MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500. (a) What is the probability that the average aptitude test score in the sample will be between 70.14 and 82.14? (b) What is the probability that the average aptitude test score in the sample will be equal to or greater than 82.68? (c part is in image)

MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500. (a) What is the probability that the average aptitude test score in the sample will be between 70.14 and 82.14? (b) What is the probability that the average aptitude test score in the sample will be equal to or greater than 82.68? (c part is in image)

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

According to an almanac, 70% of adult smokers started smoking before turning 18 years old. (a) Compute the mean and standard deviation of the random variable X, the number of smokers who started before 18 in 300 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers? Why? ..... (a) Hx = Oy = (Round to the nearest tenth as needed.) (b) What is the correct interpretation of the mean? O A. It is expected that in 50% of random samples of 300 adult smokers, 210 will have started smoking before turning 18. B. It is expected that in a random sample of 300 adult smokers, 210 will have started smoking after turning 18. C. It is expected that in a random sample of 300 adult smokers, 210 will have started smoking before turning 18. (c) Would it be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult…
2
According to an almanac, 80% of adult smokers started smoking before turning 18 years old. (a) Compute the mean and standard deviation of the random variable X, the number of smokers who started before 18 in 300 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 255 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers? Why? D= 1 (e) O, =(Round to the nearest tenth as needed.) (b) What is the correct interpretation of the mean? O A. It is expected that in a random sample of 300 adult smokers, 240 will have started smoking before turning 18. O B. It is expected that in a random sample of 300 adult smokers, 240 will have started smoking after turning 18. OC. It is expected that in 50% of random samples of 300 adult smokers, 240 will have started smoking before turning 18. (c) Would it be unusual to observe 255 smokers who started smoking before turning 18 years old in a random sample of 300 adult…
About 71% of all female heart transplant patients will survive for at least 3 years. Eighty female heart transplant patients are randomly selected. What is the probability that the sample proportion surviving for at least 3 years will be pq less than 63%? Assume the sampling distribution of sample proportions is a normal distribution. The mean of the sample proportion is equal to the population proportion and the standard deviation is equal to The probability that the sample proportion surviving for at least 3 years will be less than 63% is (Round to four decimal places as needed.) More Clear All Final Check Help Me Solve This View an Example Get More Help - 4:51 PM 73°F Rain coming 4* 10/4/2021 O Type here to search 10/05/17 EGO PrtSc Insert Delete 口回 F10 F11 F12 F7 F8 F9 F6 Esc F2 F4 F5 F1 F3 & Backspace Num Lock 5 6 2 R T. Y U Home G K L Enter F H. J + |I LEGY P * 00 44 %#3 山

The state of California has a mean annual rainfall of 22 inches, whereas the state of New York has a mean annual rainfall of 42 inches. Assume that the standard deviation for both states is 4 inches. A sample of 40 years of rainfall for California and a sample of 45 years of rainfall for New York has been taken. (a) Show the probability distribution of the sample mean annual rainfall for California. (b) What is the probability that the sample mean is within 1 inch of the population mean for California? (c) What is the probability that the sample mean is within 1 inch of the population mean for New York?.
Business weekly conducted a survey of graduates for 30 top MBA programs. On the basis of the survey assume the means annual salary 10 years after graduation is $162,000. Assume the standard deviation is $40,000. Suppose you take a simply random sample of 39 graduates. For a single randomly selected graduate, find the probability that her salary is between $156,692.00 and $161,395.00. For a simple random sample of 39 graduates, find the probability that the average salary is between $156,692.00 and $161,395.00. Is the assumption of normal necessary? Yes, or No.
The mean annual salary for nurses is $65,000 with a standard deviation of $7,000. What is he probability that a randomly selected nurse has a salary less than $60,000? A random sample of 50 nurses is selected. What is the probability that the sample mean annual salary is less than $60,000?
The median annual cost of auto insurance is $ 939 (According to DACO). Suppose the standard deviation is o = $ 241. What is the probability that in a simple random sample of auto insurance policies the sample mean differs more than $ 70.42 from the population mean if the sample size is 45? You must calculate the probability that the difference between the average cost in the sample and the average annual cost do not differ by $ 70.42. Find the probability that P (\ µ - š | 0) = P (-70.42 H-iS 70.42) Select one: a. The probability is 99% that the auto policy will not differ by $ 70.42 from the average annual cost. b. The probability is 50% that the auto policy will not differ by $ 70.42 from the average annual cost. c. The probability is 80% that the auto policy will not differ by $ 70.42 from the average annual cost. d. The probability is 70% that the auto policy will not differ by $ 70.42 from the average annual cost. e. The probability is 95% that the auto policy will not differ by $…