assume that when human resource managers are randomly selected, 41% say job applicants should follow up…

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