According to a survey by the Administrative Management Society, one-half of U.S. companies give employees 4 weeks of vacation after they have been with the company for 15 years. Find the probability that among 6 companies surveyed at random, the number that give employees 4 weeks of vacation after 15 years of employment is (a) anywhere from 2 to 5; (b) fewer than 3.

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xleb123 · Accepted Answer

To solve the problem, we need to determine the probability of a certain number of companies giving employees 4 weeks of vacation after 15 years of employment.(a) To find the probability that among 6 randomly surveyed companies, the number that give employees 4 weeks of vacation after 15 years of employment is anywhere from 2 to 5, we need to calculate the probability of each possible scenario (2, 3, 4, or 5 companies) and sum them up.The probability of a single company giving 4 weeks of vacation after 15 years is 1/2, and the probability of not giving it is 1/2. We can use the binomial probability formula to calculate the probabilities for each scenario:P(X=k)=(nk)pk(1−p)n−kwhere n is the total number of trials (6 companies), k is the number of successes (2, 3, 4, or 5 companies), and p is the probability of success (1/2).Let&#039;s calculate the probabilities for each scenario and sum them up:P(X=2)=(62)(12)2(12)4P(X=3)=(63)(12)3(12)3P(X=4)=(64)(12)4(12)2P(X=5)=(65)(12)5(12)1P(2&amp;nbsp;to&amp;nbsp;5&amp;nbsp;companies)=P(X=2)+P(X=3)+P(X=4)+P(X=5)(b) To find the probability that among 6 randomly surveyed companies, the number that give employees 4 weeks of vacation after 15 years of employment is fewer than 3, we need to calculate the probabilities of 0, 1, or 2 companies giving 4 weeks of vacation and sum them up:P(X&amp;lt;3)=P(X=0)+P(X=1)+P(X=2)Let&#039;s calculate the probabilities for each scenario and sum them up:P(X=0)=(60)(12)0(12)6P(X=1)=(61)(12)1(12)5P(X&amp;lt;3)=P(X=0)+P(X=1)+P(X=2)

According to a survey by the Administrative Management

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