Q. 1. A recent survey of local cell phone retails showed that of all cell phone sold last month, 64% had a camera, 28% had a music player and 22% had both. The probability that a cell phone sold last month had a camera or music player is?

A. 0.22

B. 0.70

**C. 0.92**

D. 0.30

E. 0.08

Answer: C. 0.92

Consider the following to answer the next three questions.

Purchase warrant? YES NO Total

Digital camera 30 42 72

Laptop computer 145 203 348

TOTAL 175 245 420

2. The probability that a consumer does not purchase an extended warranty is?

A. 0.07

B. 0.42

**C. 0.58**

D. 0.17

E. 0.83

Answer. C. 0.58

Answer.

3. The probability that consumer purchase a digital and an extended warranty is?

**A. 0.07**

B. 0.42

C. 0.58

D. 0.17

E. 0.83

Answer. A. 0.07

4. Suppose that in a certain town, 90% of the voters favor a new ball park. Find the probability that among 6 voters question, 5 of them favor the new ball park?

A. 0.590

B. 0.354

C. 0.833

D. 0.531

**E. None of the above**

Answer. E. None of the above

(the correct answer is 0.059)

5. Consider the standard normal distribution. Find the probability that Z<1.13.

A. 0.8485

B. 0.1292

C. 0.8907

**D. 0.8708**

E. None of the above

Answer. D. 0.8708

6. Consider the standard normal distribution. Find the probability that Z<-1.81 ?

A. 0.0359

**B. 0.0351**

C. -0.0351

D. 0.9649

Answer. B. 0.0351

7. Consider the standard normal distribution. Find P(Z>0.59)

A. 0.7224

B. 0.2190

C. 0.2224

**D. 0.2776**

E. None of the above

Answer. D. 0.2776

8. Consider the standard normal distribution. Find the probability P(-1.10<Z<-0.36) ?

A. 0.4951

B. 0.2237

C. 0.2239

D. -0.2237

**E. None of the above**

Answer. E. None of the above

(The correct answer is 0.8643-0.3594=0.5049)

9. Consider the standard normal curve. Find the value of Z such that the area to the Right of Z is 67% of the total area?

A. 0.44

**B. -0.44**

C. 1.00

D. -1.50

E. None of the above

Answer. B. -0.44

10. Consider the standard normal curve. Find the value of Z such that the area to the Right of Z is 2% of the total area?

**A. 2.05**

B. 2.06

C. -2.05

D. 2.10

E. None of the above

Answer. A. 2.05

11. Find the critical value Z* for the 97% confidence interval?

A. 1.88

B. 2.07

C. 1.96

**D. 2.17**

E. None of the above

Answer. D. 2.17

12. The mean amount spent by a family of four on food per month is $500 with a standard deviation of $75. Assuming a normal distribution, what is the probability that a family spends more than $410 per month?

**A. 0.1151**

B. 0.1539

C. 0.8849

D. 0.8461

E. None of the above

Answer. A. 0.1151

13. Assume that woman’s height are normally distributed with mean 63.60 inches and standard deviation 2.50 inches. Find the value of P90 the 90th percentile.

A. 66.10 in

B. 66.60 in

**C. 66.80**

D. 66.83

E. None of the above

Answer. C. 66.80

14. The scores on a certain test are normally distributed with mean 61 and standard deviation 3. What is the probability that a sample of 100 students will have a mean score of at least 61.3?

A. 0.4602

B. 0.3413

C. 0.8413

**D. 0.1587**

E. None of the above

Answer. D. 0.1587

15. A random sample of 86 light bulbs had a mean life 545 hours. From other studies it is know that the population standard deviation if such light bulbs is o=29 hours. Construct a 90% confidence interval for the mean life U, of all such light bulbs?

A. (537.7,552.3)

B. (526.9,553.1)

C. (538.9,551.1)

**D. (539.9,550.1)**

E. None of the above

Answer. D. (539.9,550.1)

16. Thirty randomly selected students too a test. If the sample mean was 77 and the sample standard deviation was 5.6, construct the 99% confidence interval for the mean score of all students?

A. (0.064,0.136)

B. (0.067,0.133)

C. (0.072,0.128)

D. (74.48,79.52)

**E. None of the above**

Answer. E. None of the above

(The correct answer is (74.36, 79.64))

17. A survey of 450 people found that 45 of them had no cell phone. Find the 99% confidence interval of the true proportion of people who did not have cell phones?

**A. (0.064,0.136)**

B. (0.067, 0.133)

C. (0.072, 0.128)

D. (0.077, 0.123)

E. None of the above

Answer. A. (0.064,0.136)

18. The 95% confidence interval of people who hate broccoli was found to be (7.4%, 13%). The sample size was 1000. Which of the following is the correct interpretation?

A. The percentage of people who hate broccoli is between 7.4% and 13.0%

**B. We are 95% confident that between 7.4% and 13.0% of the people hate broccoli**

C. The margin or error for the true percentage of people who hate broccoli is between 7.4% and 13.0%

D. All sample of size 1000 will yield a percentage of people who hate broccoli that falls within 7.4% and 13.0%

E. None of the above

Answer. E. B. We are 95% confident that between 7.4% and 13.0% of the people hate broccoli

19. All else being equal, increase the level of confidence, say from 95% to 99% will?

A. Tighten the confidence interval

B. Decrease the margin of error

C. Increase precision

**D. Increase the margin of error**

E. Both A and D

Answer. D. Increase the margin of error

20. Suppose we draw all possible sample of a given size N from a large population that is Not normally distributed. Then the distribution of the samples means will?

A. Be skewed to the right

B. Be skewed to the left

C. Be normal

**D. Approach a normal distribution as the sample size N increase**

E. None of the above

Answer. D. Approach a normal distribution as the sample size N increase

21. A manufacturer wishes to estimate the proportion of washing machines leaving the factory that is defective. How large a sample should be checked in order to be 96% confidence that the true proportion is estimated within 2%

A. 2401

B. 1915

C. 2626

**D. 2627**

E. None of the above

Answer. D. 2627

22. Please perform each of the following steps.

A. State the hypotheses and identify the claim

B. Find the critical value(s)

C. Find the test value

D. Make the decision

E. Summarize the results

Number of jobs. The U.S Bureau of labor and statistics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this average is correct, a researcher selected a sample of 8 workers between the ages of 18 and 34 and asked how many different places they had worked. The results were as follow:

8, 12, 15, 6, 1, 9, 13, 2.

At a=0.05 can it be concluded that the mean is 9.2? Use the P – value method. Give one reason why the respondents might not have given the exact number of jobs that they have reported.

Answer.

A. State the hypotheses and identify the claim

For the sample, n=8, mean=8.25, s=5.06.

H0: m=9.2 (claim)

Ha: m≠9.2

B. Find the critical value(s)

The sample is small (n=8), so t-statistic should be used; n=8, so critical value for 8-1=7 degrees of freedom, a=0.05 (two-tailed) is 2.3645. Critical p-value is 0.05.

C. Find the test value

t statistic = (8.25-9.2)/(5.06/)=-0.53

D. Make the decision

The p-value of H0 corresponding to t=-0.53 is 0.6126, which is greater than 0.05. Therefore, null hypothesis cannot be rejected.

E. Summarize the results

At a=0.05, it can be concluded that the mean is 9.2; therefore, the average reported by the U.S Bureau of labor and statistics is correct.

The respondents might not have given the exact number of jobs that they reported due to various reasons; one of possible reasons might be the difference in counting jobs by the U.S Bureau of labor and statistics and by the respondents themselves. Some respondents might also be not willing to reveal frequent job changes. In general, the answers of the respondents confirm the statistics reported by the U.S Bureau of labor and statistics.

23. NFL Salaries. An agent claims that there is no difference between the pay of safeties and linebackers in the NFL. A survey of 15 safeties found an average salary of $501.580, and a survey of 15 linebackers found an average salary of 513,360. If the standard deviation in the first sample was $20,000 and the standard deviation in the second sample is $18,000, is the agent correct? Use a=0.05.

Answer.

A. State the hypotheses and identify the claim

H0: (claim)

Ha:

B. Find the critical value(s)

The test is two-tailed, a=0.05, and the number of the degrees of freedom is 15-1=14. Critical value of t is 2.145.

C. Find the test value

The formula for the test value is:

D. Make the decision

-1.70>-2.145, and therefore null hypothesis should not be rejected.

E. Summarize the results

The agent is correct: it is possible to state that pays for safeties and linebackers are the same.