# Multiple choices | Mathematics homework help

Multiple choices | Mathematics homework help   1.      In a poll, respondents were asked whether they had ever been in a car accident. 220 respondents indicated that they had been in a car accident and 370 respondents said that they had not been in a car accident. If one of these respondents is randomly selected, what is the probability of getting someone who has been in a car accident? Round to the nearest thousandth.A. 0.384 B. 0.380 C. 0.373D. 0.370
P(accident) = 220 / (220 + 370) = 0.373
2.      If you flip a coin three times, the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the probability of getting at least one head?
A. 4/9                          B. 5/6                 C. 7/8           D. 5/8

P(at least one head) = (Number of combinations with one or more H) / (Total                                                                     number of combinations)

P(at least one head) = 7 / 8

3.      Joe dealt 20 cards from a standard 52-card deck, and the number of red cards exceeded the number of black cards by 8. He reshuffled the cards and dealt 30 cards. This time, the number of red cards exceeded the number of black cards by 10. Determine which deal is closer to the 50/50 ratio of red/black expected of fairly dealt hands from a fair deck and why.
A. The first series is closer because 1/10 is farther from 1/2 than is 1/8.
B. The series closer to the theoretical 50/50 cannot be determined unless the number of red and black cards for each deal is given.
C. The second series is closer because 20/30 is closer to 1/2 than is 14/20.
D. The first series is closer because the difference between red and black is smaller than the difference in the second series.

1st deal: 14 red and 6 black     P(red) = 14/20 = 0.70
2nd deal: 20 red and 10 black  P(red) = 20/30 = 0.67 1/5 > 1/11, the first series is closer.
D. The series closer to the theoretical 50/50 cannot be determined unless the total number of rolls for both series is given.
12.  Jody checked the temperature 12 times on Monday, and the last digit of the temperature was odd six times more than it was even. On Tuesday, she checked it 18 times and the last digit was odd eight times more than it was even. Determine which series is closer to the 50/50 ratio of odd/even expected of such a series of temperature checks.
A. The Monday series is closer because 1/6 is closer to 1/2 than is 1/8.
B. The Monday series is closer because 6/12 is closer to 0.5 than is 8/18.
C. The Tuesday series is closer because the 13/18 is closer to 0.5 than is 9/12.
D. The series closest to the theoretical 50/50 cannot be determined without knowing the number of odds and evens in each series.

Monday: 9 odd, 3 even           P(odd) = 9/12 = 0.75
Tuesday: 13 odd, 5 even         P(odd) = 13/18 ≈ 0.722 < 0.75

13.  Suppose you pay \$1.00 to roll a fair die with the understanding that you will get back \$3.00 for rolling a 5 or a 2, nothing otherwise. What is your expected value?
A. \$1.00           B. \$0.00              C. \$3.00        D. −\$1.00

E(x) = (1/3)(\$3) – \$1 = \$1 – \$1 = \$0

14.  On a multiple choice test, each question has 6 possible answers. If you make a random guess on the first question, what is the probability that you are correct?
A. 1/5         B. 1/6         C. 1/4        D. 2/5

P(correct) = Number of correct answers / Number of possible choices

P(correct) = 1/6

15.  A study of two types of weed killers was done on two identical weed plots. One weed killer killed 15% more weeds than the other. This difference was significant at the 0.05 level. What does this mean?
A. The improvement was due to the fact that there were more weeds in one study.
B. The probability that the difference was due to chance alone is greater than 0.05.
C. The probability that one weed killer performed better by chance alone is less than 0.05.
D. There is not enough information to make any conclusion.

16.  The distribution of B.A. degrees conferred by a local college is listed below, by major.
Major                    Frequency
English                 2073
Mathematics        2164
Chemistry            318
Physics                856
Liberal Arts          1358
Engineering          868
9313
What is the probability that a randomly selected degree is not in Business?
A. 0.7800          B. 0.8200         C. 0.8300         D. 0.9200

P(not in Business) = (Number not in Business) / Total number of degrees

P(not in Business) = (9313 – 1676) / 9313

P(not in Business) = 0.8200

17.  A 28-year-old man pays \$125 for a one-year life insurance policy with coverage of \$140,000. If the probability that he will live through the year is 0.9994, to the nearest dollar, what is the man’s expected value for the insurance policy?
A. \$139,916        B. −\$41         C. \$84         D. −\$124

E(x) = (1 – 0.9994)(\$140,000) – \$125 = -\$41

18.  A bag contains 4 red marbles, 3 blue marbles, and 7 green marbles. If a marble is randomly selected from the bag, what is the probability that it is blue?
A. 2/11        B. 3/11       C. 5/14        D. 3/14

P(blue) = Number of blue marbles / Total number of marbles

P(blue) = 3 / (4 + 3 + 7) = 3/14

19.  The data set represents the income levels of the members of a country club. Estimate the probability that a randomly selected member earns at least \$98,000.
112,000 126,000 90,000 133,000 94,000 112,000 98,000 82,000 147,000 182,000 86,000 105,000
140,000 94,000 126,000 119,000 98,000 154,000 78,000 119,000
A. 0.4         B. 0.6        C. 0.66         D. 0.7

P(salary ≥ \$98,000) = Number of salaries ≥ 98,000 / Total number of salaries

P(salary ≥ \$98,000) = 14/20 = 0.7

20.  A study of students taking Statistics 101 was done. Four hundred students who studied for more than 10 hours averaged a B. Two hundred students who studied for less than 10 hours averaged a C. This difference was significant at the 0.01 level. What does this mean?
A. The probability that the difference was due to chance alone is greater than 0.01.
B. There is less than a 0.01 chance that the first group’s grades were better by chance alone.
C. The improvement was due to the fact that more people studied.
D. There is not enough information to make any conclusion

21.  Sample size = 400, sample mean = 44, sample standard deviation = 16. What is the margin of error?
A. 1.4               B. 1.6               C. 2.2             D. 2.6

Assuming a 95% confidence level, and noting that this class seems to use  z = 2     for this confidence level instead of the more accurate 1.96 value.

E = z*s/√n = 2 * 16 / √400 = 1.6

22.  Select the best fit line on the scatter diagram below.

D. None of the lines is the line of best

23.              A researcher wishes to estimate the mean amount of money spent per month on food by households in a certain neighborhood. She desires a margin of error of \$30. Past studies suggest that a population standard deviation of \$248 is reasonable. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy.
A. 274        B. 284      C. 264        D. 272

Again, assuming a 95% confidence interval, with z = 2.00 instead of 1.96:

n = (z * s / E)2 = (2 * 248 / 30)2 = 273.35 à 274

24.           Suggest the cause of the correlation among the data.

The graph shows strength of coffee (y) and number of scoops used to make 10 cups of coffee (x). Identify the probable cause of the correlation.

A. The variation in the x variable is a direct cause of the variation in the y variable.

B. There is no correlation between the variables.

C. The correlation is due to a common underlying cause.

D. The correlation between the variables is coincidental

25.  Suggest the cause of the correlation among the data.

The graph shows strength of coffee (y) and number of scoops used to make 10 cups of coffee (x). Identify the probable cause of the correlation.

A.The variation in the x variable is a direct cause of the variation inthe y variable.

B. There is no correlation between the variables.

C. The correlation is due to a common underlying cause.

D. The correlation between the variables is coincidental.

26.A population proportion is to be estimated. Estimate the minimum sample size needed to achieve a margin of error E = 0.01with a 95% degree of confidence.
A. 7,000           B. 8,000              C. 9,000       D. 10,000

n = p(1-p)(z/E)2 = (0.5)(0.5)(2/0.01)2 = 10,000

27. Select the best estimate of the correlation coefficient for the data depicted in the scatter diagram.

A. -0.9

B. 0.1

C. 0.5

D. 0.9

The slope of the regression line is negative, so the coefficient of correlation must
also be negative.

28. A sample of nine students is selected from among the students taking a particular exam. The nine students were asked how much time they had spent studying for the exam and the responses (in hours) were as follows:18, 7, 10, 13, 12, 16, 5, 20, 21

Estimate the mean study time of all students taking the exam. Round your answer to the nearest tenth of an hour if necessary.     A. 13 hours     B. 12.2 hours         C. 13.6 hours
D. It is not possible to estimate the population mean from this sample data

Mean = (18 + 7 + 10 + 13 + 12 + 16 + 5 + 20 + 21) / 9 = 13.6
29. The scatter plot and best-fit line show the relation among the data for the price of a stock (y) and employment (x) in arbitrary units. The correlation coefficient is 0.8. Predict the stock price for an employment value of 6.

A. 8.8

B. 6.2

C. 8.2

D. None of the values are correct

Draw a vertical line through “6” on the horizontal axis, and extend the line vertically until it intersects the line on the graph. Then draw a horizontal line through the point of intersection and extend that line to the left until it intersects the vertical axis. Read the value from the vertical axis, which is about 6.2.

30.The graph shows a measure of fitness (y) and miles walked weekly. Identify the probable cause of the correlation.

A. The correlation is coincidental.

B. There is a common underlying cause of the correlation.

C. There is no correlation between the variables.

D. Walking is a direct cause of the fitness.

31. Eleven female college students are selected at random and asked their heights. The heights (in inches) are as follows:67, 59, 64, 69, 65, 65, 66, 64, 62, 64, 62Estimate the mean height of all female students at this college. Round your answer to the nearest tenth of an inch if necessary.

A. It is not possible to estimate the population mean from this sample data

B. 64.3 inches

C. 64.9 inches

D. 63.7 inches

Mean = (67 + 59 + 64 + 69 + 65 + 65 + 66 + 64 + 62 + 64 + 62) / 11 = 64.3
32. Select the best fit line on the scatter diagram below.

A. A

B. B

C. C

D. All of the lines are equally good

33. In a poll of 400 voters in a certain state, 61% said that they opposed a voter ID bill that might hinder some legitimate voters from voting. The margin of error in the poll was reported as 4 percentage points (with a 95% degree of confidence). Which statement is correct?
A. The reported margin of error is consistent with the sample size.
B. There is not enough information to determine whether the margin of error is consistent with the sample size.
C. The sample size is too small to achieve the stated margin of error.
D. For the given sample size, the margin of error should be smaller than stated

Required sample size = (phat)(1 – phat)(z/E)2 = (0.61)(1 – 0.61)(2/0.04)2 = 594.75
34.The scatter plot and best-fit line show the relation among the number of cars waiting by a school (y) and the amount of time after the end of classes (x) in arbitrary units. The correlation coefficient is -0.55. Determine the amount of variation in the number of cars not explained by the variation time after sch

A. 55%

B. 70%

C. 30%

D. 45%

Coefficient of non-determination = 1 – r2 = 1 – (-0.55)2 = 0.6975 → 70%

35. Which graph has two groups of data, correlations within each group, but no correlation among all the data?

A.

B.

C.

D

36. The scatter plot and best-fit line show the relation among the number of cars waiting by a school (y) and the amount of time after the end of classes (x) in arbitrary units. The correlation coefficient is -0.55. Use the line of best fit to predict the number of cars at time 4 after the end of classes.

A. 7.0

B. 6.0

C. 8.0

D. 3.5

Draw a vertical line through the value 4 on the horizontal axis and extend the line vertically until it intersects the line shown on the graph. Then draw a horizontal line through the point of intersection and extend it to the left until it intersects the vertical axis. Read the value where the horizontal line crosses the vertical axis. The value is approximately 7.

37. A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 560 college students showed that 27% of them had, or intended to, cheat on examinations. Find the 95% confidence interval.
A. 0.2323 to 0.3075
B. 0.2325 to 0.3075
C. 0.2325 to 0.3185
D. 0.2323 to 0.3185

Lower limit = phat – z*sqrt(phat*(1 – phat)/n)
Lower limit = 0.27 – 2*sqrt(0.27*0.73/560)
Lower limit = 0.2325

Upper limit = phat+ z*sqrt(phat*(1 – phat)/n)
Upper limit = 0.27 + 2*sqrt(0.27*0.73/560)
Upper limit = 0.3075

38. Among a random sample of 150 employees of a particular company, the mean commute distance is 29.6 miles. This mean lies 1.2 standard deviations above the mean of the sampling distribution. If a second sample of 150 employees is selected, what is the probability that for the second sample, the mean commute distance will be less than 29.6 miles?
A. 0.8849           B. 0.5          C. 0.1131        D. 0.1151

P(second sample mean ≤ 29.6) = P(z ≤ 1.2) = 0.8849

39. The scatter plot and best-fit line show the relation between the price per item (y) and the availability of that item (x) in arbitrary units. The correlation coefficient is -0.95. Determine the amount of variation in pricing explained by the variation in availability.

A. 5%

B. 10%

C. 95%

D. 90%

Coefficient of determination = r2 = (-0.95)2 = 0.9025 ≈ 90%

40. Of the 6796 students in one school district, 1537 cannot read up to grade level. Among a sample of 812 of the students from this school district, 211 cannot read up to grade level. Find the sample proportion of students who cannot read up to grade level       A. 0.14        B. 0.26      C. 211      D. 0.23

phat = x/n = 211 / 812 = 0.26  1.      In a poll, respondents were asked whether they had ever been in a car accident. 220 respondents indicated that they had been in a car accident and 370 respondents said that they had not been in a car accident. If one of these respondents is randomly selected, what is the probability of getting someone who has been in a car accident? Round to the nearest thousandth.A. 0.384 B. 0.380 C. 0.373D. 0.370
P(accident) = 220 / (220 + 370) = 0.373
2.      If you flip a coin three times, the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. What is the probability of getting at least one head?
A. 4/9                          B. 5/6                 C. 7/8           D. 5/8

P(at least one head) = (Number of combinations with one or more H) / (Total                                                                     number of combinations)

P(at least one head) = 7 / 8

3.      Joe dealt 20 cards from a standard 52-card deck, and the number of red cards exceeded the number of black cards by 8. He reshuffled the cards and dealt 30 cards. This time, the number of red cards exceeded the number of black cards by 10. Determine which deal is closer to the 50/50 ratio of red/black expected of fairly dealt hands from a fair deck and why.
A. The first series is closer because 1/10 is farther from 1/2 than is 1/8.
B. The series closer to the theoretical 50/50 cannot be determined unless the number of red and black cards for each deal is given.
C. The second series is closer because 20/30 is closer to 1/2 than is 14/20.
D. The first series is closer because the difference between red and black is smaller than the difference in the second series.

1st deal: 14 red and 6 black     P(red) = 14/20 = 0.70
2nd deal: 20 red and 10 black  P(red) = 20/30 = 0.67 1/5 > 1/11, the first series is closer.
D. The series closer to the theoretical 50/50 cannot be determined unless the total number of rolls for both series is given.
12.  Jody checked the temperature 12 times on Monday, and the last digit of the temperature was odd six times more than it was even. On Tuesday, she checked it 18 times and the last digit was odd eight times more than it was even. Determine which series is closer to the 50/50 ratio of odd/even expected of such a series of temperature checks.
A. The Monday series is closer because 1/6 is closer to 1/2 than is 1/8.
B. The Monday series is closer because 6/12 is closer to 0.5 than is 8/18.
C. The Tuesday series is closer because the 13/18 is closer to 0.5 than is 9/12.
D. The series closest to the theoretical 50/50 cannot be determined without knowing the number of odds and evens in each series.

Monday: 9 odd, 3 even           P(odd) = 9/12 = 0.75
Tuesday: 13 odd, 5 even         P(odd) = 13/18 ≈ 0.722 < 0.75

13.  Suppose you pay \$1.00 to roll a fair die with the understanding that you will get back \$3.00 for rolling a 5 or a 2, nothing otherwise. What is your expected value?
A. \$1.00           B. \$0.00              C. \$3.00        D. −\$1.00

E(x) = (1/3)(\$3) – \$1 = \$1 – \$1 = \$0

14.  On a multiple choice test, each question has 6 possible answers. If you make a random guess on the first question, what is the probability that you are correct?
A. 1/5         B. 1/6         C. 1/4        D. 2/5

P(correct) = Number of correct answers / Number of possible choices

P(correct) = 1/6

15.  A study of two types of weed killers was done on two identical weed plots. One weed killer killed 15% more weeds than the other. This difference was significant at the 0.05 level. What does this mean?
A. The improvement was due to the fact that there were more weeds in one study.
B. The probability that the difference was due to chance alone is greater than 0.05.
C. The probability that one weed killer performed better by chance alone is less than 0.05.
D. There is not enough information to make any conclusion.

16.  The distribution of B.A. degrees conferred by a local college is listed below, by major.
Major                    Frequency
English                 2073
Mathematics        2164
Chemistry            318
Physics                856
Liberal Arts          1358
Engineering          868
9313
What is the probability that a randomly selected degree is not in Business?
A. 0.7800          B. 0.8200         C. 0.8300         D. 0.9200

P(not in Business) = (Number not in Business) / Total number of degrees

P(not in Business) = (9313 – 1676) / 9313

P(not in Business) = 0.8200

17.  A 28-year-old man pays \$125 for a one-year life insurance policy with coverage of \$140,000. If the probability that he will live through the year is 0.9994, to the nearest dollar, what is the man’s expected value for the insurance policy?
A. \$139,916        B. −\$41         C. \$84         D. −\$124

E(x) = (1 – 0.9994)(\$140,000) – \$125 = -\$41

18.  A bag contains 4 red marbles, 3 blue marbles, and 7 green marbles. If a marble is randomly selected from the bag, what is the probability that it is blue?
A. 2/11        B. 3/11       C. 5/14        D. 3/14

P(blue) = Number of blue marbles / Total number of marbles

P(blue) = 3 / (4 + 3 + 7) = 3/14

19.  The data set represents the income levels of the members of a country club. Estimate the probability that a randomly selected member earns at least \$98,000.
112,000 126,000 90,000 133,000 94,000 112,000 98,000 82,000 147,000 182,000 86,000 105,000
140,000 94,000 126,000 119,000 98,000 154,000 78,000 119,000
A. 0.4         B. 0.6        C. 0.66         D. 0.7

P(salary ≥ \$98,000) = Number of salaries ≥ 98,000 / Total number of salaries

P(salary ≥ \$98,000) = 14/20 = 0.7

20.  A study of students taking Statistics 101 was done. Four hundred students who studied for more than 10 hours averaged a B. Two hundred students who studied for less than 10 hours averaged a C. This difference was significant at the 0.01 level. What does this mean?
A. The probability that the difference was due to chance alone is greater than 0.01.
B. There is less than a 0.01 chance that the first group’s grades were better by chance alone.
C. The improvement was due to the fact that more people studied.
D. There is not enough information to make any conclusion

21.  Sample size = 400, sample mean = 44, sample standard deviation = 16. What is the margin of error?
A. 1.4               B. 1.6               C. 2.2             D. 2.6

Assuming a 95% confidence level, and noting that this class seems to use  z = 2     for this confidence level instead of the more accurate 1.96 value.

E = z*s/√n = 2 * 16 / √400 = 1.6

22.  Select the best fit line on the scatter diagram below.

D. None of the lines is the line of best

23.              A researcher wishes to estimate the mean amount of money spent per month on food by households in a certain neighborhood. She desires a margin of error of \$30. Past studies suggest that a population standard deviation of \$248 is reasonable. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy.
A. 274        B. 284      C. 264        D. 272

Again, assuming a 95% confidence interval, with z = 2.00 instead of 1.96:

n = (z * s / E)2 = (2 * 248 / 30)2 = 273.35 à 274

24.           Suggest the cause of the correlation among the data.

The graph shows strength of coffee (y) and number of scoops used to make 10 cups of coffee (x). Identify the probable cause of the correlation.

A. The variation in the x variable is a direct cause of the variation in the y variable.

B. There is no correlation between the variables.

C. The correlation is due to a common underlying cause.

D. The correlation between the variables is coincidental

25.  Suggest the cause of the correlation among the data.

The graph shows strength of coffee (y) and number of scoops used to make 10 cups of coffee (x). Identify the probable cause of the correlation.

A.The variation in the x variable is a direct cause of the variation inthe y variable.

B. There is no correlation between the variables.

C. The correlation is due to a common underlying cause.

D. The correlation between the variables is coincidental.

26.A population proportion is to be estimated. Estimate the minimum sample size needed to achieve a margin of error E = 0.01with a 95% degree of confidence.
A. 7,000           B. 8,000              C. 9,000       D. 10,000

n = p(1-p)(z/E)2 = (0.5)(0.5)(2/0.01)2 = 10,000

27. Select the best estimate of the correlation coefficient for the data depicted in the scatter diagram.

A. -0.9

B. 0.1

C. 0.5

D. 0.9

The slope of the regression line is negative, so the coefficient of correlation must
also be negative.

28. A sample of nine students is selected from among the students taking a particular exam. The nine students were asked how much time they had spent studying for the exam and the responses (in hours) were as follows:18, 7, 10, 13, 12, 16, 5, 20, 21

Estimate the mean study time of all students taking the exam. Round your answer to the nearest tenth of an hour if necessary.     A. 13 hours     B. 12.2 hours         C. 13.6 hours
D. It is not possible to estimate the population mean from this sample data

Mean = (18 + 7 + 10 + 13 + 12 + 16 + 5 + 20 + 21) / 9 = 13.6
29. The scatter plot and best-fit line show the relation among the data for the price of a stock (y) and employment (x) in arbitrary units. The correlation coefficient is 0.8. Predict the stock price for an employment value of 6.

A. 8.8

B. 6.2

C. 8.2

D. None of the values are correct

Draw a vertical line through “6” on the horizontal axis, and extend the line vertically until it intersects the line on the graph. Then draw a horizontal line through the point of intersection and extend that line to the left until it intersects the vertical axis. Read the value from the vertical axis, which is about 6.2.

30.The graph shows a measure of fitness (y) and miles walked weekly. Identify the probable cause of the correlation.

A. The correlation is coincidental.

B. There is a common underlying cause of the correlation.

C. There is no correlation between the variables.

D. Walking is a direct cause of the fitness.

31. Eleven female college students are selected at random and asked their heights. The heights (in inches) are as follows:67, 59, 64, 69, 65, 65, 66, 64, 62, 64, 62Estimate the mean height of all female students at this college. Round your answer to the nearest tenth of an inch if necessary.

A. It is not possible to estimate the population mean from this sample data

B. 64.3 inches

C. 64.9 inches

D. 63.7 inches

Mean = (67 + 59 + 64 + 69 + 65 + 65 + 66 + 64 + 62 + 64 + 62) / 11 = 64.3
32. Select the best fit line on the scatter diagram below.

A. A

B. B

C. C

D. All of the lines are equally good

33. In a poll of 400 voters in a certain state, 61% said that they opposed a voter ID bill that might hinder some legitimate voters from voting. The margin of error in the poll was reported as 4 percentage points (with a 95% degree of confidence). Which statement is correct?
A. The reported margin of error is consistent with the sample size.
B. There is not enough information to determine whether the margin of error is consistent with the sample size.
C. The sample size is too small to achieve the stated margin of error.
D. For the given sample size, the margin of error should be smaller than stated

Required sample size = (phat)(1 – phat)(z/E)2 = (0.61)(1 – 0.61)(2/0.04)2 = 594.75
34.The scatter plot and best-fit line show the relation among the number of cars waiting by a school (y) and the amount of time after the end of classes (x) in arbitrary units. The correlation coefficient is -0.55. Determine the amount of variation in the number of cars not explained by the variation time after sch

A. 55%

B. 70%

C. 30%

D. 45%

Coefficient of non-determination = 1 – r2 = 1 – (-0.55)2 = 0.6975 → 70%

35. Which graph has two groups of data, correlations within each group, but no correlation among all the data?

A.

B.

C.

D

36. The scatter plot and best-fit line show the relation among the number of cars waiting by a school (y) and the amount of time after the end of classes (x) in arbitrary units. The correlation coefficient is -0.55. Use the line of best fit to predict the number of cars at time 4 after the end of classes.

A. 7.0

B. 6.0

C. 8.0

D. 3.5

Draw a vertical line through the value 4 on the horizontal axis and extend the line vertically until it intersects the line shown on the graph. Then draw a horizontal line through the point of intersection and extend it to the left until it intersects the vertical axis. Read the value where the horizontal line crosses the vertical axis. The value is approximately 7.

37. A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 560 college students showed that 27% of them had, or intended to, cheat on examinations. Find the 95% confidence interval.
A. 0.2323 to 0.3075
B. 0.2325 to 0.3075
C. 0.2325 to 0.3185
D. 0.2323 to 0.3185

Lower limit = phat – z*sqrt(phat*(1 – phat)/n)
Lower limit = 0.27 – 2*sqrt(0.27*0.73/560)
Lower limit = 0.2325

Upper limit = phat+ z*sqrt(phat*(1 – phat)/n)
Upper limit = 0.27 + 2*sqrt(0.27*0.73/560)
Upper limit = 0.3075

38. Among a random sample of 150 employees of a particular company, the mean commute distance is 29.6 miles. This mean lies 1.2 standard deviations above the mean of the sampling distribution. If a second sample of 150 employees is selected, what is the probability that for the second sample, the mean commute distance will be less than 29.6 miles?
A. 0.8849           B. 0.5          C. 0.1131        D. 0.1151

P(second sample mean ≤ 29.6) = P(z ≤ 1.2) = 0.8849

39. The scatter plot and best-fit line show the relation between the price per item (y) and the availability of that item (x) in arbitrary units. The correlation coefficient is -0.95. Determine the amount of variation in pricing explained by the variation in availability.

A. 5%

B. 10%

C. 95%

D. 90%

Coefficient of determination = r2 = (-0.95)2 = 0.9025 ≈ 90%

40. Of the 6796 students in one school district, 1537 cannot read up to grade level. Among a sample of 812 of the students from this school district, 211 cannot read up to grade level. Find the sample proportion of students who cannot read up to grade level       A. 0.14        B. 0.26      C. 211      D. 0.23

phat = x/n = 211 / 812 = 0.26

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