Question

assume that when human resource managers are randomly selected, 41% 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 binomial distribution parameters

Let \(n = 10\) (number of trials, i.e., number of managers selected), \(p=0.41\) (probability of success, i.e., a manager saying job - applicants should follow up within two weeks), and \(q = 1 - p=1 - 0.41 = 0.59\). 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 \(P(X = 0)\)

\[

$$\begin{align*} C(10,0)&=\frac{10!}{0!(10 - 0)!}=1\\ P(X = 0)&=C(10,0)\times(0.41)^{0}\times(0.59)^{10}\\ &=1\times1\times(0.59)^{10}\\ &\approx0.0034 \end{align*}$$

\]

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

\[

$$\begin{align*} C(10,1)&=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10\\ P(X = 1)&=C(10,1)\times(0.41)^{1}\times(0.59)^{9}\\ &=10\times0.41\times(0.59)^{9}\\ &\approx0.0223 \end{align*}$$

\]

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

\[

$$\begin{align*} C(10,2)&=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45\\ P(X = 2)&=C(10,2)\times(0.41)^{2}\times(0.59)^{8}\\ &=45\times0.1681\times(0.59)^{8}\\ &\approx0.0735 \end{align*}$$

\]

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

\[

$$\begin{align*} P(X\lt3)&=P(X = 0)+P(X = 1)+P(X = 2)\\ &=0.0034 + 0.0223+0.0735\\ &=0.0992 \end{align*}$$

\]

Answer:

0.0992