Question

assume that when human resource managers are randomly selected, 36% say job applicants should follow up within two weeks. if 9 human resource managers are randomly selected, find the probability that exactly 6 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 in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

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

Here, \(n = 9\) (number of human - resource managers selected), \(k = 6\) (number of managers who say job applicants should follow up within two weeks), \(p=0.36\) (probability that a manager says job applicants should follow up within two weeks), and \(1 - p = 1-0.36 = 0.64\)

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

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

\(p^{k}=(0.36)^{6}\approx0.002176782\)
\((1 - p)^{n - k}=(0.64)^{3}=0.64\times0.64\times0.64 = 0.262144\)

Step5: Calculate the probability \(P(X = 6)\)

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Answer:

\(0.0479\)