Question

assume that when human - resource managers are randomly selected, 61% say job applicants should follow up within two weeks. if 20 human - resource managers are randomly selected, find the probability that exactly 12 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

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, $n = 20$, $k = 12$, and $p=0.61$.

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

$C(20,12)=\frac{20!}{12!(20 - 12)!}=\frac{20!}{12!8!}=\frac{20\times19\times18\times17\times16\times15\times14\times13}{8\times7\times6\times5\times4\times3\times2\times1}=125970$.

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

$p^{k}=(0.61)^{12}\approx0.003277$, and $1 - p = 1-0.61 = 0.39$, $(1 - p)^{n - k}=(0.39)^{8}\approx0.000133$.

Step4: Calculate the probability $P(X = 12)$

$P(X = 12)=C(20,12)\times p^{12}\times(1 - p)^{8}=125970\times0.003277\times0.000133\approx0.1596$.

Answer:

$0.1596$