Chapter 5 The Standard Deviation as a Ruler and the Normal...

Chapter 5 The Standard Deviation as aRuler and the Normal Model

z-scores

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-Score Formula

z =y− y

s=

x− µσ

z-scores



X ∼ N(µ, σ)


z-Score Formula

z =y− y

s=

x− µσ

z-scores



X ∼ N(µ, σ)


z-Score Formula

z =y− y

s=

x− µσ

z-Score Examples

ExampleFind the z-score for x = 15 if µ = 12 and σ = 2.3.

z =15− 12

2.3=

32.3

= 1.30

Note: We always express z-scores with 2 decimal places because ofthe table we use to look up probabilities.

ExampleSuppose the average number of home runs for a player on the RedSox roster last season was 22 with a standard deviation of 4.2. DustinPedroia hit 15 home runs last season. How many standard deviationsis his total from the mean?

z =15− 22

4.2=−74.2

= −1.67

z-Score Examples


z =15− 12

2.3=

32.3

= 1.30



z =15− 22

4.2=−74.2

= −1.67

z-Score Examples


z =15− 12

2.3=

32.3

= 1.30



z =15− 22

4.2=−74.2

= −1.67

z-Score Examples


z =15− 12

2.3=

32.3

= 1.30



z =15− 22

4.2=−74.2

= −1.67

Who Is Better?

ExampleSuppose we had 2 students. Student A got a 90 and a 75 on twoexams. Student B got a 85 and a 88 on the same two exams. If examA had a mean of 82 with a standard deviation of 10 and exam 2 had amean of 70 with a standard deviation of 12, which student performedbetter?

First instinct?

You want to say that the student with the higher average performedbetter, but that neglects how the class did on each.

Student A: z1 = .8, z2 = .42. So, the total deviation is 1.22.Student B: z1 = .3, z2 = 1.5. So, the total deviation is 1.80.So, student B actually did better even though student A had a slightlyhigher average.

Who Is Better?


First instinct?



Who Is Better?


First instinct?



Who Is Better?


First instinct?


Student A: z1 = .8, z2 = .42. So, the total deviation is 1.22.

Student B: z1 = .3, z2 = 1.5. So, the total deviation is 1.80.So, student B actually did better even though student A had a slightlyhigher average.

Who Is Better?


First instinct?


Student A: z1 = .8, z2 = .42. So, the total deviation is 1.22.Student B: z1 = .3, z2 = 1.5. So, the total deviation is 1.80.

So, student B actually did better even though student A had a slightlyhigher average.

Who Is Better?


First instinct?



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.

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.

68-95-99.7 Rule

When we are dealing with exactly 1, 2 or 3 standard deviations fromthe mean, we know the percentages of observations that lie in a givenrange. This approximation tool is known as the Empirical Rule, or the68− 95− 99.7 Rule.

We do have to be careful in making sure we have a symmetric,unimodal distribution if we are not told that the distribution isNormal. The probabilities associated here are only valid for Normaldistributions.

68-95-99.7 Rule

When we are dealing with exactly 1, 2 or 3 standard deviations fromthe mean, we know the percentages of observations that lie in a givenrange. This approximation tool is known as the Empirical Rule, or the68− 95− 99.7 Rule.

We do have to be careful in making sure we have a symmetric,unimodal distribution if we are not told that the distribution isNormal. The probabilities associated here are only valid for Normaldistributions.

68-95-99.7 Rule

We know that in a Normal distribution with mean µ and standarddeviation σ, 68% of all observations lie within one standard deviationof the mean.

µ+ σµ− σ

68%

34% 34%

68-95-99.7 Rule

We know that in a Normal distribution with mean µ and standarddeviation σ that 95% of all observations lie with 2 standard deviations.

µ+ 2σµ− 2σ

95%

47.5% 47.5%

68-95-99.7 Rule

Finally, we know that in a Normal distribution with mean µ andstandard deviation σ that 99.7% of all observations lie within 3standard deviations of the mean.

µ+ 3σµ− 3σ

99.7%

49.85% 49.85%

Empirical Rule Example

Example

IQ scores are Normally distributed with X ∼ N(100, 15). Find1 P(X ≥ 115)2 P(X ≤ 130)3 P(85 ≤ X ≤ 115)4 P(70 ≤ X ≤ 115)5 P(115 ≤ X ≤ 130)

P(X ≥ 115)

115

What do we know?

P(X ≥ 115)

115

50% 34%

What part do we need?

P(X ≥ 115)

115

50% 34%

16%

P(X ≥ 115) = 16%

P(X ≥ 115)

115

50% 34%

16%

P(X ≥ 115) = 16%

P(X ≤ 130)

130

What do we know?

P(X ≤ 130)

130

50% 47.5%

P(X ≤ 130) = 97.5%

P(X ≤ 130)

130

50% 47.5%

P(X ≤ 130) = 97.5%

P(85 ≤ X ≤ 115)

11585

This is a symmetric interval, so ...

P(85 ≤ X ≤ 115)

11585

This is a symmetric interval, so ...

P(85 ≤ X ≤ 115)

11585

68%

P(85 ≤ X ≤ 115) = 68%

P(70 ≤ X ≤ 115)

11570

What do we know?

P(70 ≤ X ≤ 115)

11570

47.5% 34%

P(70 ≤ X ≤ 115) = 81.5%

P(70 ≤ X ≤ 115)

11570

47.5% 34%

P(70 ≤ X ≤ 115) = 81.5%

P(115 ≤ X ≤ 130)

130115

What do we know?

P(115 ≤ X ≤ 130)

130115

34%

47.5%

P(115 ≤ X ≤ 130) = 13.5%

P(115 ≤ X ≤ 130)

130115

34%

47.5%

P(115 ≤ X ≤ 130) = 13.5%

Back to z-Scores

What do we do if we want to work with a number of standarddeviations from the mean that is not exactly 1, 2, or 3 away? Back toour 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

Calculator answer: normalcdf(-100000000,159,161.5,6.3) returns.3457

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


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

Calculator: P=.1505

Another Example

So, we need

P(40 ≤ X ≤ 50) =

P(.88 ≤ z ≤ 1.75)

= .9599− .8106

= .1493

Calculator: P=.1505

Another Example

So, we need

P(40 ≤ X ≤ 50) =

P(.88 ≤ z ≤ 1.75)

= .9599− .8106

= .1493

Calculator: P=.1505

A Different Kind of Normal Example

ExampleAdmissions only considers to 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 to 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 = .85

Then, we go to our z-score formula

z =x− µσ⇒ .85 =

x− 615107

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

.85 =x− 615

107.85(107) = x− 615

.85(107) + 615 = x

So, x = 705.95, or more appropriately, x = 706.



80%⇒ z = .85


z =x− µσ⇒ .85 =

x− 615107


.85 =x− 615

107.85(107) = x− 615

.85(107) + 615 = x




80%⇒ z = .85


z =x− µσ⇒ .85 =

x− 615107


.85 =x− 615

107.85(107) = x− 615

.85(107) + 615 = x




80%⇒ z = .85


z =x− µσ⇒ .85 =

x− 615107


.85 =x− 615

107.85(107) = x− 615

.85(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


Percentiles

These basically work the same way, except the answer is always thepercent less than the given value. That is, Px is the percent ofobservations that are less than x.

From the previous examples, the only one that we could directlytranslate to percentiles is P(x ≤ 130). We are 2 standard deviationsabove, so this can be expressed as P97.5.

Percentiles

These basically work the same way, except the answer is always thepercent less than the given value. That is, Px is the percent ofobservations that are less than x.

From the previous examples, the only one that we could directlytranslate to percentiles is P(x ≤ 130). We are 2 standard deviationsabove, so this can be expressed as P97.5.

ExampleFor our IQ example, find the value of

1 P50 =

100 (median)2 P99.85 = 145

What would happen if we tried to do this the long way?

z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475

The value is close, and in this situation we would have roundedto 145, but the table is more accurate than the Empirical Rule.

3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05


1 P50 = 100 (median)2 P99.85 =

145What would happen if we tried to do this the long way?

z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475


3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05


1 P50 = 100 (median)2 P99.85 = 145


z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475


3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05


1 P50 = 100 (median)2 P99.85 = 145


z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475


3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05


1 P50 = 100 (median)2 P99.85 = 145


z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475


3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05


1 P50 = 100 (median)2 P99.85 = 145


z = 2.965⇒ 2.965 =x− 100

15⇒ x = 144.475


3 P37

z = −.33⇒ −.33 =x− 100

15⇒ x = 95.05

More Percentiles

Example

Find the percentile of a student who ranks 30th in their class of 200students.

If a student is 30th in the class, then there are 170 below them.

170200

= .85

So, the student is in the 85th percentile.

More Percentiles

Example



170200

= .85


More Percentiles

Example



170200

= .85


One More Percentile Problem

Example

What is the class rank of a student in the 45th percentile of a class of200?

.45× 200 = 90

So, there are 90 students below this student in question.

Therefore, they are 110th in the class.


Example


.45× 200 = 90




Example


.45× 200 = 90



Changing the Median

Changing the median does not have the same major effect as changingthe mean.

ExampleSuppose we have a 5-number summary for salaries (in thousands ofdollars) at a small company as follows:

Minimum 28Q1 37M 45Q3 48

Maximum 65

Suppose the company suddenly had a cash windfall and the ownerdecided to give everyone a $5000 raise. What effect does this have onthese 5 values?

IQR? 48− 37 = 11

Changing the Median




Maximum 65


IQR? 48− 37 = 11

Changing the Median




Maximum 65


IQR?

48− 37 = 11

Changing the Median




Maximum 65


IQR? 48− 37 = 11

Changing the Median

If we give a $5000 raise across the board, it will increase each valueby exactly the same amount.

Old Salary New SalaryMinimum 28 33

Q1 37 42M 45 50Q3 48 53

Maximum 65 70

The values all changed, but the IQR is still 53− 42 = 11. Why?

Changing the Median

If we give a $5000 raise across the board, it will increase each valueby exactly the same amount.

Old Salary New SalaryMinimum 28 33

Q1 37 42M 45 50Q3 48 53

Maximum 65 70

The values all changed, but the IQR is still 53− 42 = 11. Why?

Date post:	18-Oct-2020
Category:	Documents
Upload:	others
View:	2 times
Download:	0 times

Chapter 5 The Standard Deviation as a Ruler and the Normal...

Documents