Question

assume that when human resource managers are randomly selected, 43% say job applicants should follow up within two weeks. if 25 human resource managers are randomly selected, find the probability that exactly 16 of them say job applicants should follow up within two weeks.
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the 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, $n = 25$ (the number of human - resource managers), $k = 10$ (the number of managers who say job applicants should follow up within two weeks), and $p=0.43$ (the probability that a single manager says job applicants should follow up within two weeks), and $1 - p = 1-0.43 = 0.57$.

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

$C(25,10)=\frac{25!}{10!(25 - 10)!}=\frac{25!}{10!15!}=\frac{25\times24\times\cdots\times16}{10\times9\times\cdots\times1}=3268760$.

Step4: Calculate $p^{k}$ and $(1 - p)^{n - k}$

$p^{k}=(0.43)^{10}\approx0.000306$ and $(1 - p)^{n - k}=(0.57)^{15}\approx0.00707$.

Step5: Calculate the probability $P(X = k)$

$P(X = 10)=C(25,10)\times(0.43)^{10}\times(0.57)^{15}$
$P(X = 10)=3268760\times0.000306\times0.00707\approx0.0679$.

Answer:

0.0679