Midterm-2-A-Answers

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University of Maryland, Baltimore County *

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121

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Statistics

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Jan 9, 2024

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Name: Student ID: STAT 121, Midterm-2, Version A Completely fill in the appropriate bubble on the bubble sheet provided, and also circle your answer choices on this test sheet. Your name, student ID and midterm version (A or B) should also be bubbled-in on the bubble sheet. Alphabetical part of the student ID should be entered first. Please return both the midterm sheet and the bubble sheet. Each question is worth 1 point. Due to rounding, some of your answers could be slightly different from the choices given. Please select the closest answer. 1. The probability that a certain basketball player makes a free throw is 0.40. Assume free throws are independent. If the player attempts 4 free throws , what is the probability that he does not make any of them? A. 0.0256 B. 0.9744 C. 0.8704 D. 0.1296 2. Referring to the previous problem, if the player attempts 3 free throws , what is the probability that he makes at least 1 of them? A. 0.064 B. 0.784 C. 0.936 D. 0.216 E. 0.8704 3. The probability that a teenager suffers from depression is 0.20. You randomly select 3 teenagers. What is the probability that all of them suffer from depression? A. 0.60 B. 0.512 C. 0.488 D. 0.992 E. 0.008 (For Questions 4-7). 1000 adults were asked whether they favored a law that would provide support for higher education. In addition, each person was classified as likely to vote or not likely to vote in an upcoming election. Favor Oppose Undecided Likely to vote 370 260 80 Not likely to vote 130 140 20 4. What is the probability that a randomly selected adult is likely to vote? A. 0.74 B. 0.71 C. 0.37 D. 0.26 E. 0.63 5. What is the probability that a randomly selected adult opposes the law? A. 0.26 B. 0.65 C. 0.40 D. 0.26 6. What is the probability that a randomly selected adult is likely to vote and opposes the law? A. 0.65 B. 0.284 C. 0.71 D. 0.26 E. 1.11 7. What is the probability that a randomly selected adult is likely to vote or opposes the law? A. 0.26 B. 0.85 C. 0.284 D. 0.15 E. 1.11

(For Questions 8-10) . For a certain population group, it is known that the scores on an IQ test are normally distributed with mean 100 and standard deviation 15. 8. Compute the 58 th percentile of the IQ scores A. 103 B. 97 C. 115 D. 108.7 E. 110.8 9. Compute the 42 nd percentile of the IQ scores A. 103 B. 115 C. 97 D. 85 E. 106.3 10. What is the median of the IQ scores? A. 50 B. 100 C. 115 D. 85 E. Can’t be determined 11. The amount of time full-time employed Americans spend daily on working and work related activities is normally distributed with a mean of 6.5 hours and a standard deviation of 0.5 hours. What proportion of full-time employed Americans spend more than 7 hours daily on working and work related activities? A. 0.8413 B. 0.6915 C. 0.1587 D. 0.3085 12. The time spent daily on personal care (including sleep) by employed American women is normally distributed with a mean of 9.5 and a standard deviation of 1. What amount of time must an employed American woman spend daily on personal care (including sleep) to be in the top 20% among all similar women. A. 10.08 B. 8.66 C. 9.52 D. 10.34 13. Find the z-scores that include the middle 72% of the area under the standard normal curve. A. -1.13 and 1.13 B. -0.58 and 0.58 C. -1.08 and 1.08 D. -0.24 and 0.24 E. 0.54 (For Questions 14-15). In a certain population of adults, 40% are homeowners. A random sample of 400 individuals is randomly selected from this population. 14. The sampling distribution of the sample proportion who are homeowners would have a SD equal to A. 0.0245 B. 0.0006 C. 0.24 D. 0.49 15. Determine the probability that the proportion of homeowners in the random sample of 400 is less than 0.36 A. 0.9484 B. 0.4325 C. 0.0516 D. 0.5675 (For Questions 16-18). The body mass index (BMI) of all US residents living in the Southeast averages 33 with a standard deviation of 6. Suppose we randomly sample 36 residents of the Southeast United States and determine the mean BMI.

16. The sampling distribution of the sample mean BMI will have a standard deviation of A. 6 B. 1 C. 0.167 D. Can’t be determined from the given information 17. What is the probability that the sample mean BMI in the sample 36 residents is greater than 34? A. 0.5000 B. 0.8413 C. 0.7967 D. 0.2033 E. 0.1587 18. True or false? The probability that the sample mean BMI is greater than 33 is the same as the probability that the sample mean BMI is less than 33 A. True B. False (For Questions 19-22). The result of a poll, based on the opinions of a random sample of 1200 adult Americans, noted that 540 adults in the sample approved of President Biden. 19. What is the standard error of the sample proportion of adult Americans who approve of President Biden? A. 0.00021 B. 0.0144 C. 0.2475 D. 0.4975 20. What is the margin of error for computing a 90% confidence interval for the proportion of all adult Americans who approve of President Biden? A. 0.0282 B. 0.0239 C. 0.0237 D. 0.4071 21. Compute a 90% confidence interval for the proportion of all adult Americans who approve of President Biden A. (0.422, 0.478) B. (0.043, 0.857) C. (0.436, 0.464) D. (0.426, 0.474) 22. Suppose you want to estimate the proportion of all adult Americans who approve of President Biden with a margin of error of 0.01 and using a 90% confidence interval. What is the required sample size? Use 𝑝𝑝̂ obtained from the opinion of the random sample of 1200 adult Americans, with 540 adults in the sample approving President Biden. A. 6766 B. 9508 C. 9604 D. 6706 E. 6698 (For Questions 23-25). A food chemist analyzed the calorie content for a popular type of chocolate cookie. For a sample of 30 cookies, the sample mean calorie content and the sample SD were 𝑥𝑥̅ = 115 and 𝑠𝑠 = 3 . 23. What is the critical value needed to compute a 99% confidence interval for the population mean calorie content in this type of cookie? A. 2.750 B. 2.576 C. 1.96 D. 2.756 24. What is the margin of error computing a 99% confidence interval for the population mean calorie content? A. 1.510 B. 1.506 C. 1.411 D. 1.074 25. Compute a 99% confidence interval for the population mean calorie content. A. (113.494, 116.506) B. (113.589, 116.411) C. (113.490, 116.510) D. (113.926, 115.074)

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Midterm-2-A-Answers

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