Question

assume that when human resource managers are randomly selected, 46% say job applicants should follow up within two weeks. if 7 human resource managers are randomly selected, find the probability that exactly 4 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 formula parameters

This is a binomial - probability problem. 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 $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 7$, $k = 4$, and $p=0.46$.

Step2: Calculate the combination $C(n,k)$

$C(7,4)=\frac{7!}{4!(7 - 4)!}=\frac{7!}{4!3!}=\frac{7\times6\times5}{3\times2\times1}=35$.

Step3: Calculate $p^{k}$ and $(1 - p)^{n - k}$

$p^{k}=(0.46)^{4}=0.46\times0.46\times0.46\times0.46\approx0.0447$.
$1 - p = 1-0.46 = 0.54$, and $(1 - p)^{n - k}=(0.54)^{3}=0.54\times0.54\times0.54\approx0.1575$.

Step4: Calculate the probability $P(X = k)$

$P(X = 4)=C(7,4)\times p^{4}\times(1 - p)^{3}=35\times0.0447\times0.1575\approx0.2437$.

Answer:

$0.2437$