Question

assume that when human resource managers are randomly selected, 45% say job applicants should follow up within two weeks. if 10 human resource managers are randomly selected, find the probability that fewer than 3 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 \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 10\), \(p=0.45\), and we want \(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\)

Step2: Calculate \(P(X = 0)\)

For \(k = 0\), \(C(10,0)=\frac{10!}{0!(10-0)!}=1\), \(P(X = 0)=C(10,0)\times(0.45)^{0}\times(1 - 0.45)^{10-0}=1\times1\times(0.55)^{10}\approx0.0003405\)

Step3: Calculate \(P(X = 1)\)

For \(k = 1\), \(C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10\), \(P(X = 1)=C(10,1)\times(0.45)^{1}\times(0.55)^{9}=10\times0.45\times(0.55)^{9}\approx0.004161\)

Step4: Calculate \(P(X = 2)\)

For \(k = 2\), \(C(10,2)=\frac{10!}{2!(10-2)!}=\frac{10\times9}{2\times1}=45\), \(P(X = 2)=C(10,2)\times(0.45)^{2}\times(0.55)^{8}=45\times0.2025\times(0.55)^{8}\approx0.020725\)

Step5: Sum the probabilities

\(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.0003405 + 0.004161+0.020725=0.0252265\approx0.0252\)

Answer:

\(0.0252\)