Ch. 6.3 #7-49 odd

To find the area between the z=0 and z=0.75, look up the area in table E corresponding to z=0.75. Therefore, area that corresponds to this z value is 0.2734.

To find the area between the z=0 and z=-2.07, look up the area in the table E corresponding to z=2.07 because the normal distribution is symmetric about the mean, the area to the left of the mean is the same as the area to the right of the mean. Therefore, the area that corresponds to this value is 0.4808.

Since this area is a tail area, find the area between z=0 and z=0.23 and subtract it from 0.5000. Hence, 0.5000 - 0.0910 = 0.4090.

Table E gives us the area for positive z values. Because of the symmetric property of normal distribution, the area to the left of -1.43 is the same as the area to the right of +1.43. Now, find

the area between z=0 and z=1.43 which is 0.4236 and subtract it from 0.5000. Therefore, the area is equal to 0.0764.

To find the area between z=0.79 and z=1.28, first look up the area from z=0 to z=1.28 which is 0.3997, and the area from z=0 to z=0.79 which is 0.2852, then subtract the smaller area from the larger area. Hence, 0.3997 – 0.2852 = 0.1145.

First find the area between z=0 and z=1.83 which is 0.4664, then find the area between z=0 and z=1.56 which is 0.4406, and subtract the smaller area from the larger area to get the answer. Hence, 0.4664 – 0.4406 = 0.0258.

Since the two area are on opposite sides of z=0, one must find both areas and add them. First find the area between z=0 and z=2.47 which is 0.4932, then find the area between z=0 and z=-1.03 which is 0.3485, and add them. Hence, 0.4932 + 0.3485 = 0.8417.

Because Table E only gives us the area between z=0 and z=2.11, we must add 0.5000 to the table area, because 0.5000 is the total area that rests to the left of z=0. Hence, 0.4826 + 0.5000 = 0.9826.

Find the area between z=0 and z=-0.15 which is 0.0596, then add 0.5000 to it. Hence, 0.0596 + 0.5000 = 0.5596.

The area to the right of z=1.92 is 0.5000 - 0.4726 = 0.0274, the area to the left of z=-0.44 is 0.5000 – 0.1700 = 0.3300. The total area, then, is 0.0274 + 0.3300 = 0.3574.

Look up the area in table E corresponding to z=0.67. It is 0.2486.

Look up in table E for the area corresponding to the z=1.57. We do this since the area to the right of the mean is the same as the area to the left of the mean because the symmetry of the normal distribution. Thus, the area that corresponds to this z value is 0.4418.

Use table E to look up for the area that corresponds to z=2.83. Once you have found this area which is 0.4977, since we are interested on the right hand side of z=2.83 subtract 0.4977 from 0.5000. Thus, the area that corresponds to z>2.83 is 0.5000-0.4977=0.0023

To find the area that corresponds to z=-1.21, look up the area that corresponds to z=1.21 because we know that a normal distributed graph is symmetric about the mean (the area to the right of the

mean is the same as the area to the left of the mean). The area that corresponds to z=1.21 is 0.3869. We are interested in the area to the left of z=-1.21. For this reason, subtract 0.3869 from 0.5000.

Then P(z<-1.21) = 0.5000-0.3869 = 0.1131

Use table E. First find the area that corresponds to z=1.74 which is 0.4591, then find the area that corresponds to z=2.46 (by symmetry) which is 0.4931. Since one of the z values lies above the mean and the other below the mean, we add their corresponding areas 0.4591+0.4931=0.9522. Then, the ( 2.46 1.74) 0.9522P z

Use table E. Find the area corresponding to the z=1.46 which is 0.4279 and the area corresponding to z=2.97 which is 0.4985. Since both z values are on the same side of the mean, we subtract the smaller area from the larger. Then

(1.46 2.97) 0.4985 0.4279 0.0706P z .

Using table E, find the area that corresponds to z=1.42 which is 0.4222, and since we are interested in the area to the left hand side of z=1.42 we add its area to 0.5000. Then

( 1.42) 0.5000 0.4222 0.9222P z

For exercises 40 through 45, find the z value that corresponds to the given area.

41)

Look up for the z value that corresponds to area of 0.4738. The z value corresponding to this area is -1.94

43)

Since the given area is a left tail area, it means that to get the corresponding z value first we need to subtract 0.0166 from 0.5000. Hence 0.5000-0.0166=0.4834. Now use table E to look up for the z value whose area is 0.4834. For this area z=2.13, however since this is a left tail area

z=-2.13.

45) For this exercise, the area of 0.8962 lies to the right of the z value. To find the z value, first subtract 0.5000 from 0.8962. 0.8962-0.5000=0.3962. Now look up table E and find the z value that corresponds to this area. For this area z= 1.26, but since the z value is below the mean then it has to be negative, then z=-1.26.

47) Find the z values to the left of the mean so that

a) 98.87% of the area under the distribution curve lies to the right of it.

Similarly to exercise # 45, first subtract 0.5000 from 0.9887.

0.9887-0.5000=0.4887. Now, find the z value whose area is 0.4887 which is z=2.28, but since the z values lies below the mean then it has to be negative, thus z= -2.28

b) 82.12% of the area under the distribution curve lies to the right of it.

0.8212-0.5000=0.3212

z-value corresponding to area of 0.3212 is 0.92, but since the z value lies below the mean, then it becomes z=-0.92.

c) 60.64% of the area under the distribution curve lies to the right of it.

0.6064-0.5000=0.1064

The z value whose area is 0.1064 is 0.27, thus the z value to the left of the mean is z=-0.27.

49) Find two z values, one positive and one negative, so that the areas in the two tails total the following values.

a) 5%

0.05/2=0.025

Therefore each tail has area of 0.025. Use table E to find the z value whose area to the right of it is 0.025. So subtract 0.025 from 0.5000, 0.5000-0.025=0.475. The z value whose area is 0.475 is 1.96. Now by symmetry, the z value above the mean is z=1.96, and the z value below the mean is z=-1.96.

b) 10%

.010/2=0.05

Each tail has an area of 0.05. Similarly as exercise (a), subtract 0.05 from 0.5000.

0.5000-0.05=0.4500

The z value whose area is 0.4500 is 1.65. Thus by symmetry, the z value above the mean is z=1.65, and the z value below the mean is z=-1.65.

c) 1%

0.01/2=0.005

Each tail has an area of 0.005.

0.5000-0.005=0.495

The z value whose area is 0.495 is 2.58. Now by symmetry, the z value above the mean is z=2.58, and the z value below the mean is z=-2.58.

Section 6-4 Pg. 312 Exercises 1, 2, 9, 11, 15, 21, 22, 23, 26, 28, 30

1. Admission Charges for Movies The average admission charge for a movie is $5.39. If the distribution of admission charges is normal with a standard deviation of $0.79, what is the probability that a randomly selected admission charge is less than $3.00?

$3.00 $5.39 $2.393.04

$0.79 $0.79

Xz

then look up z = -3.04 in Table E and you get that the area = 0.4988

( 3.04) 0.5 0.4988 0.0012P z

-3.04 0

2. Teachers’ Salaries The average salary for first-year teachers is $27,989. If the distribution is

approximately normal with $3250 , what is the probability that a randomly selected first-year teacher makes these salaries?

a) Between $20,000 and $30,000 a year

$20,000 $27,989 $7,9892.46

$3250 $3250

Xz

then look up z = -2.46 in Table E and you get that the area = 0.4931

$30,000 $27,989 $2,0110.62

$3250 $3250

Xz

then look up z = 0.62 in Table E and you get that the area = 0.2324

( 2.46 0.62) 0.4931 0.2324 0.7255P z

-2.46 0 0.62

b) Less than $20,000 a year

$20,000 $27,989 $7,9892.46

$3250 $3250

Xz

( 2.46) 0.5 0.4931 0.0069P z

-2.46 0

9. Microwave Oven Lines A survey found that people keep their microwave ovens an average of 3.2 years. The standard deviation is 0.56 year. If a person decides to buy a new microwave oven, find the probability that he or she owned the old oven for the following amount of time. Assume the variable is normally distributed.

a) Less than 1.5 years

1.5 3.2 1.73.04

0.56 0.56

Xz

then look up z = -3.04 in Table E and you get that the area = 0.4988

( 3.04) 0.5 0.4988 0.0012P z

-3.03 0

b) Between 2 and 3 years

2 3.2 1.22.14

0.56 0.56

Xz

then look up z = -2.14 in Table E and you get that the area = 0.4838

3 3.2 0.20.36

0.56 0.56

Xz

then look up z = 0.36 in Table E and you get that the area = 0.1406

( 2.14 0.36) 0.4838 0.1406 0.3432P z

-2.14 -0.36

c) More than 3.2 years

3 3.2 00

0.56 0.56

Xz

( 0) 0.5P z

0

d) What percent of microwave ovens would be replaced if a warranty of 18 months were given?

For an 18 month (1.5 year) warranty,

1.5 3.2 1.73.04

0.56 0.56

Xz

then look up z = -3.04 in Table E and you get that the area = 0.4988

( 3.04) 0.5 0.4988 0.0012P z Hence, 12% of ovens would be replaced.

11. Credit Card Debt 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 $1000

1000 3262 22622.06

1100 1100

Xz

hen look up z = -2.06 in Table E and you get that the area = 0.4803

( 2.06) 0.5 0.4803 0.9803P z

-2.06 0

b) That the senior owes more than $4000

4000 3262 7380.67

1100 1100

Xz

hen look up z = 0.67 in Table E and you get that the area = 0.2486

( 0.67) 0.5 0.2486 0.2514P z

0 0.67

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

3000 3262 2620.24

1100 1100

Xz

then look up z = -0.24 in Table E and you get that the area = 0.0948

4000 3262 7380.67

1100 1100

Xz

then look up z = 0.67 in Table E and you get that the area = 0.2486

( 0.24 0.67) 0.0948 0.2486 0.3434P z

-0.24 0.67

15. Waiting to Be Seated The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes. Assume the variable is normally distributed. When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the following time.

a) Between 15 and 22 minutes

15 23.5 8.52.36

3.6 3.6

Xz

then look up z = -2.36 in Table E and you get that the area = 0.4909

22 23.5 1.50.42

3.6 3.6

Xz

then look up z = -0.42 in Table E and you get that the area = 0.1628

( 2.36 0.42) 0.4909 0.1628 0.3281P z

-2.36 -0.42

b) Less than 18 minutes or more than 25 minutes

18 23.5 5.51.53

3.6 3.6

Xz

then look up z = -1.53 in Table E and you get that the area = 0.4370

25 23.5 1.50.42

3.6 3.6

Xz

then look up z = 0.42 in Table E and you get that the area = 0.1628

( 1.53 0.42) (0.5 0.4370) (0.5 0.1628) 0.063 0.3372 0.4002P z or z

-1.53 0.42

c) Is it likely that a person will be seated in less than 15 minutes?

15 23.5 8.52.36

3.6 3.6

Xz

then look up z = -2.36 in Table E and you get that the area = 0.4909

( 2.36) 0.5 0.4909 0.0091P z

Since the probability is small, it is not likely that a person would be seated in less than 15 minutes.

21. Cost of Personal Computers The average price of a personal computer (PC) is $949. If the

computer prices are approximately normally distributed and $100 , what is the probability that a randomly selected PC costs more than $1200? The least expensive 10% of personal computers cost less than what amount?

Let X be the amount a least expensive personal computer can cost.

1200 949 2512.51

100 100

Xz

then look up z = 2.51 in Table E and you get that the area = 0.4940

( 2.51) 0.5 0.4940 0.006P z

2.51

For the least expensive 10%, the area is 0.4 on the left side of the curve. Thus,

0.4000 1.28

1.28(100) 949 $821

z value whose area is is z

x

22. Reading Improvement Program to help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 5% of the students in the district, based on the scores on a reading achievement exam. If the average score for the students in the district is 122.6, find the cutoff score that will make a student eligible for the program. The standard deviation is 18. Assume the variable is normally distributed.

122.6, 18

1 0.05 0.95

0.95 0.5 0.45

,

0.4500

1.64

*

1.64*18 122.6 93

In table E find the z valuethat

corresponds to an area of

z

x z

x

Therefore, a cutoff score of 93 will make a student eligible for the program.

23. Used Car Prices an automobile dealer finds that the average price of a previously owned vehicle is $8256. He decides to sell cars that will appeal to the middle 60% of the market in terms of price. Find the maximum and minimum prices of the cars the dealer will sell. The standard deviation is $1150, and the variable is normally distributed.

$8256, $1150

0.6 / 2 0.3

0.3 0.84,0.84

* 0.84*1150 8256 $7290

* 0.84*1150 8256 $9222

z value whose area is is z

Min z

Max z

26. High School Competency Test a mandatory competency test for high school sophomores has a normal distribution with a mean of 400 and a standard deviation of 100.

a) The top 3% of students receive $500. What is the minimum score you would need to receive this

award?

400, 100

1 0.03 0.97

0.97 0.5 0.47

0.47 1.88

* 1.88*100 400 588

z value whose area is is z

Min z

b) The bottom 1.5% of students must go to summer school. What is the minimum score you would need to stay out of this group?

400, 100

1 0.015 0.985

0.985 0.5 0.485

0.485 2.17

* 2.17*100 400 183

z value whose area is is z

Min z

28. Cost of Groceries If a one-person household spends an average of $40 per week on groceries; find the maximum and minimum dollar amounts spent per week for the middle 50% of one-person households. Assume that the standard deviation is $5 and the variable is normally distributed.

$40, $5

0.5 / 2 0.25

0.2500 0.66,0.66

* 0.66*5 40 $36.7

* 0.66*5 40 $43.3

z value whose area is is z

Min z

Max z

30. Security Officer Stress Tolerance to qualify for security officers’ training, recruits is tested for stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8. If only the top 15% of recruits are selected, find the cutoff score.

62, 8

1 0.15 0.85

0.85 0.5 0.35

0.3500 1.04

* 1.04*8 62 70.32

z value whose area is is z

Cutoff z

6.5 (p.325) #9, 13, 15, 17, 21, 23

9. Cost of Dog Ownership The average yearly cost per household of owning a dog is $186.80. Suppose that we randomly select 50 households that own a dog. What is the probability that the sample mean for these 50 households is less than $175.00? Assume σ= $32.

µ=186.80, n=50, σ= 32, P( x < 175.00)

xZ

x

n

175.00 186.80

32

50

175.00 186.8032

7.07

11.80

4.526

= 2.61

( 2.61) 0.4955P z

Since we want the area to the left of z=-2.61, subtract 0.4955 from 0.5000

=0.0045 or 0.45%

0

12. Teachers’ Salaries The average teacher’s salary in North Dakota is $29,863. Assume a normal distribution with α =$5100.

a. What is the probability that a randomly selected teacher’s salary is greater than $40,000?

40,000 29,8631.99

5100( 1.99) .4767

( 1.99) 0.5 0.4767 0.0233

xz

P z

P z

The probability that a randomly selected teacher’s salary is greater than $40,000 is 2.33%

b. What is the probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000?

30,000 29,8630.24

5100( 0.24) 0.0948

( 0.24) 0.5 0.0948 0.4052

xz

P z

P z

The probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000 is about 40%.

13. Price of Bacon 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 that the mean of the sample will be less than $2.00.

µ=2.02, n=40, σ= 0.08, P( x < 2.0)

2.00 2.020.08

40

Z

0.02

0.08

6.324

0.02

0.01265

1.58

( 1.58)P z 0.4429

Therefore, the P( x < 2.0)= 0.5 – 0.4429= 0.0571 or 5.71%

15. Sodium in Frozen Food The average number of milligrams (mg) of sodium in a certain brand of low-salt microwave frozen dinners is 660mg, and the standard deviation is 35 mg. Assume the variable is normally distributed.

a. If a single is selected, find the probability that the sodium content will be more than 670 mg.

µ=660, σ= 35, P( x < 670)

670 660

35Z

10

35 0.285 0.29

( 0.29) 0.1141P z

P( x < 670) = 0. 5 – 0.1141 = 0.3859 or 38.59%

b. If a sample of 10 dinners is selected, find the probability that the mean of the sample will be larger than 670 mg.

µ=660, n=10, σ= 35, P( x <670)

670 60035

10

Z

1035

3.16

10

11.08 0.902 0.90

( 0.90) 0.3159P z

P( x <670) = 0.5 – 0.3159 = 0.1841 or 18.41%

c) Why is the probability for part a greater than that for part b?

Because the individual values are more variable than means.

17. Water Use The Old Farmer’s Almanac reports that the average person uses 123 gallons of water daily. If the standard deviation is 21 gallons, find the probability that the mean of a randomly selected sample of 15 people will be between 120 and 126 gallons. Assume the variable is normally distributed.

µ=123, n=15, σ= 21, normally distributed

P(120< x <126)

126 12321

15

Z

3

21

3.87

3

5.43

0.55

( 0.55)P z 0.2088

By symmetry, 0.2088 x 2 = 0.4176 or 41.76%

21. Time to Complete an Exam 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.

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

µ=46.2, σ= 8, P( x <43) normally distributed

43 46.2

8Z

3.2

8

0.4

( 0.4)P z 0.1554

P( x <43) = 0.5 – 1.554 = 0.3446 or 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.

µ=46.2, σ= 8, n=50 P( x <43)

43 46.28

50

z

3.28

7.07

3.2

1.13

2.83

( 2.83)P z = 0.4977

P( x <43)=0.5 – 0.4977= 0.0023 or 0.23%

23. Cholesterol Content The average cholesterol content 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.

120 123 126

0.4

µ=215, σ= 15, P( x >220), normally distributed

220 215

15Z

5

15 0.33

( 0.33)P z = 0.1293

P( x >220) = 0.5 – 0.1293= 0.3707 or 37.07%

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

µ=215, n=25, σ= 15, P( x >220)

220 21515

25

z

5

15

5

5

3 1.67

( 1.67)P z = 0.4525

P( x >220) = 0.5 – 0.4525 = 0.0475 or 4.75%

Ch 6.6 p.333 #3a, b, c, d, 7, 8, 11

3. Check each binomial distribution to see whether it can be approximated by a normal distribution (i.e., are np ≥ 5 and nq ≥ 5?).

a. n=20, p=0.5

q = 1–q = 1 – 0.5 = 0.5

np = (20)(0.5) =10

nq = (20)(0.5) = 10

Yes, because np ≥ 5 and nq ≥ 5.

b. n=10, p=0.6

q = 1–q = 1 – 0.6 = 0.4

np = (10)(0.6) =6

nq = (10)(0.4) = 4

No because np ≥ 5 and nq ≤ 5.

c. n=40, p=0.9

q = 1–q = 1 – 0.9 = 0.1

np = (40)(0.9) =36

nq = (40)(0.1) = 4

No because np ≥ 5 and nq ≥ 5.

d. n=50, p=0.2

q = 1–q = 1 – 0.2 = 0.8

np = (50)(0.2) =10

nq = (50)(0.8) = 40

Yes, because np ≥ 5 and nq ≥ 5..

7. College Enrollment 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 old and older, what is the probability that more than 175 have at least some college education?

n=300, p=0.509

q = 1–q = 1 – 0.509 = 0.491

Since np = (300)(0.509) = 152.7 and nq = (300)(0.491) = 147.3, the normal distribution can be used to approximate the binomial distribution.

µ = np = (300)(0.509) = 152.7

σ = npq = (300)(0.509)(0.491) = 74.9757 = 8.65885

For binomial distribution, P(X >175)

For normal distribution, P(X >175 + 0.5) = P(X > 175.5)

The z value is z = 175.5 152.7

8.65885

= 2.63

From table E (Normal Distribution ), z=2.63 → 0.4957

0.5 – 0.4957 = 0.0043 or 0.43%

8. Personal Computers According to recent surveys, 53% of households have personal computers. If a random sample of 175 households is selected, what is the probability that more than 75 but fewer than 110 have a personal computer?

n=175, p=0.53

q = 1–q = 1 – 0.53 = 0.47

Since np = (175)(0.53) = 92.75 and nq = (175)(0.47) = 82.25, the normal distribution can be used to approximate the binomial distribution.

µ = np = (175)(0.53) = 92.75

σ = npq = (175)(0.53)(0.47) = 43.5925 = 6.60246

For binomial distribution, P(X > 75) and P(X < 110)

For normal distribution, P(75 + 0.5 < X < 110 – 0.5)

= P(75.5 < X < 109.5)

z1 = 109.5 92.75

6.60246

= 2.54 z2 =

75.5 92.75

6.60246

= –2.61

From table E (Normal Distribution), z1=2.54 → 0.4945, and z2=2.61 → 0.4955

0.4955 + 0.4945 = 0.99 or 0.99%

0.4957

152.7 175.5

92.75 109.575.5

0.49550.4945

11. Female Scientists and Engineers 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?

n=300, p=0.833

q = 1–q = 1 – 0.833 = 0.167

Since np = (300)(0.833) = 249.9 and nq = (300)(0.167) = 50.1, the normal distribution can be used to approximate the binomial distribution.

µ = np = (300)(0.833) = 249.9

σ = npq = (300)(0.833)(0.167) = 41.7333 = 6.4601

When P(X > 50) for women for binomial, this means P(X < 250) for men.

For normal distribution, P(X < 250 – 0.5)

= P(X < 249.5)

z = 249.5 249.9

6.4601

= –0.06

From table E (Normal Distribution), z=0.06 → 0.0239

0.5 –0.0239 = 0.4761 or 47.61%

249.9249.5

0.4761