Question

assume that when human resource managers are randomly selected, 53% say job applicants should follow up within two weeks. if 12 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 distribution parameters

This is a binomial - distribution problem. Let \(n = 12\) (the number of trials, i.e., the number of human - resource managers selected), \(p=0.53\) (the probability that a single human - resource manager says job applicants should follow up within two weeks), and \(q = 1 - p=1 - 0.53 = 0.47\). We want to find \(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\).
The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

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

\[

$$\begin{align*} C(12,0)&=\frac{12!}{0!(12 - 0)!}=1\\ P(X = 0)&=C(12,0)\times(0.53)^{0}\times(0.47)^{12}\\ &=1\times1\times(0.47)^{12}\\ &\approx0.00013 \end{align*}$$

\]

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

\[

$$\begin{align*} C(12,1)&=\frac{12!}{1!(12 - 1)!}=\frac{12!}{1!11!}=12\\ P(X = 1)&=C(12,1)\times(0.53)^{1}\times(0.47)^{11}\\ &=12\times0.53\times(0.47)^{11}\\ &\approx0.0014 \end{align*}$$

\]

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

\[

$$\begin{align*} C(12,2)&=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66\\ P(X = 2)&=C(12,2)\times(0.53)^{2}\times(0.47)^{10}\\ &=66\times(0.53)^{2}\times(0.47)^{10}\\ &\approx0.0075 \end{align*}$$

\]

Step5: Calculate \(P(X\lt3)\)

\[

$$\begin{align*} P(X\lt3)&=P(X = 0)+P(X = 1)+P(X = 2)\\ &\approx0.00013 + 0.0014+0.0075\\ &\approx0.0090 \end{align*}$$

\]

Answer:

0.0090