Question

assume that when human resource managers are randomly selected, 50% say job applicants should follow up within two weeks. if 14 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 with \(n = 14\) (number of trials, i.e., number of managers selected), \(p=0.5\) (probability of success, i.e., a manager saying job - applicants should follow up within two weeks), and we want to find \(P(X < 3)=P(X = 0)+P(X = 1)+P(X = 2)\). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

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

\[

$$\begin{align*} C(14,0)&=\frac{14!}{0!(14 - 0)!}=1\\ P(X = 0)&=C(14,0)\times(0.5)^{0}\times(1 - 0.5)^{14-0}\\ &=1\times1\times(0.5)^{14}\\ &=\frac{1}{16384}\approx0.000061 \end{align*}$$

\]

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

\[

$$\begin{align*} C(14,1)&=\frac{14!}{1!(14 - 1)!}=\frac{14!}{1!×13!}=14\\ P(X = 1)&=C(14,1)\times(0.5)^{1}\times(1 - 0.5)^{14 - 1}\\ &=14\times0.5\times(0.5)^{13}\\ &=14\times\frac{1}{2}\times\frac{1}{8192}\\ &=\frac{14}{16384}\approx0.000854 \end{align*}$$

\]

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

\[

$$\begin{align*} C(14,2)&=\frac{14!}{2!(14 - 2)!}=\frac{14\times13\times12!}{2\times1\times12!}=91\\ P(X = 2)&=C(14,2)\times(0.5)^{2}\times(1 - 0.5)^{14 - 2}\\ &=91\times(0.5)^{2}\times(0.5)^{12}\\ &=91\times\frac{1}{4}\times\frac{1}{4096}\\ &=\frac{91}{16384}\approx0.005554 \end{align*}$$

\]

Step5: Calculate \(P(X < 3)\)

\[

$$\begin{align*} P(X < 3)&=P(X = 0)+P(X = 1)+P(X = 2)\\ &=\frac{1 + 14+91}{16384}\\ &=\frac{106}{16384}\\ &\approx0.0065 \end{align*}$$

\]

Answer:

0.0065