Question

assume that when human resource managers are randomly selected, 52% say job applicants should follow up within two weeks. if 6 human resource managers are randomly selected, find the probability that exactly 2 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 binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$

Here, $n = 6$ (the number of human - resource managers selected), $k = 2$ (the number of managers who say job applicants should follow up within two weeks), and $p=0.52$ (the probability that a single manager says job applicants should follow up within two weeks), and $1 - p = 1-0.52 = 0.48$.

Step3: Calculate the combination $C(n,k)$

$C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6!}{2!4!}=\frac{6\times5}{2\times1}=15$.

Step4: Calculate the probability

$P(X = 2)=C(6,2)\times(0.52)^{2}\times(0.48)^{4}=15\times0.2704\times0.05308416\approx0.2161$.

Answer:

$0.2161$