Chapter 6 The Normal Distribution 1 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission...

Chapter 6The Normal Distribution

1Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Outline

6The Normal Distribution

6-1 Normal Distributions

6-2 Applications of the Normal Distribution

6-3 The Central Limit Theorem

6-4 The Normal Approximation to the BinomialDistribution

Objectives


1 Identify distributions as symmetric or skewed.

2 Identify the properties of a normal distribution.

3 Find the area under the standard normal distribution, given various z values.

4 Find probabilities for a normally distributed variable by transforming it into a standard normal variable.

5 Find specific data values for given percentages, using the standard normal distribution.

Objectives


6 Use the central limit theorem to solve problems involving sample means for large samples.

7 Use the normal approximation to compute probabilities for a binomial variable.

6.1 Normal Distributions• Many continuous variables have distributions that

are bell-shaped and are called approximately approximately normally distributed variablesnormally distributed variables.

• The theoretical curve, called the bell curvebell curve or the Gaussian distributionGaussian distribution, can be used to study many variables that are not normally distributed but are approximately normal.

Bluman, Chapter 6 5

Normal Distributions

2 2( ) (2 )

2

Xe

y

The mathematical equation for the normal distribution is:

2.718

3.14

population mean

population standard deviation

where

e

Bluman, Chapter 6 6

Normal Distributions• The shape and position of the normal distribution

curve depend on two parameters, the meanmean and the standard deviationstandard deviation.

• Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable’s mean and standard deviation.

Bluman, Chapter 6 7

Normal Distributions

Bluman, Chapter 6 8

Normal Distribution Properties• The normal distribution curve is bell-shaped.• The mean, median, and mode are equal and located

at the center of the distribution.• The normal distribution curve is unimodalunimodal (i.e., it

has only one mode).• The curve is symmetrical about the mean, which is

equivalent to saying that its shape is the same on both sides of a vertical line passing through the center.

Bluman, Chapter 6 9

Normal Distribution Properties• The curve is continuous—i.e., there are no gaps or

holes. For each value of X, there is a corresponding value of Y.

• The curve never touches the x-axis. Theoretically, no matter how far in either direction the curve extends, it never meets the x-axis—but it gets increasingly closer.

Bluman, Chapter 6 10

Normal Distribution Properties• The total area under the normal distribution curve is

equal to 1.00 or 100%. • The area under the normal curve that lies within

– one standard deviation of the mean is approximately 0.68 (68%).

– two standard deviations of the mean is approximately 0.95 (95%).

– three standard deviations of the mean is approximately 0.997 ( 99.7%).


Normal Distribution Properties


Standard Normal Distribution• Since each normally distributed variable has its own

mean and standard deviation, the shape and location of these curves will vary. In practical applications, one would have to have a table of areas under the curve for each variable. To simplify this, statisticians use the standard normal distribution.

• The standard normal distributionstandard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.


z value (Standard Value)The z value is the number of standard deviations that a particular X value is away from the mean. The formula for finding the z value is:

value mean

standard deviationz

Xz


Area under the Standard Normal Distribution Curve

1. To the left of any z value:

Look up the z value in the table and use the area given.



2. To the right of any z value:

Look up the z value and subtract the area from 1.



3. Between two z values:

Look up both z values and subtract the corresponding areas.


Chapter 6Normal Distributions

Section 6-1Example 6-1

Page #304


Example 6-1: Area under the Curve

Find the area to the left of z = 2.06.

The value in the 2.0 row and the 0.06 column of Table E is 0.9803. The area is 0.9803.




Page #304



Find the area to the right of z = –1.19.

The value in the –1.1 row and the .09 column of Table E is 0.1170. The area is 1 – 0.1170 = 0.8830.




Page #305



Find the area between z = +1.68 and z = –1.37.

The values for z = +1.68 is 0.9535 and for

z = –1.37 is 0.0853. The area is 0.9535 – 0.0853 = 0.8682.




Page #306


Example 6-4: Probability

a. Find the probability: P(0 < z < 2.32)

The values for z = 2.32 is 0.9898 and for z = 0 is 0.5000. The probability is 0.9898 – 0.5000 = 0.4898.


0.750

Find the area under the normal distribution curve.

Between z = 0 and z = 0.75

area = 0.2734

Exercise #7

0 0.79 1.28

0.3997 – 0.2852 = 0.1145

Find the area under the normal distribution curve.between z = 0.79 and z = 1.28. The area is found by looking up the values 0.79 and 1.28 in table E and subtracting the areas as shown in Block 3 of the Procedure Table.

Exercise #15

0 2.83

Find probabilities, using the standard normal distributionP(z > 2.83).The area is found by looking up z = 2.83 in Table E then subtracting the area from 0.5 as shown in Block 2 of the Procedure Table.

0.5 – 0.4977 = 0.0023

Exercise #31

Find the z value that corresponds to the given area.

0.8962 – 0.5 = 0.3962

0z

0.8962

Exercise #45

Find the z value that corresponds to the given area.

Using Table E, find the area 0.3962 and read the correct z value [corresponding to this area] to get 1.26. Finally, because the z value lies to the left of 0, z = – 1.26.

0z

0.8962



Page #307


Example 6-5: ProbabilityFind the z value such that the area under the standard normal distribution curve between 0 and the z value is 0.2123.

Add 0.5000 to 0.2123 to get the cumulative area of 0.7123. Then look for that value inside Table E.


Example 6-5: Probability

The z value is 0.56.

Add .5000 to .2123 to get the cumulative area of .7123. Then look for that value inside Table E.


6.2 Applications of the Normal Distributions

• The standard normal distribution curve can be used to solve a wide variety of practical problems. The only requirement is that the variable be normally or approximately normally distributed.

• For all the problems presented in this chapter, you can assume that the variable is normally or approximately normally distributed.


Applications of the Normal Distributions

• To solve problems by using the standard normal distribution, transform the original variable to a standard normal distribution variable by using the z value formula.

• This formula transforms the values of the variable into standard units or z values. Once the variable is transformed, then the Procedure Table (Sec. 6.1) and Table E in Appendix C can be used to solve problems.




Page #315


Example 6-6: Summer SpendingA survey found that women spend on average $146.21 on beauty products during the summer months.

Assume the standard deviation is $29.44.

Find the percentage of women who spend less than $160.00. Assume the variable is normally distributed.


Example 6-6: Summer SpendingStep 1: Draw the normal distribution curve.


Example 6-6: Summer SpendingStep 2: Find the z value corresponding to $160.00.

Table E gives us an area of .6808.

68% of women spend less than $160.

160.00 146.210.47

29.44

Xz

Step 3: Find the area to the left of z = 0.47.



Section 6-2Example 6-7a

Page #315


Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at random, find the probability of its generating between 27 and 31 pounds per month. Assume the variable is approximately normally distributed.

Step 1: Draw the normal distribution curve.

Example 6-7a: Newspaper Recycling


Example 6-7a: Newspaper RecyclingStep 2: Find z values corresponding to 27 and 31.

Table E gives us an area of 0.9332 – 0.3085 = 0.6247. The probability is 62%.

27 280.5

2

z

Step 3: Find the area between z = -0.5 and z = 1.5.

31 281.5

2

z




Page #317


Americans consume an average of 1.64 cups of coffee per day. Assume the variable is approximately normally distributed with a standard deviation of 0.24 cup.

If 500 individuals are selected, approximately how many will drink less than 1 cup of coffee per day?

Example 6-8: Coffee Consumption



Example 6-8: Coffee Consumption


Step 4: To find how many people drank less than 1 cup of coffee, multiply the sample size 500 by 0.0038 to get

1.9.

Since we are asking about people, round the answer to 2 people. Hence, approximately 2 people will drink less

than 1 cup of coffee a day.

Example 6-8: Coffee ConsumptionStep 2: Find the z value for 1.

1 1.642.67

0.24z

Step 3: Find the area to the left of z = –2.67. It is 0.0038.




Page #318


To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.


Example 6-9: Police Academy


The cutoff, the lowest possible score to qualify, is 226.

Example 6-8: Police AcademyStep 2: Subtract 1 – 0.1000 to find area to the left, 0.9000. Look

for the closest value to that in Table E.

200 1.28 20 225.60 X z Step 3: Find X.




Page #319


For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study.


Example 6-10: Systolic Blood Pressure


Area to the left of the positive z: 0.5000 + 0.3000 = 0.8000.

Using Table E, z 0.84.

Area to the left of the negative z: 0.5000 – 0.3000 = 0.2000.

Using Table E, z –0.84.

The middle 60% of readings are between 113 and 127.

Example 6-10: Systolic Blood Pressure

X = 120 + 0.84(8) = 126.72

X = 120 – 0.84(8) = 113.28


Normal Distributions• A normally shaped or bell-shaped distribution is

only one of many shapes that a distribution can assume; however, it is very important since many statistical methods require that the distribution of values (shown in subsequent chapters) be normally or approximately normally shaped.

• There are a number of ways statisticians check for normality. We will focus on three of them.


Checking for Normality• Histogram

• Pearson’s Index PI of Skewness

• Outliers

• Other Tests– Normal Quantile Plot– Chi-Square Goodness-of-Fit Test– Kolmogorov-Smikirov Test– Lilliefors Test




Page #320


A survey of 18 high-technology firms showed the number of days’ inventory they had on hand. Determine if the data are approximately normally distributed.

5 29 34 44 45 63 68 74 74

81 88 91 97 98 113 118 151 158

Method 1: Construct a Histogram.

The histogram is approximately bell-shaped.

Example 6-11: Technology Inventories


Method 2: Check for Skewness.

The PI is not greater than 1 or less than –1, so it can be concluded that the distribution is not significantly skewed.

Method 3: Check for Outliers.

Five-Number Summary: 5 - 45 - 77.5 - 98 - 158

Q1 – 1.5(IQR) = 45 – 1.5(53) = –34.5

Q3 + 1.5(IQR) = 98 + 1.5(53) = 177.5

No data below –34.5 or above 177.5, so no outliers.


3 79.5 77.53( )PI 0.148

40.5

X MD

s

79.5, 77.5, 40.5 X MD s


A survey of 18 high-technology firms showed the number of days’ inventory they had on hand. Determine if the data are approximately normally distributed.

5 29 34 44 45 63 68 74 74

81 88 91 97 98 113 118 151 158

Conclusion:

•The histogram is approximately bell-shaped.

•The data are not significantly skewed.

•There are no outliers.

Thus, it can be concluded that the distribution is approximately normally distributed.



Section 6-4

Applications of the Normal Distribution


a. Greater than 700,000.

b. Between 500,000 and 600,000.

The average daily jail population in the United States is 618,319. If the distribution is normal and the standard deviation is 50,200, find the probability that on a randomly selected day, the jail population is…

Section 6-4 Exercise #3

a. Greater than 700,000

P (z > 1.63) = 0.5 – 0.4484

= 0.0516 or 5.16%

z = 700,000– 618,319

50,200 = 1.63

0 1.63

z = X –

b. Between 500,000 and 600,000.

z = 500,000–618,319

50,200 = – 2.36

area = 0.4909

z =

600,000– 618,31950,200 = – 0.36

area = 0.1406

z = X –

area = 0.4909 area = 0.1406

P( – 2.36 < z < – 0.36)

= 0.3503 or 35.03%

= 0.4909 – 0.1406

– 2 .36 – 0.36

b. Between 500,000 and 600,000.

The average credit card debt for college seniors is $3262. If the debt is normally distributed with a standard deviation of $1100, find these probabilities.

a. That the senior owes at least $1000b. That the senior owes more than $4000c. That the senior owes between $3000 and $4000


z = X –

a. That the senior owes at least $1000

z =

1000 – 32621100

= – 2.06

area = 0.4803

= 0.9803 or 98.03% P(z – 2.06) = 0.5 + 0.4803

0.9803 or 98.03%

– 2.06 0

z = X –

b. That the senior owes more than $4000

z =

4000 – 32621100 = 0.67

= 0.2514 or 25.14% P (z > 0.67) = 0.5 – 0.2486

area = 0.2486

0.2514 or 25.14%

0.67 0

z = X –

c. That the senior owes between $3000 and $4000.

z =

3000 – 32621100 = – 0.24

area = 0.0948

= 0.3434 or 34.34% P( – 0.24 < z < 0.67) = 0.0948 + 0.2486

0.3434 or 34.34%

0.67– 0.24 0

An advertising company plans to market a product to low-income families. A study states that for a p

$2ar

4,ticu

596lar

area, the average income per family is and the standard devia $6256tion is . If the company plans to target the bottom of the families based on income, find the cut off income. Assume the variable is normally distribu

18%

ted.


The bottom 18% means that 32% of the area is between z and 0. The corresponding z score will be . – 0.92

$18,840.48 $24,596

0.320.18

X = – 0.92(6256) + 24,596

= $18,840.48

Section 6-5

The Central Limit Theorem


The average price of a pound of sliced bacon is $2.02. Assume the standard deviation is $0.08. If a random sample of 40 one-pound packages is selected, find the probability the the mean of the sample will be less than $2.00.

z = X –n

= – 1.58

= 2.00– 2.02

0.08

40

P(z < –1.58) = 0.5 – 0.4429

area = 0.4429

= 0.0571or 5.71%


$2.00 $2.02

The average price of a pound of sliced bacon is $2.02. Assume the standard deviation is $0.08. If a random sample of 40 one-pound packages is selected, find the probability the the mean of the sample will be less than $2.00.

0.0571 or 5.71%

The average time it takes a group of adults to complete a certain achievement test is 46.2 minutes. The standard deviation is 8 minutes. Assume the variable is normally distributed.


Average time = 46.2 minutes, Standard deviation = 8 minutes, variable is normally distributed.

a. Find the probability that a randomly selected adult will complete the test in less than 43 minutes.b. Find the probability that, if 50 randomly selected adults take the test, the mean time it takes the group to complete the test will be less than 43 minutes.

c. Does it seem reasonable that an adult would finish the test in less than 43 minutes? Explain.d. Does it seem reasonable that the mean of 50 adults could be less than 43 minutes? Explain.

Average time = 46.2 minutes, Standard deviation = 8 minutes, variable is normally distributed.

a. Find the probability that a randomly selected adult will complete the test in less than 43 minutes.

z = X –

= 43 – 46.28

= – 0.4

area = 0.1554

P(z < – 0.4) = 0.5 – 0.1554 = 0.3446or 34.46%

43 46.2

0.3446or 34.46%

b. Find the probability that, if 50 randomly selected adults take the test, the mean time it takes the group to complete the test will be less than 43 minutes.

z = 43 – 46.2850

= – 2.83

area = 0.4977

P(z < – 2.83) = 0.5 – 0.4977 = 0.0023or 0.23%

43 46.2

0.0023or 0.23%

c. Does it seem reasonable that an adult would finish the test in less than 43 minutes? Explain.

Yes, since it is within one standard deviation of the mean.

d. Does it seem reasonable that the mean of 50 adults could be less than 43 minutes? Explain.

It is very unlikely, since the probability would be less than 1%.

The average cholesterol of a certain brand of eggs is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed.

a. If a single egg is selected, find the probability that the cholesterol content will be greater than 220 milligrams.

b. If a sample of 25 eggs is selected, find the probability that the mean of the sample will be larger than 220 milligrams.


z = X –

a. If a single egg is selected, find the probability that the cholesterol content will be greater than milligr220 ams.

=

220–21515

area = 0.1293

= 0.33

P(z > 0.33) = 0.5 – 0.1293 = 0.3707or 37.07%

220215

0.3707or 37.07%

z = X –

n

= 220– 215

15

25

= 1.67

area = 0.4525

b. If a sample of eggs is selected, find the probability that

the mean of the sample will be larger than milligr

25

220 ams.

215 220

P(z > 1.67) = 0.5 – 0.4525

= 0.0475 or 4.75%

Section 6-6

The Normal Approximation toThe Binomial Distribution


Two out of five adult smokers acquired the habit by age 14. If 400 smokers are randomly selected, find the probability that 170 or more acquired the habit by age 14.

p =

25 = 400(0.4) = 0.4 = 160

= (400)(0.4)(0.6) = 9.8

z =

169.5 – 1609.8

area = 0.3340

= 0.97


169.5160

P(X > 169.5) = 0.5 – 0.3340 = 0.1660

The percentage of Americans 25 years or older who have at least some college education is 50.9%. In a random sample of 300 Americans 25 years and older, what is the probability that more than 175 have at least some college education?

= 300(0.509)

= (300)(0.509)(0.491)

= 152.7

= 8.66

z =

175.5 – 152.78.66

area = 0.4957 = 2.63

P (X > 175.5) = 0.5 – 0.4957 = 0.0043


175.5152.7

P(X > 175.5) = 0.5 – 0.4957 = 0.0043

Women comprise 83.3% of all elementary school teachers. In a random sample of 300 elementary school teachers, what is the probability that more than 50 are men?

= 300(0.167)

= (300)(0.167)(0.833)

= 50.1

= 6.46

z =

50.5 – 50.16.46 area = 0.0239 = 0.06

P (X > 50.5) = 0.5 – 0.0239 = 0.4761


50.550.1

P(X > 50.5) = 0.5 – 0.0239 = 0.4761

Chapter 6 The Normal Distribution 1 Copyright © 2013 The McGraw-Hill Companies, Inc. Permission...

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