assume that when human - resource managers are randomly selected, 56% say job applicants should follow up…

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