 7 Random Variables andProbability Distributions

§ 1 What Is A Random Variable?

Defining the Random Variable

We have talked of random variables and stressed the idea ofrandomness being important. But how can we define a randomvariable?

A basic example to illustrate:

ExampleSuppose you flip a coin two times. What are the possible outcomes?{HH,HT,TH,TT}

Let the variable X represent the number of heads that result from thisexperiment.

What values can X take on? 0, 1, or 2

Here, X is a random variable because its value is determined by theoutcome of a statistical experiment.











ExampleSuppose you flip a coin two times. What are the possible outcomes?

{HH,HT,TH,TT}























What values can X take on?

0, 1, or 2

















So who wants to take a shot at defining the random variable?

DefinitionA random variable is a variable whose value is defined by thenumerical outcome of a random phenomenon.

DefinitionA random variable is defined as a function that associates a realnumber (the probability value) to an outcome of an experiment.

DefinitionA probability distribution of a random variable X tell us the values Xcan take on and how to assign probabilities to those values.
















Types of Random Variables

DefinitionA discrete random variable is a variable that represents numbersfound by counting.

An example of this is our coin flip example.

Probability Distribution

Heads Probability0 1

41 2

42 1

4






41 2

42 1

4






41 2

42 1

4


ExampleSuppose we are interested in looking at statistics test scores from acertain college from a sample of 100 students. The random variable Xwould be the test scores, which could range from 0% (didn’t study atall) to 100% (excellent student). However, since test scores vary quitea bit and they may even have decimal places in their scores, we can’tpossibly denote all the test scores using discrete numbers. So in thiscase, we use intervals of scores to denote the various values of myrandom variable.

DefinitionA continuous random variable X is a random variable found frommeasuring where X is continuous because it represents an infinitenumber of values on the number line.


ExampleSuppose we are interested in looking at statistics test scores from acertain college from a sample of 100 students. The random variable Xwould be the test scores, which could range from 0% (didn’t study atall) to 100% (excellent student). However, since test scores vary quitea bit and they may even have decimal places in their scores, we can’tpossibly denote all the test scores using discrete numbers. So in thiscase, we use intervals of scores to denote the various values of myrandom variable.

DefinitionA continuous random variable X is a random variable found frommeasuring where X is continuous because it represents an infinitenumber of values on the number line.



Test Score Frequency (percentage)0% to < 20% 520% to < 40% 2040% to < 60% 3060% to < 80% 3580% to 100% 10

We need ranges here because the probability of achieving any onesingle grade from an interval is 0.



Test Score Frequency (percentage)0% to < 20% 520% to < 40% 2040% to < 60% 3060% to < 80% 3580% to 100% 10

We need ranges here because the probability of achieving any onesingle grade from an interval is 0.

Keys to Remember

1 We cannot know which data element is being selected ahead oftime.

2 We must know which values can be taken on by the randomvariable.

3 We must know the probability of each value (or range of values)ahead of time.

4 Each data element has the same chances of being selected.

Keys to Remember





Keys to Remember





Keys to Remember





§ 2 The Mean of a Random Variable

What is the Mean of X?

We have seen this before a bit but have not yet formally defined thisidea. This is the expected value.

DefinitionThe mean of a discrete random variable X is a weighted average of thepossible values that the random variable can take.

Mean of a Discrete Random Variable

E(X) = µ =n∑

i=1

[xi · P(xi)]





E(X) = µ =n∑

i=1

[xi · P(xi)]





E(X) = µ =

n∑i=1

[xi · P(xi)]

The Mean of X

What we need to keep in mind is that this is the long term average forthe variable.

Mean of a Continuous Random Variable

E(X) =∫

xx · f (x) dx

We will only work with discrete random variables in this class.

The Mean of X



E(X) =∫

xx · f (x) dx


The Mean of X



E(X) =∫

xx · f (x) dx


Mean of a Random Variable

Since we have seen this aspect, we will just do one example here as areminder.

ExampleSuppose we have the following probability distribution for a game.

Outcome -$1.00 $0.00 $3.00 $5.00Probability .30 .40 .20 .10

Why is this a discrete random variable?


Since we have seen this aspect, we will just do one example here as areminder.



Why is this a discrete random variable?




What is the mean of the random variable?

E(X1) = .3(−1) + .4(0) + .2(3) + .1(5) = .8

How do we interpret this value?





E(X1) = .3(−1) + .4(0) + .2(3) + .1(5) = .8






E(X1) = .3(−1) + .4(0) + .2(3) + .1(5) = .8


Properties of the Mean of a Random Variable

ExampleSuppose the casino didn’t like that this was a losing game so itdecided to subtract 1 from all of the outcomes. What is the new mean?

Outcome -$2.00 -$1.00 $2.00 $4.00Probability .30 .40 .20 .10

E(X2) = .3(−2) + .4(−1) + .2(2) + .1(4) = −.20

= E(X1)− 1




E(X2) = .3(−2) + .4(−1) + .2(2) + .1(4) = −.20

= E(X1)− 1




E(X2) = .3(−2) + .4(−1) + .2(2) + .1(4) = −.20

= E(X1)− 1

Properties of Random Variables

ExampleSuppose the casino decides that the game does not have impressiveenough top prize with the lower payouts, so it decides to double alloutcomes. What is the new mean?


E(X3) = .3(−4) + .4(−2) + .2(4) + .1(8) = −.40

= 2E(X2)




E(X3) = .3(−4) + .4(−2) + .2(4) + .1(8) = −.40

= 2E(X2)




E(X3) = .3(−4) + .4(−2) + .2(4) + .1(8) = −.40

= 2E(X2)


Scaling Random VariablesIf a random variable X is adjusted by multiplying by the value b andadding the value a, then the mean is affected as follows:

µa+bx = a + bµX

Sums of Random VariablesThe mean of the sum of two random variables X and Y is the sum oftheir means:

µX+Y = µX + µY


Scaling Random VariablesIf a random variable X is adjusted by multiplying by the value b andadding the value a, then the mean is affected as follows:

µa+bx = a + bµX

Sums of Random VariablesThe mean of the sum of two random variables X and Y is the sum oftheir means:

µX+Y = µX + µY

§ 3 Variance and Standard Deviation

Variance of a Random Variable

DefinitionThe variance of a discrete random variable X measures the spread, orvariability, of the distribution.

Variance of a Random Variable X

Var(X) = σ2

= σ2X

=

n∑i=1

(xi − µ)2 · P(xi)

Variance of a Random Variable

DefinitionThe variance of a discrete random variable X measures the spread, orvariability, of the distribution.


Var(X) = σ2

= σ2X

=

n∑i=1

(xi − µ)2 · P(xi)

Variance

This formula can be difficult to work with, especially for large datasets, so we have a computational formula that is easier to work with.


Var(X) =n∑

i=1

[x2i · P(xi)]− µ2

Variance

This formula can be difficult to work with, especially for large datasets, so we have a computational formula that is easier to work with.


Var(X) =n∑

i=1

[x2i · P(xi)]− µ2

Back to the Original Game

ExampleFind the variance of the original gambling game.


σ2 = [(−1)2(.3) + 02(.4) + 32(.2) + 52(.1)]− .82

= 3.96




σ2 = [(−1)2(.3) + 02(.4) + 32(.2) + 52(.1)]− .82

= 3.96




σ2 = [(−1)2(.3) + 02(.4) + 32(.2) + 52(.1)]− .82

= 3.96

Properties of Variance

ExampleConsider the outcomes from this game at our third iteration:


Find the variance.

σ2 = [(−4)2(.3) + (−2)2(.4) + 42(.2) + 82(.1)]− (−.4)2

= 15.84

= 4(3.96)

= 22(3.96)




Find the variance.

σ2 = [(−4)2(.3) + (−2)2(.4) + 42(.2) + 82(.1)]− (−.4)2

= 15.84

= 4(3.96)

= 22(3.96)




Find the variance.

σ2 = [(−4)2(.3) + (−2)2(.4) + 42(.2) + 82(.1)]− (−.4)2

= 15.84

= 4(3.96)

= 22(3.96)




Find the variance.

σ2 = [(−4)2(.3) + (−2)2(.4) + 42(.2) + 82(.1)]− (−.4)2

= 15.84

= 4(3.96)

= 22(3.96)


Scaling Random VariablesIf a random variable X is adjusted by multiplying by the value b andadding the value a, then the variance is affected as follows:

σ2a+bX = b2σ2

Since the spread of the distribution is not affected by adding orsubtracting a constant, the value a is not considered. But since thevariance is the sum of squares, we must square b when adjusting thevariance.


Scaling Random VariablesIf a random variable X is adjusted by multiplying by the value b andadding the value a, then the variance is affected as follows:

σ2a+bX = b2σ2

Since the spread of the distribution is not affected by adding orsubtracting a constant, the value a is not considered. But since thevariance is the sum of squares, we must square b when adjusting thevariance.

Variance Properties

One more property that may surprise you ...

Sums/Differences of VariancesFor independent random variables X and Y , the variances of the sumsor differences are:

σ2X+Y = σ2

X + σ2Y (1)

σ2X−Y = σ2

X + σ2Y (2)

If the variables are not independent, the variance of one is affected bythe other and therefore we cannot use our formula to calculate thevariances.

For example, if X is the amount of money a group of individualsspend on lunch and Y is the amount of money a group of individualsspend on dinner. It is not reasonable to assume these are independentrandom variable and we can’t calculate the variance using our method.

Variance Properties



σ2X+Y = σ2

X + σ2Y (1)

σ2X−Y = σ2

X + σ2Y (2)



Variance Properties



σ2X+Y = σ2

X + σ2Y (1)

σ2X−Y = σ2

X + σ2Y (2)



Variance Properties



σ2X+Y = σ2

X + σ2Y (1)

σ2X−Y = σ2

X + σ2Y (2)



Variance Properties



σ2X+Y = σ2

X + σ2Y (1)

σ2X−Y = σ2

X + σ2Y (2)



Standard Deviation of a Random Variable

We have a similar relationship between variance and standarddeviation here as we did before.

Standard Deviation of a Random VariableThe standard deviation can be interpreted as the typical amount thatvalues of the variable deviate from the mean.

σ =√σ2 =

√√√√ n∑i=1

(xi − µ)2 · P(xi)

Standard Deviation of a Random Variable

We have a similar relationship between variance and standarddeviation here as we did before.

Standard Deviation of a Random VariableThe standard deviation can be interpreted as the typical amount thatvalues of the variable deviate from the mean.

σ =√σ2 =

√√√√ n∑i=1

(xi − µ)2 · P(xi)

Back to z-Scores

Who remembers the formula?

z-ScoresIf x is an observation from a distribution with mean µ and standarddeviation σ, the standardized value of x is

z =x− µσ

Standard NormalThe standard Normal distribution is the Normal distribution N(0, 1)with mean 0 and standard deviation 1.If a variable x has any Normal distribution N(µ, σ) with mean µ andstandard deviation σ, the then standard variable

z =x− µσ

has the standard Normal distribution.

Back to z-Scores



z =x− µσ


z =x− µσ


Back to z-Scores



z =x− µσ


z =x− µσ


A Classroom Experiment

I am going to steal an idea from the book because I think it is a goodexercise.

First, we need to create a probability distribution table for all possiblesums we can attain when rolling two dice.

X 2 3 4 5 6 7 8 9 10 11 12P(X) 1

36236

336

436

536

636

536

436

336

236

136




X 2 3 4 5 6 7 8 9 10 11 12P(X) 1

36236

336

436

536

636

536

436

336

236

136




X 2 3 4 5 6 7 8 9 10 11 12P(X) 1

36236

336

436

536

636

536

436

336

236

136

A Classroom Example

Find the mean of the random variable.

µ = 2(

136

)+ 3

(2

36

)+ 4

(336

)+ 5

(436

)+ 6

(5

36

)+ 7

(636

)+8(

536

)+ 9

(4

36

)+ 10

(336

)+ 11

(236

)+ 12

(136

)= 7

A Classroom Example


µ = 2(

136

)+ 3

(2

36

)+ 4

(336

)+ 5

(436

)+ 6

(5

36

)+ 7

(636

)+8(

536

)+ 9

(4

36

)+ 10

(336

)+ 11

(236

)+ 12

(136

)

= 7

A Classroom Example


µ = 2(

136

)+ 3

(2

36

)+ 4

(336

)+ 5

(436

)+ 6

(5

36

)+ 7

(636

)+8(

536

)+ 9

(4

36

)+ 10

(336

)+ 11

(236

)+ 12

(136

)= 7

A Classroom Example

Find the variance of the random variable.

σ2 =

[22(

136

)+ 32

(236

)+ 42

(336

)+ 52

(436

)+ 62

(536

)+72

(636

)+ 82

(536

)+ 92

(4

36

)+ 102

(336

)+112

(2

36

)+ 122

(136

)]−72

= 5.83

σ = 2.415

A Classroom Example


σ2 =

[22(

136

)+ 32

(236

)+ 42

(336

)+ 52

(436

)+ 62

(536

)+72

(636

)+ 82

(536

)+ 92

(4

36

)+ 102

(336

)+112

(2

36

)+ 122

(136

)]−72

= 5.83

σ = 2.415

A Classroom Example


σ2 =

[22(

136

)+ 32

(236

)+ 42

(336

)+ 52

(436

)+ 62

(536

)+72

(636

)+ 82

(536

)+ 92

(4

36

)+ 102

(336

)+112

(2

36

)+ 122

(136

)]−72

= 5.83

σ = 2.415

A Classroom Example


σ2 =

[22(

136

)+ 32

(236

)+ 42

(336

)+ 52

(436

)+ 62

(536

)+72

(636

)+ 82

(536

)+ 92

(4

36

)+ 102

(336

)+112

(2

36

)+ 122

(136

)]−72

= 5.83

σ = 2.415

A Classroom Example

Now to the calculator:

1 Press MATH and scroll to PRB

2 Select Option 5: randInt(

3 The syntax you want is randInt(1,6,100) STO 2nd L1

4 Repeat, but this time store under L2

5 Press 2nd L1+ 2nd L2 STO L3

6 Press STAT and in the CALC menu, select 1-Var Stats7 Syntex: 1-Var Stats L3 and then press ENTER

What is the mean and standard deviation? How does it compare to thetheoretic values?

A Classroom Example


1 Press MATH and scroll to PRB2 Select Option 5: randInt(






A Classroom Example








A Classroom Example








A Classroom Example








A Classroom Example






6 Press STAT and in the CALC menu, select 1-Var Stats

7 Syntex: 1-Var Stats L3 and then press ENTER


A Classroom Example








A Classroom Example








§ 4 Binomial Random Variables

An Example

ExampleThe blood type of successive children born to the same parents areindependent and have fixed probabilities that depend on the geneticmake-up of the parents. Each child born to a particular set of parentshas probability .25 of having blood type O. If these parents have 5children, what is the probability that exactly 2 of them have bloodtype O?

Thoughts on how to solve?

We need to factor in both the probability they do have blood type O,the probability they do not, and also where in the sequence of childrenthey were born, since it is a different scenario if it is, say the first twochildren v. the third and fifth.

An Example




An Example




Defining Binomial Random Variables

We must have repeated, identical trials

There must be a fixed number of trials (n)

There can only be two outcomes for a trial (success and failure)

The probability of success is the same for all trials (p)

The probability of failure is the same for all trials (q = 1− p)

The trials must be independent

The random variable X is the number of successes in n trials.This is called the binomial random variable.

















































Defining Binomial Probabilities

Binomial ProbabilityIf X has the binomial distribution with n observations and probabilityp of success on each observation, the possible values of X are0, 1, 2, . . . , n. If k is any one of these values, the binomial probabilityof k successes is

P(X = k) =n Ckpk(1− p)n−k

In our example, we have

P(X = 2) =5 C2(.25)2(.75)3

= .2673

Note: nCk here is referred to as a binomial coefficient.





P(X = 2) =5 C2(.25)2(.75)3

= .2673






P(X = 2) =5 C2(.25)2(.75)3

= .2673






P(X = 2) =5 C2(.25)2(.75)3

= .2673


Examples

ExampleYou roll a standard die 3 times. What is the probability of rollingexactly one 5?

P(X = 1) = 3C1

(16

)1 (56

)2

=75216≈ .3472

Examples


P(X =

1) = 3C1

(16

)1 (56

)2

=75216≈ .3472

Examples


P(X = 1) =

3C1

(16

)1 (56

)2

=75216≈ .3472

Examples


P(X = 1) = 3C1

(16

)1 (56

)2

=75216≈ .3472

Examples


P(X = 1) = 3C1

(16

)1

(56

)2

=75216≈ .3472

Examples


P(X = 1) = 3C1

(16

)1 (56

)2

=75216≈ .3472

Examples


P(X = 1) = 3C1

(16

)1 (56

)2

=75216≈ .3472

Examples

ExampleYou roll a standard die 3 times. What is the probability of rollingexactly three 6’s?

P(X = 3) = 3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples


P(X =

3) = 3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples


P(X = 3) =

3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples


P(X = 3) = 3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples


P(X = 3) = 3C3

(16

)3

(56

)0

=1

216≈ .00463

Examples


P(X = 3) = 3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples


P(X = 3) = 3C3

(16

)3 (56

)0

=1

216≈ .00463

Examples

ExampleIn the old days, there was a probability of .8 of success in any attemptto make a telephone call. Calculate the probability of having 7successful calls in 10 attempts.

P(X = 7) = 10C7(.8)7(.2)3

≈ .20133

Examples


P(X = 7) =

10C7(.8)7(.2)3

≈ .20133

Examples


P(X = 7) = 10C7(.8)7(.2)3

≈ .20133

Examples


P(X = 7) = 10C7(.8)7(.2)3

≈ .20133

Examples

ExampleA (blindfolded) marksman finds that on the average he hits the target4 times out of 5. If he fires 4 shots, what is the probability of

1 more than 2 hits?2 at least 3 misses?

P(X > 2)

P(X = 3) =

4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) =

4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) =

P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X > 2)

P(X = 3) = 4C3(.8)3(.2)1

= .4096

P(X = 4) = 4C4(.8)4(.2)0

= .4096

P(X) = P(X = 3) + P(X = 4)

= .81921

P(X ≤ 1)

P(X =

1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) =

4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X =

0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) =

4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) =

P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

P(X ≤ 1)

P(X = 1) = 4C1(.8)1(.2)3

= .0256

P(X = 0) = 4C0(.8)0(.2)4

= .0016

P(X) = P(X = 1) + P(X = 0)

= .0272

Mean, Variance and Standard Deviation

The FormulasIf X is a binomial random variable with probability p of success forthe fixed n trials, then

µ = np

σ2 = npq

σ =√

npq



µ = np

σ2 = npq

σ =√

npq



µ = np

σ2 = npq

σ =√

npq

Blood Type Example

Back to our family with 5 children. What is the mean and standarddeviation of the binomial distribution?

µ = 5(.25) = 1.25

σ =√

5(.25)(.75) = .9682

What do these values represent?

If we look at a great many couples with similar genetic make-up, weexpect on average for 1.25 children to be of blood type O out of 5children. We expect the actual number of children they have withblood type O to be within .9682 of the mean.

Blood Type Example


µ =

5(.25) = 1.25

σ =√

5(.25)(.75) = .9682



Blood Type Example


µ = 5(.25) = 1.25

σ =√

5(.25)(.75) = .9682



Blood Type Example


µ = 5(.25) = 1.25

σ =√

5(.25)(.75) = .9682



Blood Type Example


µ = 5(.25) = 1.25

σ =√

5(.25)(.75) = .9682



Blood Type Example


µ = 5(.25) = 1.25

σ =√

5(.25)(.75) = .9682



Using the Calculator

We can find binomial probabilities using the TI-Series graphingcalculator as well.

For out last example.

Press 2nd and VARS to get into the DISTR menu

Find binomcdf:Syntex; binomcdf(5,.25,2)This syntax is binomcdf(n,p,k)

You should get the same answer we did earlier.

CDF stands for cumulative distribution function. Anyone want to takea shot at defining this?












Find binomcdf:

Syntex; binomcdf(5,.25,2)This syntax is binomcdf(n,p,k)
























§ 5 The Normal Curve

z-scores

What do z-scores do for us?

z-scores standardize the data. It takes the value of the observation andconverts it to the number of standard deviations from the mean.

We use z-scores when dealing with finding probabilities associatedwith Normal distributions. At this point, we will be given the meanand standard deviation for the population, so our Normal distributionwill be expressed as

X ∼ N(µ, σ)

This is unrealistic, however ...

z-scores




X ∼ N(µ, σ)


z-scores




X ∼ N(µ, σ)


Thoughts on Rescaling and Shifting

The idea behind shifting is that we may need to change values forsome reason - change pounds to kg, inches to feet, etc. When we dothis, the relationship in the graph and between the data elements doesnot change. The graphs would look exactly the same even though thevalues themselves change.

When we rescale, we change the values because all of the data ispacked into too small a region or is spread out too much. Somethinglike this would occur when we, for example, are making a histogramand don’t pay attention to the range and make out scale way too small.We end up with all of the data in a quarter of the space. To repair this,we rewrite the graph on the correct scaled graph to spread the data outover the whole axis. The relationships, range, center, etc. are exactlythe same; the only thing that changes is that the graph is morereadable.

Thoughts on Rescaling and Shifting

The idea behind shifting is that we may need to change values forsome reason - change pounds to kg, inches to feet, etc. When we dothis, the relationship in the graph and between the data elements doesnot change. The graphs would look exactly the same even though thevalues themselves change.

When we rescale, we change the values because all of the data ispacked into too small a region or is spread out too much. Somethinglike this would occur when we, for example, are making a histogramand don’t pay attention to the range and make out scale way too small.We end up with all of the data in a quarter of the space. To repair this,we rewrite the graph on the correct scaled graph to spread the data outover the whole axis. The relationships, range, center, etc. are exactlythe same; the only thing that changes is that the graph is morereadable.

Standardizing

When we standardize, however, there are some changes. The shape ofthe distribution does not change. Since we are usually looking atNormal distributions when we are using z-scores, we will still have aNormal distribution. But the center becomes 0 and the standarddeviation becomes 1. This is the point of standardizing; no matterwhat the data elements are, we can reduce them down to a situationwhere the mean is 0 and the standard deviation is 1.

That is, z ∼ N(0, 1). We call this the Standard Normal distribution.

It should be pointed out that in order to use the Normal curve toapproximate probabilities, we need a roughly symmetric, unimodaldistribution.

Standardizing




Standardizing




The Normal Distribution Curve

µ+ σµ− σ

The Normal curve has the following properties:

1 Area under curve is 1

2 Mean equals the median3 Unimodal4 Bell-shaped5 inflection points at ±1 standard deviation


µ+ σµ− σ


1 Area under curve is 12 Mean equals the median

3 Unimodal4 Bell-shaped5 inflection points at ±1 standard deviation


µ+ σµ− σ


1 Area under curve is 12 Mean equals the median3 Unimodal

4 Bell-shaped5 inflection points at ±1 standard deviation


µ+ σµ− σ


1 Area under curve is 12 Mean equals the median3 Unimodal4 Bell-shaped

5 inflection points at ±1 standard deviation


µ+ σµ− σ


1 Area under curve is 12 Mean equals the median3 Unimodal4 Bell-shaped5 inflection points at ±1 standard deviation

How We Use The Normal Curve

yx

The probability of an observation occurring between x and y, we needa way to find this area.

Using z-Scores

Example

Suppose we are looking at IQ scores again N(100, 15) and we wantedto know the following:

1 The percent of people with IQ scores less than 122.2 The probability of randomly selecting one person and their IQ is

greater than 138.3 The probability of randomly selecting one person and their IQ is

between 92 and 113.

To solve these, we need to find z-scores and use our Normalprobability table.

P(X ≤ 122)

First, we have to find the z-score.

z =122− 100

15=

2215

= 1.47

We are looking for ‘less than’ so the visual is

1.47

P(X ≤ 122)

First, we have to find the z-score.

z =122− 100

15=

2215

= 1.47

We are looking for ‘less than’ so the visual is

1.47

P(X ≤ 122)

1.47

Now we go to our table. The table gives the probability that arandomly selected value lies to the left of the z-score. So here, sincethat is what we are looking for, we only need to look up the value inthe table.

P(X ≤ 122) = P(z ≤ 1.47) = .9292

This probability is larger than 50%, as we would expect.

P(X ≤ 122)

1.47

Now we go to our table. The table gives the probability that arandomly selected value lies to the left of the z-score. So here, sincethat is what we are looking for, we only need to look up the value inthe table.

P(X ≤ 122) = P(z ≤ 1.47) = .9292

This probability is larger than 50%, as we would expect.

P(X ≥ 138)

For the IQ greater than 138, we start in a similar way. We want to findP(X ≥ 138) and since this is not exactly 1, 2 or 3 standard deviationsfrom the mean, we need to use z-scores.

z =138− 100

15=

3815

= 2.53

Visually, we have

2.53

P(X ≥ 138)

For the IQ greater than 138, we start in a similar way. We want to findP(X ≥ 138) and since this is not exactly 1, 2 or 3 standard deviationsfrom the mean, we need to use z-scores.

z =138− 100

15=

3815

= 2.53

Visually, we have

2.53

P(X ≥ 138)

2.53

So when we look the value up in the table, we get ...

P = .9943. So isthis our answer?

We use the fact that the area under the curve is 100%.

P(X ≥ 138) = P(z ≥ 2.53) = 1− .9943 = .0057

P(X ≥ 138)

2.53

So when we look the value up in the table, we get ... P = .9943. So isthis our answer?


P(X ≥ 138) = P(z ≥ 2.53) = 1− .9943 = .0057

P(X ≥ 138)

2.53

So when we look the value up in the table, we get ... P = .9943. So isthis our answer?


P(X ≥ 138) = P(z ≥ 2.53) = 1− .9943 = .0057

P(92 ≤ X ≤ 113)

Where do we start?

We need to find 2 z-scores, one for each IQ score.

z1 =92− 100

15=−815

= −.53

z2 =113− 100

15=

1315

= .87

.87−.53

P(92 ≤ X ≤ 113)

Where do we start?We need to find 2 z-scores, one for each IQ score.

z1 =92− 100

15=−815

= −.53

z2 =113− 100

15=

1315

= .87

.87−.53

P(92 ≤ X ≤ 113)


z1 =92− 100

15=−815

= −.53

z2 =113− 100

15=

1315

= .87

.87−.53

P(92 ≤ X ≤ 113)


z1 =92− 100

15=−815

= −.53

z2 =113− 100

15=

1315

= .87

.87−.53

P(92 ≤ X ≤ 113)

.87−.53

So what we are looking for is the area between the two curves. Howdo we get there?

The values in the table give us the area to the left of the z-score.

The area to the left of z = .87 is .8078 and the area to the left of= −.53 is .2981. What do we do with these?

P(92 ≤ X ≤ 113) = P(−.53 ≤ z ≤ .87) = .8078− .2981 = .5097

P(92 ≤ X ≤ 113)

.87−.53




P(92 ≤ X ≤ 113) = P(−.53 ≤ z ≤ .87) = .8078− .2981 = .5097

P(92 ≤ X ≤ 113)

.87−.53




P(92 ≤ X ≤ 113) = P(−.53 ≤ z ≤ .87) = .8078− .2981 = .5097

P(92 ≤ X ≤ 113)

.87−.53




P(92 ≤ X ≤ 113) = P(−.53 ≤ z ≤ .87) = .8078− .2981 = .5097

WARNING

Note: when finding probabilities,NEVER SUBTRACT Z-SCORES!

For example, in this last problem, if you would have subtracted zscores, we would be looking up z = .87− (−.53) = 1.40 and thatcorresponds to a probability of .9192, nowhere near the correct value.

WARNING

Note: when finding probabilities,NEVER SUBTRACT Z-SCORES!

For example, in this last problem, if you would have subtracted zscores, we would be looking up z = .87− (−.53) = 1.40 and thatcorresponds to a probability of .9192, nowhere near the correct value.

Using The TI

We can find probabilities for problems like this using the calculator aswell.

Press 2nd DISTR and select option 2, which is normalcdf

Enter (lower bound, upper bound, mean, standard deviation)

If you want the probability of less than some value, usesomething extremely small for the lower bound to simulatenegative infinity (use -1000000000)

If you want the probability more than some value then use alarge number to simulate infinity (use 1000000000).

If you want the probability between two values, just use thevalues directly

Using The TI


Press 2nd DISTR and select option 2, which is normalcdfEnter (lower bound, upper bound, mean, standard deviation)




Using The TI






Using The TI






Using The TI






Normal Example

ExampleA car manufacturer found that a certain model was uncomfortable forwomen shorter than 159 cm. Women’s heights are N(161.5, 6.3).Find the percent of women for which the model is uncomfortable.

What do we want?

We want P(X ≤ 159).

z =159− 161.5

6.3= −.40

So,P(X ≤ 159) = P(z ≤ −.40) = .3446

Normal Example


What do we want?


z =159− 161.5

6.3= −.40

So,P(X ≤ 159) = P(z ≤ −.40) = .3446

Normal Example


What do we want?


z =159− 161.5

6.3= −.40

So,P(X ≤ 159) = P(z ≤ −.40) = .3446

Normal Example


What do we want?


z =159− 161.5

6.3= −.40

So,P(X ≤ 159) = P(z ≤ −.40) = .3446

Another Example

ExampleA banker studying customer service needs finds that the number oftimes a person uses an ATM machine in a year are Normallydistributed N(30, 11.4). Find the percent of customers who use ATMsbetween 40 and 50 times per year.

What do we want?

We want P(40 ≤ X ≤ 50).

z1 =40− 30

11.4= .88

z2 =50− 30

11.4= 1.75

Another Example


What do we want?

We want P(40 ≤ X ≤ 50).

z1 =40− 30

11.4= .88

z2 =50− 30

11.4= 1.75

Another Example


What do we want?

We want P(40 ≤ X ≤ 50).

z1 =40− 30

11.4= .88

z2 =50− 30

11.4= 1.75

Another Example


What do we want?

We want P(40 ≤ X ≤ 50).

z1 =40− 30

11.4= .88

z2 =50− 30

11.4= 1.75

Another Example

So, we need

P(40 ≤ X ≤ 50) =

P(.88 ≤ z ≤ 1.75)

= .9599− .8106

= .1493

Another Example

So, we need

P(40 ≤ X ≤ 50) =

P(.88 ≤ z ≤ 1.75)

= .9599− .8106

= .1493

A Different Kind of Normal Example

ExampleAdmissions only considers the top 20% of applicants based on GREscores. If the mean score is 615 with a standard deviation of 107,what is the cut-off score?

Here, we are looking at z-scores in reverse. We know µ and σ and canfind the percent. What we are missing is the z-score.

z

20%


ExampleAdmissions only considers the top 20% of applicants based on GREscores. If the mean score is 615 with a standard deviation of 107,what is the cut-off score?

Here, we are looking at z-scores in reverse. We know µ and σ and canfind the percent. What we are missing is the z-score.

z

20%


Since we are talking about the top 20%, we are also talking about theline that separates this from the bottom 80%. This is the value weneed to look up in the table. When we look up .8000, we will not findthe exact value; we get as close as we can.

80%⇒ z = .84

Then, we go to our z-score formula

z =x− µσ⇒ .84 =

x− 615107

Now we use our algebra skills to solve for x, which is the score weseek.

.84 =x− 615

107.84(107) = x− 615

.84(107) + 615 = x

So, x = 704.88, or more appropriately, x = 705.



80%⇒ z = .84


z =x− µσ⇒ .84 =

x− 615107


.84 =x− 615

107.84(107) = x− 615

.84(107) + 615 = x




80%⇒ z = .84


z =x− µσ⇒ .84 =

x− 615107


.84 =x− 615

107.84(107) = x− 615

.84(107) + 615 = x




80%⇒ z = .84


z =x− µσ⇒ .84 =

x− 615107


.84 =x− 615

107.84(107) = x− 615

.84(107) + 615 = x


Another Example

ExampleA psychologist wants to interview the top 1% of people based on IQscores. If IQ scores are N(100, 15), what is the score that separatesthe top 1% from the bottom 99%?

What do we do first?

We need to first look up the z-score that corresponds to the bottom99%, which is z = 2.33. Then, we solve as in the last example.

2.33 =x− 100

152.33(15) = x− 100

2.33(15) + 100 = x

So, the IQ score we seek is 134.95, or 135.

Another Example




2.33 =x− 100

152.33(15) = x− 100

2.33(15) + 100 = x


Another Example



We need to first look up the z-score that corresponds to the bottom99%, which is

z = 2.33. Then, we solve as in the last example.

2.33 =x− 100

152.33(15) = x− 100

2.33(15) + 100 = x


Another Example




2.33 =x− 100

152.33(15) = x− 100

2.33(15) + 100 = x


Another Example




2.33 =x− 100

152.33(15) = x− 100

2.33(15) + 100 = x


What About One Like This?

ExampleSuppose the age of cars on the road are Normally distributed with amean of 6.2 years and a standard deviation of 1.1 years. What is thecut-off age for the middle 20% of all cars on the road?

Similar idea to the last problem, except we are looking at twodifferent z-scores here.

z2z1

20%


ExampleSuppose the age of cars on the road are Normally distributed with amean of 6.2 years and a standard deviation of 1.1 years. What is thecut-off age for the middle 20% of all cars on the road?

Similar idea to the last problem, except we are looking at twodifferent z-scores here.

z2z1

20%


We need both z-scores so that we can find the corresponding x values.What we have to consider is what percentages correspond to each ofthese z-scores.

Since the band is 20% wide and is in the middle, then there must be10% on each side of the mean. So, z2 is at the line that is 10% belowthe mean and z2 is at the line that is 10% above the mean.

z1: the probability associated with 40% is -.25.z2: the probability associated with 60% is .25.(notice the symmetry ...)So, we have

z1 =x− µσ⇒ −.25 =

x− 6.21.1

Solving gives the lower cut-off point as 5.925.

z2 =x− µσ⇒ .25 =

x− 6.21.1

Solving this gives the upper cut-off point at 6.475.





z1 =x− µσ⇒ −.25 =

x− 6.21.1


z2 =x− µσ⇒ .25 =

x− 6.21.1





z1: the probability associated with 40% is -.25.z2: the probability associated with 60% is .25.(notice the symmetry ...)

So, we have

z1 =x− µσ⇒ −.25 =

x− 6.21.1


z2 =x− µσ⇒ .25 =

x− 6.21.1






z1 =x− µσ⇒ −.25 =

x− 6.21.1


z2 =x− µσ⇒ .25 =

x− 6.21.1






z1 =x− µσ⇒ −.25 =

x− 6.21.1


z2 =x− µσ⇒ .25 =

x− 6.21.1


Seeing the Graph

To do Normal area plots:

Turn off all STAT PLOTS and functions on the Y= screen.

From the WINDOW screen, set XMin=mean-4*standarddeviation and XMax=mean+4*standard deviation.

Set the Ymin = -.4/standard deviation and Ymax = .4/standarddeviation.

Press 2nd DISTR and scroll to the DRAW submenu of theDISTR menu

Select ShadeNormEnter the parameters (lower bound, upper bound, mean, standarddeviation) and hit ENTER

To erase the plot when you are done, press2nd DRAW ENTER

Seeing the Graph








Seeing the Graph








Seeing the Graph








Seeing the Graph






Select ShadeNorm

Enter the parameters (lower bound, upper bound, mean, standarddeviation) and hit ENTER


Seeing the Graph








Seeing the Graph








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