Question

assume that when human resource managers are randomly selected, 48% say job applicants should follow up within two weeks. if 7 human resource managers are randomly selected, find the probability that at least 5 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 $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 7$, $p=0.48$, and $1 - p = 0.52$. We want to find $P(X\geq5)=P(X = 5)+P(X = 6)+P(X = 7)$.

Step2: Calculate $P(X = 5)$

$C(7,5)=\frac{7!}{5!(7 - 5)!}=\frac{7!}{5!2!}=\frac{7\times6}{2\times1}=21$.
$P(X = 5)=C(7,5)\times(0.48)^{5}\times(0.52)^{2}=21\times(0.48)^{5}\times(0.52)^{2}$
$P(X = 5)=21\times0.0254803968\times0.2704 = 21\times0.0068808983 = 0.1444988643$.

Step3: Calculate $P(X = 6)$

$C(7,6)=\frac{7!}{6!(7 - 6)!}=\frac{7!}{6!1!}=7$.
$P(X = 6)=C(7,6)\times(0.48)^{6}\times(0.52)^{1}=7\times0.0122305894\times0.52 = 7\times0.0063609065 = 0.0445263455$.

Step4: Calculate $P(X = 7)$

$C(7,7)=\frac{7!}{7!(7 - 7)!}=1$.
$P(X = 7)=C(7,7)\times(0.48)^{7}\times(0.52)^{0}=(0.48)^{7}=0.005714475$.

Step5: Calculate $P(X\geq5)$

$P(X\geq5)=P(X = 5)+P(X = 6)+P(X = 7)$
$P(X\geq5)=0.1444988643+0.0445263455 + 0.005714475=0.1947396848\approx0.1947$.

Answer:

$0.1947$