Compare the population range to the mean of the sample ranges. choose the correct answer below.

1.

Student: Alejandra Gutierrez Date: 02/11/18

Instructor: Laurie Saylor Course: Introduction to Statistics MAT-201-OL02

Assignment: Module 4 problem set

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between and minutes. Find the probability that a given class period runs between and minutes.

48.0 53.0 51.5 51.75

Find the probability of selecting a class that runs between and minutes.51.5 51.75

(Round to three decimal places as needed.)0.05

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

1 of 48 2/11/18, 6:58 PM

2.

1: Standard Normal Table (Page 1)

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

Click to view page 1 of the table. Click to view page 2 of the table.1 2

The area of the shaded region is .0.6700 (Round to four decimal places as needed.)

z = 0.44

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

2 of 48 2/11/18, 6:58 PM

2: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

3 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

4 of 48 2/11/18, 6:58 PM

3.

4.

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

The area of the shaded region is .0.8708 (Round to four decimal places as needed.)

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

The area of the shaded region is .0.7020 (Round to four decimal places as needed.)

z= -1.13

z = − 0.89 z = 1.22

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

5 of 48 2/11/18, 6:58 PM

5.

3: Standard Normal Table (Page 1)

Assume that thermometer readings are normally distributed with a mean of 0 C and a standard deviation of 1.00 C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.)

° °

Between and 0.25 1.00

Click to view page 1 of the table. Click to view page 2 of the table.3 4

Draw a sketch. Choose the correct graph below.

A. B. C.

The probability of getting a reading between C and C is .0.25° 1.00° 0.2427 (Round to four decimal places as needed.)

z=0.25 z=1.00 z=0.25 z=1.00 z=0.25 z=1.00

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

6 of 48 2/11/18, 6:58 PM

4: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

7 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

8 of 48 2/11/18, 6:58 PM

6.

5: Standard Normal Table (Page 1)

Assume the readings on thermometers are normally distributed with a mean of 0 C and a standard deviation of 1.00 C. Find the probability that a randomly selected thermometer reads between and and draw a sketch of the region.

° ° − 1.01 − 0.24

Click to view page 1 of the table. Click to view page 2 of the table.5 6

Sketch the region. Choose the correct graph below.

A. B. C.

The probability is .0.2489 (Round to four decimal places as needed.)

-0.24-1.01 -0.24-1.01 -0.24-1.01

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

9 of 48 2/11/18, 6:58 PM

6: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

10 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

11 of 48 2/11/18, 6:58 PM

7.

8.

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the probability of a bone density test score between

and .− 1.73 1.73

Sketch the region. Choose the correct graph below.

A. B. C. D.

The probability is .0.9164 (Round to four decimal places as needed.)

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than and draw a sketch of the region.

3.83

Sketch the region. Choose the correct graph below.

A. B. C.

The probability is .0.9999 (Round to four decimal places as needed.)

-1.73 1.73 -1.73 1.73 -1.73 1.73 -1.73 1.73

3.83 -3.83 3.83

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

12 of 48 2/11/18, 6:58 PM

9.

7: Standard Normal Table (Page 1)

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

Click to view page 1 of the table. Click to view page 2 of the table.7 8

The area of the shaded region is . (Round to four decimal places as needed.)0.3694

95

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

13 of 48 2/11/18, 6:58 PM

8: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

14 of 48 2/11/18, 6:58 PM

10. Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

The area of the shaded region is . (Round to four decimal places as needed.)0.7031

92

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

15 of 48 2/11/18, 6:58 PM

11.

9: Standard Normal Table (Page 1)

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

Click to view page 1 of the table. Click to view page 2 of the table.9 10

The area of the shaded region is . (Round to four decimal places as needed.)0.8609

80 125

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

16 of 48 2/11/18, 6:58 PM

10: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

17 of 48 2/11/18, 6:58 PM

12. Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

The area of the shaded region is . (Round to four decimal places as needed.)0.3922

102 124

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

18 of 48 2/11/18, 6:58 PM

13.

11: Standard Normal Table (Page 1)

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

Click to view page 1 of the table. Click to view page 2 of the table.11 12

The indicated IQ score, x, is . (Round to one decimal place as needed.)112.6

x

0.8

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

19 of 48 2/11/18, 6:58 PM

12: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

20 of 48 2/11/18, 6:58 PM

14. Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

The indicated IQ score is . (Round to the nearest whole number as needed.)117

0.1285

x

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

21 of 48 2/11/18, 6:58 PM

15.

13: Standard Normal Table (Page 1)

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

Click to view page 1 of the table. Click to view page 2 of the table.13 14

The indicated IQ score, x, is . (Round to one decimal place as needed.)87.4

x

0.8

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

22 of 48 2/11/18, 6:58 PM

14: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

23 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

24 of 48 2/11/18, 6:58 PM

16.

15: Standard Normal Table (Page 1)

Assume that adults have IQ scores that are normally distributed with a mean of and a standard deviation . Find the probability that a randomly selected adult has an IQ less than .

μ = 105 σ = 20 133

Click to view page 1 of the table. Click to view page 2 of the table.15 16

The probability that a randomly selected adult has an IQ less than is .133 0.9192 (Type an integer or decimal rounded to four decimal places as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

25 of 48 2/11/18, 6:58 PM

17.

16: Standard Normal Table (Page 2)

Assume that adults have IQ scores that are normally distributed with a mean of and a standard deviation of . Find the probability that a randomly selected adult has an IQ greater than . (Hint: Draw a graph.)

103.4 25 136.9

The probability that a randomly selected adult from this group has an IQ greater than is .136.9 0.0901 (Round to four decimal places as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

26 of 48 2/11/18, 6:58 PM

18.

17: Standard Normal Table (Page 1)

Assume that adults have IQ scores that are normally distributed with a mean of and a standard deviation . Find the probability that a randomly selected adult has an IQ between and .

μ = 105 σ = 15 88 122

Click to view page 1 of the table. Click to view page 2 of the table.17 18

The probability that a randomly selected adult has an IQ between and is .88 122 0.7416 (Type an integer or decimal rounded to four decimal places as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

27 of 48 2/11/18, 6:58 PM

18: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

28 of 48 2/11/18, 6:58 PM

19.

20.

A survey found that women's heights are normally distributed with mean . and standard deviation . The survey also found that men's heights are normally distributed with mean . and standard deviation . Consider an executive jet that seats six with a doorway height of . Complete parts (a) through (c) below.

63.6 in 3.4 in 68.7 in 3.8 in

56 in

a. What percentage of adult men can fit through the door without bending?

The percentage of men who can fit without bending is %.0.04 (Round to two decimal places as needed.)

b. Does the door design with a height of in. appear to be adequate? Why didn't the engineers design a larger door?56

A. The door design is adequate, because although many men will not be able to fit without bending, most women will be able to fit without bending. Thus, a larger door is not needed.

B. The door design is adequate, because the majority of people will be able to fit without bending. Thus, a larger door is not needed.

C. The door design is inadequate, because every person needs to be able to get into the aircraft without bending. There is no reason why this should not be implemented.

D. The door design is inadequate, but because the jet is relatively small and seats only six people, a much higher door would require major changes in the design and cost of the jet, making a larger height not practical.

c. What doorway height would allow 40% of men to fit without bending?

The doorway height that would allow 40% of men to fit without bending is in.67.7 (Round to one decimal place as needed.)

YOU ANSWERED: C.

0.07

Assume that military aircraft use ejection seats designed for men weighing between lb and lb. If women's weights are normally distributed with a mean of lb and a standard deviation of lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications?

138.8 217 162.4 40.8

The percentage of women that have weights between those limits is %. (Round to two decimal places as needed.)

Are many women excluded with those specifications?

A. Yes, the percentage of women who are excluded, which is equal to the probability found previously, shows that about half of women are excluded.

B. No, the percentage of women who are excluded, which is the complement of the probability found previously, shows that very few women are excluded.

C. No, the percentage of women who are excluded, which is equal to the probability found previously, shows that very few women are excluded.

D. Yes, the percentage of women who are excluded, which is the complement of the probability found previously, shows that about half of women are excluded.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

29 of 48 2/11/18, 6:58 PM

21.

19: Standard Normal Table (Page 1)

Assume that human body temperatures are normally distributed with a mean of and a standard deviation of .98.21°F 0.64°F a. A hospital uses as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of is appropriate?

100.6°F 100.6°F

b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is positive, but the subject is not really sick.)

Click to view page 1 of the table. Click to view page 2 of the table.19 20

a. The percentage of normal and healthy persons considered to have a fever is %. (Round to two decimal places as needed.)

Does this percentage suggest that a cutoff of is appropriate?100.6°F

A. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever.

B. Yes, because there is a large probability that a normal and healthy person would be considered to have a fever.

C. No, because there is a small probability that a normal and healthy person would be considered to have a fever.

D. No, because there is a large probability that a normal and healthy person would be considered to have a fever.

b. The minimum temperature for requiring further medical tests should be if we want only 5.0% of healthy people to exceed it.

°F

(Round to two decimal places as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

30 of 48 2/11/18, 6:58 PM

20: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

31 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

32 of 48 2/11/18, 6:58 PM

22.

21: Standard Normal Table (Page 1)

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probability of a pregnancy lasting days or longer. b. If the length of pregnancy is in the lowest %, then the baby is premature. Find the length that separates premature babies from those who are not premature.

266 15 308 3

Click to view page 1 of the table. Click to view page 2 of the table.21 22

a. The probability that a pregnancy will last days or longer is .308 (Round to four decimal places as needed.)

b. Babies who are born on or before days are considered premature. (Round to the nearest integer as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

33 of 48 2/11/18, 6:58 PM

22: Standard Normal Table (Page 2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

34 of 48 2/11/18, 6:58 PM

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

35 of 48 2/11/18, 6:58 PM

23.

(1) equal to greater than

less than

(2) do not target target

(3) do do not

(4) an unbiased a biased

The assets (in billions of dollars) of the four wealthiest people in a particular country are . Assume that samples of size n 2 are randomly selected with replacement from this population of four values.

46, 31, 18, 12 =

a. After identifying the 16 different possible samples and finding the mean of each sample, construct a table representing the sampling distribution of the sample mean. In the table, values of the sample mean that are the same have been combined.

x Probability x Probability 46 24.5

38.5 21.5 32 18 31 15 29 12

(Type integers or fractions.)

b. Compare the mean of the population to the mean of the sampling distribution of the sample mean.

The mean of the population, , is (1) the mean of the sample means, . (Round to two decimal places as needed.)

c. Do the sample means target the value of the population mean? In general, do sample means make good estimates of population means? Why or why not?

The sample means (2) the population mean. In general, sample means (3) make good

estimates of population means because the mean is (4) estimator.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

36 of 48 2/11/18, 6:58 PM

24.

(1) 92 69 46

(2) 52.5 105 78.5

(3) 71.5 48.5 97

(4) 106 53

80.5

(5) 74 49.5 99

(6) 110 55 82

(7) 51 102 76

(8) 56 84.5 112

(9) 52 104 76.5

(10) 59 88.5 118

Assume a population of , , , and . Assume that samples of size n 2 are randomly selected with replacement from the population. Listed below are the sixteen different samples. Complete parts (a) through (c).

46 51 53 59 =

,46 46 ,46 51 ,46 53 ,46 59 ,51 46 ,51 51 ,51 53 ,51 59 ,53 46 ,53 51 ,53 53 ,53 59 ,59 46 ,59 51 ,59 53 ,59 59

a. Find the median of each of the sixteen samples, then summarize the sampling distribution of the medians in the format of a table representing the probability distribution of the distinct median values. Use ascending order of the sample medians.

Sample Median Probability Sample Median Probability (1) (2)

(3) (4)

(5) (6)

(7) (8)

(9) (10)

(Type integers or simplified fractions. Use ascending order of the sample medians.)

b. Compare the population median to the mean of the sample medians. Choose the correct answer below.

A. The population median is equal to double the mean of the sample medians. B. The population median is equal to half of the mean of the sample medians. C. The population median is not equal to the mean of the sample medians (it is also not half or double the

mean of the sample medians). D. The population median is equal to the mean of the sample medians.

c. Do the sample medians target the value of the population median? In general, do sample medians make unbiased estimators of population medians? Why or why not?

A. The sample medians target the population median, so sample medians are biased estimators, because the mean of the sample medians equals the population median.

B. The sample medians target the population median, so sample medians are unbiased estimators, because the mean of the sample medians equals the population median.

C. The sample medians do not target the population median, so sample medians are unbiased estimators, because the mean of the sample medians does not equal the population median.

D. The sample medians do not target the population median, so sample medians are biased estimators, because the mean of the sample medians does not equal the population median.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

37 of 48 2/11/18, 6:58 PM

25.

(1) 45 90 0

(2) 104 54 5

(3) 9 59 113

(4) 59 118 14

The ages (years) of three government officials when they died in office were , , and . Complete parts (a) through (d).

54 45 59

a. Assuming that 2 of the ages are randomly selected with replacement, list the different possible samples.

A. (54,45),(54,59),(45,59) B. (54,54), (54,45),(54,59),(45,45),(45,59),(59,59) C. (54,54), (54,45),(54,59),(45,54),(45,45),(45,59),(59,54),(59,45),(59,59) D. (54,45),(54,59),(45,54),(45,59),(59,54),(59,45)

b. Find the range of each of the samples, then summarize the sampling distribution of the ranges in the format of a table representing the probability distribution.

Sample Range Probability (1)

(2)

(3)

(4)

(Type an integer or a fraction.)

c. Compare the population range to the mean of the sample ranges. Choose the correct answer below.

A. The population range is equal to the age of the oldest official at the time of death and the mean of the sample ranges is equal to the youngest official at the time of death.

B. The population range is equal to the youngest official at the time of death and the mean of the sample ranges is equal to the oldest official at the time of death.

C. The population range is equal to the mean of the sample ranges. D. The population range is not equal to the mean of the sample ranges (it is also not equal to the age of

the oldest official or age of the youngest official at the time of death).

d. Do the sample ranges target the value of the population range? In general, do sample ranges make good estimators of population ranges? Why or why not?

A. The sample ranges target the population range, therefore, sample ranges do not make good estimators of population ranges.

B. The sample ranges do not target the population range, therefore, sample ranges make good estimators of population ranges.

C. The sample ranges do not target the population range, therefore, sample ranges do not make good estimators of population ranges.

D. The sample ranges target the population range, therefore, sample ranges make good estimators of population ranges.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

38 of 48 2/11/18, 6:58 PM

26.

(1) 0.25 0

0.5

(2) 0.75 0.25 0.5

(3) 1 0.5

0.75

When two births are randomly selected, the sample space for genders is bb, bg, gb, and gg. Assume that those four outcomes are equally likely. Construct a table that describes the sampling distribution of the sample proportion of girls from two births. Does the mean of the sample proportions equal the proportion of girls in two births? Does the result suggest that a sample proportion is an unbiased estimator of a population proportion? For the entire population, assume the

probability of having a boy is , the probability of having a girl is , and this is not affected by how many boys or girls

have previously been born.

1 2

1 2

Determine the probabilities of each sample proportion.

Sample proportion of girls Probability

(1)

(2)

(3) (Type integers or simplified fractions.)

Does the mean of the sample proportions equal the proportion of girls in two births?

A. Yes, both the mean of the sample proportions and the population proportion are . 1 2

B. No, the mean of the sample proportions and the population proportion are not equal.

C. Yes, both the mean of the sample proportions and the population proportion are . 1 4

D. Yes, both the mean of the sample proportions and the population proportion are . 1 3

Does the result suggest that a sample proportion is an unbiased estimator of a population proportion?

A. Yes, because the sample proportions and the population proportion are not the same. B. No, because the sample proportions and the population proportion are not the same. C. Yes, because the sample proportions and the population proportion are the same. D. No, because the sample proportions and the population proportion are the same.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

39 of 48 2/11/18, 6:58 PM

27.

28.

Assume that females have pulse rates that are normally distributed with a mean of beats per minute and a standard deviation of beats per minute. Complete parts (a) through (c) below.

μ = 74.0 σ = 12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is between beats per minute and beats per minute.

67 81

The probability is . (Round to four decimal places as needed.)

b. If adult females are randomly selected, find the probability that they have pulse rates with a mean between beats per minute and beats per minute.

16 67 81

The probability is . (Round to four decimal places as needed.)

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

A. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

B. Since the mean pulse rate exceeds 30, the distribution of sample means is a normal distribution for any sample size.

C. Since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.

D. Since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.

An elevator has a placard stating that the maximum capacity is lb passengers. So, adult male passengers can have a mean weight of up to If the elevator is loaded with adult male passengers, find the probability that it is overloaded because they have a mean weight greater than lb. (Assume that weights of males are normally distributed with a mean of and a standard deviation of .) Does this elevator appear to be safe?

2325 —15 15 2325 / 15 = 155 pounds. 15

155 165 lb 28 lb

The probability the elevator is overloaded is . (Round to four decimal places as needed.)

Does this elevator appear to be safe?

A. No, there is a good chance that randomly selected people will exceed the elevator capacity.15 B. Yes, there is a good chance that randomly selected people will not exceed the elevator capacity.15 C. No, randomly selected people will never be under the weight limit.15 D. Yes, randomly selected people will always be under the weight limit.15

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

40 of 48 2/11/18, 6:58 PM

29.

30.

(1) (2)

A ski gondola carries skiers to the top of a mountain. Assume that weights of skiers are normally distributed with a mean of lb and a standard deviation of lb. The gondola has a stated capacity of passengers, and the gondola is rated for

a load limit of lb. Complete parts (a) through (d) below. 181 35 25

3500

a. Given that the gondola is rated for a load limit of lb, what is the maximum mean weight of the passengers if the gondola is filled to the stated capacity of passengers?

3500 25

The maximum mean weight is lb. (Type an integer or a decimal. Do not round.)

b. If the gondola is filled with randomly selected skiers, what is the probability that their mean weight exceeds the value from part (a)?

25

The probability is . (Round to four decimal places as needed.)

c. If the weight assumptions were revised so that the new capacity became passengers and the gondola is filled with randomly selected skiers, what is the probability that their mean weight exceeds lb, which is the maximum mean

weight that does not cause the total load to exceed lb?

20 20 175

3500

The probability is . (Round to four decimal places as needed.)

d. Is the new capacity of passengers safe?20

Since the probability of overloading is (1) the new capacity (2) to be safe enough.

over 50%, under 5%,

appears does not appear

(1) part (b) part (a)

(2) only average individuals should be considered. the seats are occupied by individuals rather than means.

Suppose that an airline uses a seat width of in. Assume men have hip breadths that are normally distributed with a mean of in. and a standard deviation of in. Complete parts (a) through (c) below.

16.2 14.9 0.9

(a) Find the probability that if an individual man is randomly selected, his hip breadth will be greater than in.16.2

The probability is . (Round to four decimal places as needed.)

(b) If a plane is filled with randomly selected men, find the probability that these men have a mean hip breadth greater than in.

130 16.2

The probability is . (Round to four decimal places as needed.)

(c) Which result should be considered for any changes in seat design: the result from part (a) or part (b)?

The result from (1) should be considered because (2)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

41 of 48 2/11/18, 6:58 PM

31.

32.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between lb and lb. The new population of pilots has normally distributed weights with a mean of and a standard

deviation of . 140 191 148 lb

31.6 lb

a. If a pilot is randomly selected, find the probability that his weight is between lb and lb.140 191

The probability is approximately . (Round to four decimal places as needed.)

b. If different pilots are randomly selected, find the probability that their mean weight is between lb and lb.34 140 191

The probability is approximately . (Round to four decimal places as needed.)

c. When redesigning the ejection seat, which probability is more relevant?

A. Part (a) because the seat performance for a single pilot is more important. B. Part (b) because the seat performance for a sample of pilots is more important. C. Part (a) because the seat performance for a sample of pilots is more important. D. Part (b) because the seat performance for a single pilot is more important.

A boat capsized and sank in a lake. Based on an assumption of a mean weight of lb, the boat was rated to carry passengers (so the load limit was lb). After the boat sank, the assumed mean weight for similar boats was

changed from lb to lb. Complete parts a and b below.

139 70 9,730

139 175

a. Assume that a similar boat is loaded with passengers, and assume that the weights of people are normally distributed with a mean of lb and a standard deviation of lb. Find the probability that the boat is overloaded because the passengers have a mean weight greater than lb.

70 180.5 36.2

70 139

The probability is . (Round to four decimal places as needed.)

b. The boat was later rated to carry only passengers, and the load limit was changed to lb. Find the probability that the boat is overloaded because the mean weight of the passengers is greater than (so that their total weight is greater than the maximum capacity of lb).

15 2,625 175

2,625

The probability is . (Round to four decimal places as needed.)

Do the new ratings appear to be safe when the boat is loaded with passengers? Choose the correct answer below.15

A. Because is greater than , the new ratings do not appear to be safe when the boat is loaded with passengers.

180.5 175 15

B. Because there is a high probability of overloading, the new ratings appear to be safe when the boat is loaded with passengers.15

C. Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with passengers.15

D. Because the probability of overloading is lower with the new ratings than with the old ratings, the new ratings appear to be safe.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

42 of 48 2/11/18, 6:58 PM

33.

34.

An airliner carries passengers and has doors with a height of in. Heights of men are normally distributed with a mean of in and a standard deviation of in. Complete parts (a) through (d).

200 76 69.0 2.8

a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.

The probability is . (Round to four decimal places as needed.)

b. If half of the passengers are men, find the probability that the mean height of the men is less than in.200 100 76

The probability is . (Round to four decimal places as needed.)

c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?

A. The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.

B. The probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend.

C. The probability from part (a) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.

D. The probability from part (b) is more relevant because it shows the proportion of male passengers that will not need to bend.

d. When considering the comfort and safety of passengers, why are women ignored in this case?

A. There is no adequate reason to ignore women. A separate statistical analysis should be carried out for the case of women.

B. Since men are generally taller than women, it is more difficult for them to bend when entering the aircraft. Therefore, it is more important that men not have to bend than it is important that women not have to bend.

C. Since men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.

Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry passengers, and a flight has fuel and baggage that allows for a total passenger load of

lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight

of the passengers is greater than lb. What is the probability that the aircraft is overloaded? Should the pilot

take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of lb and a standard deviation of .

39 6,435

= 165 6,435 lb

39

182.1 39.1

The probability is approximately . (Round to four decimal places as needed.)

Should the pilot take any action to correct for an overloaded aircraft?

A. No. Because the probability is high, the aircraft is safe to fly with its current load. B. Yes. Because the probability is high, the pilot should take action by somehow reducing the weight of the

aircraft.

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

43 of 48 2/11/18, 6:58 PM

35.

36.

Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

166

Probability that fewer than voted41

The probability that fewer than of eligible voters voted is .41 166 (Round to four decimal places as needed.)

Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

186

Probability that exactly voted45

The probability that exactly of eligible voters voted is .45 186 (Round to four decimal places as needed.)

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

44 of 48 2/11/18, 6:58 PM

37. A gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the gender-selection technique, births consisted of baby girls and baby boys. In analyzing these results, assume that boys and girls are equally likely.

963 486 477

a. Find the probability of getting exactly girls in births.486 963 b. Find the probability of getting or more girls in births. If boys and girls are equally likely, is girls in births unusually high?

486 963 486 963

c. Which probability is relevant for trying to determine whether the technique is effective: the result from part (a) or the result from part (b)? d. Based on the results, does it appear that the gender-selection technique is effective?

a. The probability of getting exactly girls in births is .486 963 (Round to four decimal places as needed.)

b. The probability of getting or more girls in births is .486 963 (Round to four decimal places as needed.)

If boys and girls are equally likely, is girls in births unusually high?486 963

A. Yes, because girls in births is not far from what is expected, given the probability of having a girl or a boy.

486 963

B. No, because girls in births is not far from what is expected, given the probability of having a girl or a boy.

486 963

C. Yes, because girls in births is far from what is expected, given the probability of having a girl or a boy.

486 963

D. No, because girls in births is far from what is expected, given the probability of having a girl or a boy.

486 963

c. Which probability is relevant for trying to determine whether the technique is effective, the result from part (a) or the result from part (b)?

A. The results from part (a) and part (b) are equal, so they are equally relevant. B. The result from part (a) is more relevant, because one wants the probability of a result that is exactly

equal to the one obtained. C. The result from part (b) is more relevant, because one wants the probability of a result that is at least as

extreme as the one obtained. D. Neither of the results are relevant.

d. Based on the results, does it appear that the gender-selection technique is effective?

A. , because the probability of having or more girls in births is unlikely, and thus, is attributable to random chance. Yes 486 963 not

not B. , because the probability of having or more girls in births is unlikely, and thus,

is attributable to random chance. No 486 963 not

C. , because the probability of having or more girls in births is unlikely, and thus, is attributable to random chance. Yes 486 963

not D. , because the probability of having or more girls in births is unlikely, and thus, is attributable

to random chance. No 486 963

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

45 of 48 2/11/18, 6:58 PM

38.

39.

(1) No, Yes,

(2) greater less

(3) 0.95. 0.5. 0.05.

Based on a smartphone survey, assume that % of adults with smartphones use them in theaters. In a separate survey of adults with smartphones, it is found that use them in theaters.

40 207 69

a. If the % rate is correct, find the probability of getting or fewer smartphone owners who use them in theaters.40 69 b. Is the result of significantly low?69

a. If the % rate is correct, the probability of getting or fewer smartphone owners who use them in theaters is 40 69 (Round to four decimal places as needed.)

b. Is the result of significantly low?69

(1) because the probability of this event is (2) than the probability cutoff that

corresponds to a significant event, which is (3)

(1) Yes, No,

(2) less greater

(3) 0.5. 0.05. 0.95.

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of peas, with of them having red flowers. If we assume, as the scientist did, that under these circumstances, there is a probability that a pea will have a red flower, we would expect that (or about ) of the peas would have red flowers, so the result of peas with red flowers is more than expected.

915 718 3 / 4

686.25 686 718

a. If the scientist's assumed probability is correct, find the probability of getting or more peas with red flowers.718 b. Is peas with red flowers significantly high?718 c. What do these results suggest about the scientist's assumption that of peas will have red flowers?3 / 4

a. If the scientist's assumed probability is correct, the probability of getting or more peas with red flowers is 718 (Round to four decimal places as needed.)

b. Is peas with red flowers significantly high?718

(1) because the probability of this event is (2) than the probability cutoff that

corresponds to a significant event, which is (3)

c. What do these results suggest about the scientist's assumption that of peas will have red flowers?3 / 4

A. Since the result of peas with red flowers significantly high, it strong evidence against the scientist's assumption that of peas will have red flowers.

718 is is not 3 / 4

B. The results do not indicate anything about the scientist's assumption. C. Since the result of peas with red flowers significantly high, it strong evidence against the

scientist's assumption that of peas will have red flowers. 718 is not is

3 / 4 D. Since the result of peas with red flowers significantly high, it strong evidence against the

scientist's assumption that of peas will have red flowers. 718 is is

3 / 4 E. Since the result of peas with red flowers significantly high, it strong evidence against

the scientist's assumption that of peas will have red flowers. 718 is not is not

3 / 4 F. Since the result of peas with red flowers significantly high, it strong evidence supporting the

scientist's assumption that of peas will have red flowers. 718 is is

3 / 4

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

46 of 48 2/11/18, 6:58 PM

40.

41.

(1) No, Yes,

(2) greater less

(3) 0.05. 0.5. 0.95.

Assume that % of people have sleepwalked. Assume that in a random sample of adults, have sleepwalked.19.2 1542 336 a. Assuming that the rate of % is correct, find the probability that or more of the adults have sleepwalked.19.2 336 1542 b. Is that result of or more significantly high?336 c. What does the result suggest about the rate of %?19.2

a. Assuming that the rate of % is correct, the probability that or more of the adults have sleepwalked is 19.2 336 1542 (Round to four decimal places as needed.)

b. Is that result of or more significantly high?336

(1) because the probability of this event is (2) than the probability cutoff that

corresponds to a significant event, which is (3)

c. What does the result suggest about the rate of %?19.2

A. Since the result of adults that have sleepwalked significantly high, it strong evidence against the assumed rate of %.

336 is not is not 19.2

B. Since the result of adults that have sleepwalked significantly high, it strong evidence against the assumed rate of %.

336 is is 19.2

C. The results do not indicate anything about the scientist's assumption. D. Since the result of adults that have sleepwalked significantly high, it strong evidence

against the assumed rate of %. 336 is is not

19.2 E. Since the result of adults that have sleepwalked significantly high, it strong evidence

against the assumed rate of %. 336 is not is

19.2 F. Since the result of adults that have sleepwalked significantly high, it strong evidence

supporting the assumed rate of %. 336 is is

19.2

In a survey of people, people said they voted in a recent presidential election. Voting records show that % of eligible voters actually did vote. Given that % of eligible voters actually did vote, (a) find the probability that among

randomly selected voters, at least actually did vote. (b) What do the results from part (a) suggest?

1287 848 63 63

1287 848

(a) P(X ) (Round to four decimal places as needed.)≥ 848 =

(b) What does the result from part (a) suggest?

A. People are being honest because the probability of P(x ) is less than 5%.≥ 848 B. Some people are being less than honest because P(x ) is less than 5%.≥ 848 C. People are being honest because the probability of P(x ) is at least 1%.≥ 848 D. Some people are being less than honest because P(x ) is at least 1%.≥ 848

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

47 of 48 2/11/18, 6:58 PM

42. In a study of cell phone users, it was found that developed cancer of the brain or nervous system. Assuming that cell phones have no effect, there is a probability of a person developing cancer of the brain or nervous system. We therefore expect about cases of such cancer in a group of people. Estimate the probability of

or fewer cases of such cancer in a group of people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous system?

317,649 122 0.000408

130 317,649 122 317,649

(a) P(x ) (Round to four decimal places as needed.)≤ 122 =

(b) What does the result from part (a) suggest about the media reports?

A. The media reports appear to be correct because one would expect that less than cell phone users would develop cancer and the study offers evidence to support this.

130 some

B. The media reports appear to be correct because one would expect that more than cell phone users would develop cancer and the study offers evidence to support this.

122 some

C. The media reports appear to be incorrect because one would expect that more than cell phone users would develop cancer. In fact, the study may offer evidence to suggest that cell phone use decreases the probability of developing cancer.

130 some

Module 4 problem set-Alejandra Gutierrez https://xlitemprod.pearsoncmg.com/api/v1/print/math

48 of 48 2/11/18, 6:58 PM