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Chapter 2

The Normal Distribution

Lesson 2-1

Density Curve

Review

Graph the data

Calculate a numerical summary of the data

Describe the shape, center, spread and

outliers of the data

Histogram with Curve

Area Under the Curve

area = proportion (or percent) = probability

Density Curve

Is always on or above the horizontal

axis, and

The area underneath is exactly 1.

Median and Mean of a

Density curve

Median is the equal-areas point, the point that divides the area under the curve in half

Mean is the balance point, at which the curve would balance if made of solid material.

Median and mean are the same for a symmetric density curve

Mean of a skewed curve is pulled away from the median in the direction of the long tail.

Mean and Median

Symmetric Skewed to the Right

Notation

Actual Observation

Mean x

Standard Deviation s

Idealized Distribution

Mean

Standard Deviation

Example – Page 83, #2.2

The density curve of a uniformed distribution. The curve

takes the constant value 1 over the interval from 0 to 1

and is zero outside the range of values. This means

that the data describe by this distribution takes values that

are uniformly spread between 0 and 1. Use the areas

under this density curve to answer the following questions

0 1


0 1

A) Why is the total area under this curve equal to 1?

1

1

The area under the curve is a rectangle with

height 1 and width 1.


0 1

B) What percent of the observation lie above 0.8?

(1 0.80) 0.20

(0.20)(1) 0.20 20%A.20


0 1

C) What percent of the observation lies below 0.60?

.60

(.60 0) .60 60%


0 1

.75.25

D) What percent of the observation lie between 0.25 and

0.75

(.75 .25) .50 50%


0 1

E) What is the mean μ of this distribution?

Mean = ½ or 0.50, the “balance point” of the density

curve.

Example, Page 84, #2.4

The following figure displays three density curves, each

with three points indicated. At which of these points on

each curve do the mean and median fall?


Median B

Mean CMedian A

Mean A

Median B

Mean A

Example – Density Curve

I’m thinking about a density curve that consists of a straight

line segment from the point (0, 2/3) to the point (1,4/3) in

the x-y plane.

A). Sketch this density curve

13

23

1

43

12

14

341


B). What percent of the observation lie below ½?

13

23

1

43

12

14

341

Area of the Rectangle

1 2 12 3 3

A bh

Area of the Triangle

1 12 3 1

122 2

bhA

51 1 41.67%3 12 12TA


C). What percent of the observation lie below 1?

13

23

1

43

12

14

341

1 100%TA


C). What percent of the observation lie between ½ and 1?

13

23

1

43

12

14

341

1 1 7 58.3%2 12 12TA

Area of the Rectangle

1 112 2

A bh

Area of the Triangle

1 13 2 1

122 2

bhA

Lesson 2-1

Normal Distribution

Normal Distribution

Symmetric, single peak

and bell shape

Tails fall quickly – so we

do no expect outliers

Mean and median lie

together at the peak in the

center of the curve

Standard deviation

determines the shape of

the curve

Standard Devotion

Inflection Point Inflection Point

Why the Normal Distribution?

Are good descriptions for some distribution

of real data

SAT Score

Characteristics of a biological population

Tossing coins many times

Works well for other roughly symmetric

distributions

68 – 95 – 99.7 Rule


The distribution of heights of adult American men is

approximately normal with mean 69 inches and

standard deviation 2.5 inches. Draw a normal curve

on which the mean and standard deviation are

correctly located.

69

2.5


69

2.5

69 71.566.5

2.5


Scores on the Wechsler Adult Intelligence Scale (WAIS,

a standard “IQ test”) for 20 to 34 age group are

approximately normally distributed with μ = 110 and

σ = 25. Use the 68 – 95 – 99.7 rule to answer these

questions.


A)About what percent of people in this age group

have scores about 110?

13585 110

about 68%


B) About what percent have scores above 160?

135

11

0

about 2.35%

160


C) In what range do the middle 95% of IQ scores lie?

135

11

0

between 60 to 160

or

160

85

60

110 50


135

11

0

160

85

60

Wechsler Adult Intelligence Scale (WAIS) scores for

young adults are N(110, 25).


135

11

0

160

85

60

A. If someone’s score were reported as the 16th percentile

about what score would that individual have.

approximately 85


135

11

0

160

85

60

B. If someone’s score were reported as the 84th percentile

and 97.5th percentile about what score would that

individual have.

approximately 135 and

160

Lesson 2-2

The Standard Normal Distribution

How do I find the area under

the curve?

Use calculus (find the integral). Too

advance for the class.

Use a series of tables to find areas for

every possible mean and standard

deviation. Infinitely many tables.

Z-Scores

Z-Scores

Assume that your variable is normally

distributed.

Use your mean and standard deviation to

convert your data into z-scores such that

the new distribution has a mean of 0 and a

standard deviation of 1

Standard Normal Distribution

The 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 and standard deviation σ then the standardized

variable is

xz

Standard Normal Distribution


Three landmarks of baseball achievement are Ty Cobb’s

batting average of .420 in 1911, Ted Williams’s .406 in

1941, and George Brett’s 0.390 in 1980. These batting

averages cannot be compared directly because the

distribution of major league batting averages has

changed over the years. The distributions are quite

symmetric and (except for outliers such as Cobb, Williams

and Brett) reasonably normal. While the mean batting

average has been held roughly constant by rule changes

and the balance between hitting and pitching, the

standard deviation has dropped over time. Here are

the facts:


Decade Mean Std. Dev.

1910s .266 .0371

1940s .267 .0326

1970s .261 .0317

Compute the standardized batting averages for Cobb,

Williams, and Brett to compare how far each stood above

his peers.


Decade Mean Std. Dev.

1910s .266 .0371

1940s .267 .0326

1970s .261 .0317

xz

Cobb:

Williams

Brett.420 .2664.15

.0371z

.406 .2674.26

.0326z

.390 .2614.07

.0317z

Williams’s z-score is the

highest.


Use Table A to find the proportion of observation from a

Standard normal distribution that falls in each of the

following regions. In each case sketch a standard normal

Curve and shade the area representing the region

2.25

2.25

1.77

2.25 1.77

z

z

z

z

a)

b)

c)

d)


2.25za)

-2.25 0


2.25za)

The area to the left

of -2.25 is 0.0122

-2.25 0


2.25zb)

Find the area to the left of -2.25

2.25 0.0122z

1 0.0122 0.9878

Area greater than -2.25 = 1 – area below -2.25

-2.25 0


1.77zc)

1 1.77A z

1.770


1.77zc)

1 1.77A z

1 0.9616

0.0384

Area greater than 1.77 = 1 – area below 1.77

1.770


2.25 1.77zd)

0-2.25 1.77

1.77 2.25A z z

0.9616 0.0

0.

2

4

12

949

A

Area between -2.25 and 1.77 = area below 1.77 – area below -2.25

TI – 83 (Area)

2.25z

2nd Vars

TI – 83 (Area)

2.25z 2.25 1.77z


Use Table A to find the value of z of a standard normal

Variable that satisfies each of the following conditions.

Use the value of z from Table A that comes closet to

satisfying the condition. In each case sketch a standard

curve with your value of z marked on the axis.

a) The point z with 25% of the observation falling below it.

b) The point z with 40% of the observation falling above it.



0.25A

?z



0.67z

0.25A


0.40A

?z


1 0.40 .60

Find the area below 0.40


0.40A

0.25z


1 0.40 .60

Find the area below 0.40

TI – 83 (Z – Value)

2nd Vars

25% below

TI – 83 (Z – Value)

40% Above


Scores on the Wechsler Adult Intelligence Scale (a

standard IQ test) for the 20 to 34 age group are

approximately normally distributed with μ = 110 and

σ = 25

A. What percent of people age 20 to 34 have IQ

scores above 100?

B. What percent have scores above 150?

C. How high an IQ score is needed to be the highest 25%?


μ = 110 and σ = 25

A. What percent of people age 20 to 34 have IQ

scores about 100?

110100

100 1100.40

25

x μz

σ

Z-0.40 065.5%


μ = 110 andσ = 25

B. What percent have scores above 150?

110Z1.600

150

150 1101.6

25

x μz

σ

5.5%


μ = 110 andσ = 25

C. How high an IQ score is needed to be the highest 25%?

110 ?

A = 0.25

1100.68

25

127

x μz

σ

x

x


The annual rate of return on stock indexes (which

combine many individual stocks) is approximately

normal. Since 1945, the Standard & Poor’s 500 Index

has had a mean yearly return of 12%, with a standard

deviation of 16.5%. Take this normal distribution to be

the distribution of yearly returns over a long period.

A. In what range do the middle 95% of all yearly returns lie?

B. The market is down for the year if the return on the index is

less than zero. In what proportion of years is the market

down?


μ = 12% and σ = 16.5%

A. In what range do the middle 95% of all yearly returns

lie?

12% 45%-21%

12% 2(16.5%)

-21% to 45%


μ = 12% and σ = 16.5%

B. The market is down for the year if the return on the

index is less than zero. In what proportion of years is

the market down?

12%0%

Z0

0 120.727

16.5

x μz

σ

-0.730.2335


μ = 12% and σ = 16.5%

C. In what proportion of years does the index gain 25%

or more?

12 25

0 Z

25 120.7878

16.5

x μz

σ

0.790.2154

Lesson 2-2

Accessing Normality

Assessing Normality

Construct a frequency histogram or a

stemplot or boxplot

Check to see if graph is approximately

bell-shape and symmetric about the mean

Construct a normal probability plot

Use TI to see if plotted points lie close to a

straight line


Repeated careful measurements of the same physical

quantity often have a distribution that is close to normal.

Here are Henry Cavendish’s 29 measurements of the

density of the earth, made in 1798. (The data gives the

density of the earth as multiple of the density of water.)

5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65

5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39

5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85


5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65

5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39

5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85

A. Construct a stemplot to show that the data are

reasonably symmetric


A. Construct a stemplot

to show that the data are

reasonably symmetric

48 8

49

50 7

51 0

52 6 7 9 9

53 0 4 4 6 9

54 2 4 6 7

55 0 3 5 7 8

56 1 2 3 5 8

57 5 9

58 5

4.88

48|8


B. Now check how closely

they follow the 68-95-99.7 rule.

Find and s, then count the

number of observations that

fall between , between

, between .

Compare the percents of the

29 observations in each of

these intervals with the

68-95-99.7 rule.

48 8

49

50 7

51 0

52 6 7 9 9

53 0 4 4 6 9

54 2 4 6 7

55 0 3 5 7 8

56 1 2 3 5 8

57 5 9

58 5

4.88

48|8

x

x s

2x s 3x s


48 8

49

50 7

51 0

52 6 7 9 9

53 0 4 4 6 9

54 2 4 6 7

55 0 3 5 7 8

56 1 2 3 5 8

57 5 9

58 5

4.88

48|8

5.4

5

5.8

9

5.6

7

5.2

3

5.0

1

x x s 2x sx s2x s

11

37.9%

4

13.8%

0

0%

1

0%

2

6.9%

11

37.9%


C. Use your calculator to construct a normal probability

plot for Cavendish’s density of the earth data, write a

brief statement about the normality of the data. Does

the normal probability reinforce your findings in (a).


The linearity of the normal probability indicates an

approximately normal distribution.

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