chapter 6 the normal distribution section 6-3 the standard normal distribution

Chapter 6The Normal Distribution

Section 6-3

The Standard Normal Distribution

Section 6-3

Exercise #7

Chapter 6The Normal Distribution

0.750

Find the area under the normal distribution curve.

Between z = 0 and z = 0.75

area = 0.2734

Section 6-3

Exercise #15

Chapter 6The Normal Distribution

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.

Section 6-3

Exercise #31

Chapter 6The Normal Distribution

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

Section 6-3

Exercise #45

Chapter 6The Normal Distribution

Find the z value that corresponds to the given area.

0.8962 – 0.5 = 0.3962

0z

0.8962

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

Section 6 - 4

Exercise #3

Chapter 6The 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…

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.

Section 6-4

Exercise #11

Chapter 6The Normal Distribution

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%

a. That the senior owes at least $1000

– 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%b. That the senior owes more than $4000

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%

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

0.67– 0.24 0

Section 6-4

Exercise #27

Chapter 6The Normal Distribution

An advertising company plans to market a product to low-income families. A study states that for a particular area, the average income per family is $24,596 and the standard deviation is $6256. If the company plans to target the bottom 18% of the families based on income, find the cutoff income. Assume the variable is normally distributed.

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

Exercise #13

Chapter 6The Normal Distribution

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%

Section 6-5

Exercise #21

Chapter 6The Normal Distribution

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%

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

43 46.2

0.3446or 34.46%

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

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%

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

43 46.2

0.0023or 0.23%

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

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.

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

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%.

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

Section 6-5

Exercise #23

Chapter 6The Normal Distribution

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 –

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.

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

=

220–21515

area = 0.1293

= 0.33

P(z > 0.33) = 0.5 – 0.1293

= 0.3707or 37.07%

220215

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.

0.3707or 37.07%

z = X –

n

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.

= 220– 215

15

25

= 1.67

area = 0.4525

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

215 220

P(z > 1.67) = 0.5 – 0.4525

= 0.0475 or 4.75%

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.

Section 6-6

Exercise #5

Chapter 6The Normal 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

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.

169.5160

P(X > 169.5) = 0.5 – 0.3340 = 0.1660

Section 6-6

Exercise #7

Chapter 6The Normal Distribution

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

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?

175.5152.7

P(X > 175.5) = 0.5 – 0.4957 = 0.0043

Section 6-6

Exercise #11

Chapter 6The Normal Distribution

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

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?

50.550.1

P(X > 50.5) = 0.5 – 0.0239 = 0.4761