Question

assume that when human resource managers are randomly selected, 63% say job applicants should follow up within two weeks. if 8 human resource managers are randomly selected, find the probability that exactly 5 of them say job applicants should follow up within two weeks. the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Determine the values of \(n\), \(k\), and \(p\)

Here, \(n = 8\) (number of human - resource managers selected), \(k=5\) (number of managers who say job applicants should follow up within two weeks), and \(p = 0.63\) (probability that a manager says job applicants should follow up within two weeks). Then \(1-p=1 - 0.63=0.37\)

Step3: Calculate the combination \(C(n,k)\)

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Step4: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.63)^{5}\approx0.09924\)
\((1 - p)^{n - k}=(0.37)^{3}\approx0.05065\)

Step5: Calculate the probability \(P(X = 5)\)

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Answer:

\(0.2811\)