Question

assume that when human resource managers are randomly selected, 57% say job applicants should follow up within two weeks. if 9 human resource managers are randomly selected, find the probability that at least 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 on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 9\), \(p=0.57\), and we want \(P(X\geq6)=P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)\)

Step2: Calculate \(P(X = 6)\)

\(C(9,6)=\frac{9!}{6!(9 - 6)!}=\frac{9\times8\times7}{3\times2\times1}=84\)
\(P(X = 6)=C(9,6)\times(0.57)^{6}\times(1 - 0.57)^{9 - 6}\)
\(=84\times(0.57)^{6}\times(0.43)^{3}\)
\(=84\times0.03429\times0.079507\)
\(\approx0.2319\)

Step3: Calculate \(P(X = 7)\)

\(C(9,7)=\frac{9!}{7!(9 - 7)!}=\frac{9\times8}{2\times1}=36\)
\(P(X = 7)=C(9,7)\times(0.57)^{7}\times(0.43)^{2}\)
\(=36\times0.01954\times0.1849\)
\(\approx0.1319\)

Step4: Calculate \(P(X = 8)\)

\(C(9,8)=\frac{9!}{8!(9 - 8)!}=9\)
\(P(X = 8)=C(9,8)\times(0.57)^{8}\times(0.43)^{1}\)
\(=9\times0.01114\times0.43\)
\(\approx0.0429\)

Step5: Calculate \(P(X = 9)\)

\(C(9,9)=\frac{9!}{9!(9 - 9)!}=1\)
\(P(X = 9)=(0.57)^{9}\)
\(\approx0.0065\)

Step6: Sum up the probabilities

\(P(X\geq6)=P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)\)
\(=0.2319+0.1319+0.0429+0.0065\)
\(=0.4132\)

Answer:

\(0.4132\)