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 $C(n,k)=\frac{n!}{k!(n - k)!}$.

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

Here, $n = 20$ (number of human - resource managers selected), $k = 12$ (number of managers who say job applicants should follow up within two weeks), and $p=0.61$ (probability that a single manager says job applicants should follow up within two weeks), and $1 - p = 0.39$.

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

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

Step4: Calculate the binomial probability

$P(X = 12)=C(20,12)\times(0.61)^{12}\times(0.39)^{8}$
$(0.61)^{12}\approx0.00397$
$(0.39)^{8}\approx0.00013$
$P(X = 12)=125970\times0.00397\times0.00013$
$P(X = 12)\approx0.1657$

Answer:

$0.1657$