Question

assume that when human - resource managers are randomly selected, 56% 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 (square) (round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial distribution parameters

Let \(n = 14\) (number of trials, i.e., number of managers selected), \(p=0.56\) (probability that a manager says job - applicants should follow up within two weeks), and \(q = 1 - p=1 - 0.56 = 0.44\). We want to find \(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\), where the binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), and \(C(n,k)=\frac{n!}{k!(n - k)!}\).

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

\[

$$\begin{align*} C(14,0)&=\frac{14!}{0!(14 - 0)!}=1\\ P(X = 0)&=C(14,0)\times(0.56)^{0}\times(0.44)^{14}\\ &=1\times1\times(0.44)^{14}\\ &\approx0.000004 \end{align*}$$

\]

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

\[

$$\begin{align*} C(14,1)&=\frac{14!}{1!(14 - 1)!}=\frac{14!}{1!13!}=14\\ P(X = 1)&=C(14,1)\times(0.56)^{1}\times(0.44)^{13}\\ &=14\times0.56\times(0.44)^{13}\\ &\approx0.00006 \end{align*}$$

\]

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

\[

$$\begin{align*} C(14,2)&=\frac{14!}{2!(14 - 2)!}=\frac{14\times13}{2\times 1}=91\\ P(X = 2)&=C(14,2)\times(0.56)^{2}\times(0.44)^{12}\\ &=91\times0.56^{2}\times(0.44)^{12}\\ &\approx0.0004 \end{align*}$$

\]

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

\[

$$\begin{align*} P(X\lt3)&=P(X = 0)+P(X = 1)+P(X = 2)\\ &\approx0.000004 + 0.00006+0.0004\\ &\approx0.0005 \end{align*}$$

\]

Answer:

0.0005