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Chapter 6Normal Probability Distributions

Overview

The Standard Normal Distribution

Normal Distributions: Finding Probabilities

Normal Distributions: Finding Values

The Central Limit Theorem

2

Continuous random variable

Normal distributionCurve is bell shaped

and symmetric

µScore

Overview

x - µ 2

y =

12

e 2

( )

3

Density Curve (or probability density   function) the graph of a continuous probability   distribution

Definitions

1. The total area under the curve must equal 1.

2. Every point on the curve must have a vertical height that is 0 or greater.

4

Because the total area under the density curve is equal to 1,

there is a correspondence between area and probability.

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Heights of Adult Men and Women

Women:µ = 63.6 = 2.5 Men:

µ = 69.0 = 2.8

69.063.6Height (inches)

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DefinitionStandard Normal Deviation

a normal probability distribution that has a

mean of 0 and a standard deviation of 1

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x - 3s x - 2s x - s x x + 2s x + 3sx + s

68% within1 standard deviation

34% 34%

95% within 2 standard deviations

99.7% of data are within 3 standard deviations of the mean

0.1% 0.1%

2.4% 2.4%

13.5% 13.5%

The Empirical RuleStandard Normal Distribution: µ = 0 and = 1

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Probability of Half of a Distribution

0

0.5

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P(a < z < b) denotes the probability that the z score is between a and

b

P(z > a) denotes the probability that the z score is greater than

a

P (z < a) denotes the probability that the z score is less than a

Notation

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Finding the Area to the Right of z = 1.27

0

This area is 0.1020

z = 1.27

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To find:

z Scorethe distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of theTable

Area

the region under the curve; refer to the values in the body of the Table

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Table E Standard Normal Distribution

µ = 0 = 1

0 x

z

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DefinitionStandard Normal Deviation

a normal probability distribution that has a

mean of 0 and a standard deviation of 1

0 1 2 3-1-2-3 0 z = 1.58

Area = 0.3413 Area found in

Table A2

0.4429

Score (z )

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Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.

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Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.

The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.4429.

0 1.58

Area = 0.4429

P ( 0 < x < 1.58 ) = 0.4429

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Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0 degrees.

The probability that the chosen thermometer will measure freezing water between -2.43 and 0 degrees is 0.4925.

-2.43 0

Area = 0.4925P ( -2.43 < x < 0 ) = 0.4925

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Finding the Area Between z = 1.20 and z = 2.30

0

Area A is 0.1044

z = 1.20

Az = 2.30

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Other Normal Distributions

If 0 or 1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution.

x - µz =

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Converting to Standard Normal Distribution

x(a)

P

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Converting to Standard Normal Distribution

x 0 z

x - z =

(a) (b)

P P

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Probability of Weight between 143 pounds and 201 pounds

143 201 z

0 2.00

z = 201 - 143

29 = 2.00

s =29x = 143

Weight

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Probability of Weight between 143 pounds and 201 pounds

143 201 z

0 2.00

x = 143

OR - 47.72% of women have weights between 143 lb and 201 lb.

Weight

s =29

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Finding a probability when given a z-score using the TI83

• 2nd

• Distributions

• 2:normalcdf

• (lower bound, upper bound, mean, s.d.)

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Finding a z - score when given a probabilityUsing the Table

1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline.

2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table E and identify the corresponding z score.

3. If the z score is positioned to the left of the centerline, make it a negative.

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0

0.45

1.645

0.50

95% 5%

5% or 0.05

Finding z Scores when Given Probabilities

Finding the 95th Percentile(z score will be positive)

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0

0.40-1.28

0.10

90%

Finding the 10th Percentile

Bottom 10%

10%

Finding z Scores when Given Probabilities

(z score will be negative)

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Finding z Scores when Given ProbabilitiesUsing the TI83

• 2nd

• Distributions

• 3:Invnorm

• (%,mean,s.d)

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1. Don’t confuse z scores and areas.

    Z scores are distances along the horizontal  scale, but areas are regions under the  normal curve.

Table A-2 lists z scores in the left column and across the top row, but  areas are found in the body of the table.

2. Choose the correct (right/left) side of the graph.

3. A z score must be negative whenever it is

     located to the left of the centerline of 0.

Cautions to keep in mind

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0

0.45

1.645

0.50

95% 5%

5% or 0.05

Finding z Scores when Given Probabilities

Finding the 95th Percentile(z score will be positive)

30

0

0.40-1.28

0.10

90%

Finding the 10th Percentile

Bottom 10%

10%

Finding z Scores when Given Probabilities

(z score will be negative)

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Procedure for Finding Values Using the Table and By Formula

1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought.

2. Use Table E to find the z score corresponding to the region bounded by x and the centerline of 0. Cautions:

    Refer to the BODY of Table E to find the closest area, then identify     the corresponding z score.

    Make the z score negative if it is located to the left of the centerline.

3. Using the Formula, enter the values for µ, , and the z score found in step 2, then solve for x.

x = µ + (z • )

4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem.

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143

40%

x = ?

50%

90%10%

Finding P10 for Weights of Women

Weight

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143

0.40

x = 106

0.500.10

Finding P10 for Weights of Women

0-1.28

The weight of 106 lb (rounded) separates the lowest 10% from the highest 90%.

Weight

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143

0.40

x = 180

0.500.10

Forgot to make z score negative???

01.28

UNREASONABLE ANSWER!

Weight

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REMEMBER!

Make the z score negative if the value is located to the left (below) the mean. Otherwise, the z score will be positive.

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Sampling Distribution of the mean the probability distribution of sample means, with all samples     

having the same sample size n.

Definition

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Central Limit Theorem

1. The random variable x has a distribution (which may or may not be normal) with mean µ and

standard deviation .

2. Samples all of the same size n are randomly

selected from the population of x values.

Given:

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Central Limit Theorem

1. The distribution of sample x will, as the sample size increases, approach a normal distribution.

2. The mean of the sample means will be the population mean µ.

3. The standard deviation of the sample means

will approach

n

Conclusions:

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Practical Rules Commonly Used:

1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better

as the sample size n becomes larger.

2. If the original population is itself normally distributed, then the sample means will be normally distributed for

any sample size n (not just the values of n larger than 30).

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Notationthe mean of the sample means

the standard deviation of sample mean

(often called standard error of the mean)

µx = µ

x = n

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0

10

20

0 1 2 3 4 5 6 7 8 9

Distribution of 200 digits

Distribution of 200 digits from Social Security Numbers

(Last 4 digits from 50 students)F

req

uen

cy

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1

5

9

5

9

4

7

9

5

7

8

3

8

1

3

2

7

1

3

6

3

8

2

3

6

1

5

3

4

6

4

6

8

5

5

2

6

4

9

4.75

4.25

8.25

3.25

5.00

3.50

5.25

4.75

5.00

2

6

2

2

5

0

2

7

8

5

3

7

7

3

4

4

4

5

1

3

6

7

3

7

3

3

8

3

7

6

2

6

1

9

5

7

8

6

4

0

7

4.00

5.25

4.25

4.50

4.75

3.75

5.25

3.75

4.50

6.00

SSN digits x

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0

10

0 1 2 3 4 5 6 7 8 9

Distribution of 50 Sample Means for 50 Students

Fre

qu

ency

5

15

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As the sample size increases, the sampling distribution of sample means approaches a

normal distribution.

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Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability

that her weight is greater than 150 lb.b.) if 36 different women are randomly selected, find the

probability that their mean weight is greater than 150 lb.

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Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, the probability that

her weight is greater than 150 lb. is 0.4052.

= 143 150= 29

0 0.24

0.0948

0.5 - 0.0948 = 0.4052

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Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, the probability that their mean weight is greater than 150 lb is 0.0735.

x = 143 150x= 4.83333

0 1.45

0.4265

0.5 - 0.4265 = 0.0735

z = 150-143 = 1.45 29

36

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Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,

a.) if one woman is randomly selected, find the probability    that her weight is greater than 150 lb.

P(x > 150) = 0.4052

b.) if 36 different women are randomly selected, their mean    weight is greater than 150 lb.

P(x > 150) = 0.0735

It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean.

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Finding a z - score for a group Using TI83

• 2nd

• Distributions

• 2:normalcdf

• (lower bound, upper bound, mean, s.e)– Replace s.d. with standard error