Question

assume that when human - resource managers are randomly selected, 53% say job applicants should follow up within two weeks. if 5 human - resource managers are randomly selected, find the probability that at least 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 parameters

Let $n = 5$ (number of human - resource managers selected), $p=0.53$ (probability that a manager says job applicants should follow up within two weeks), and $q = 1 - p=1 - 0.53 = 0.47$. We want to find $P(X\geq4)=P(X = 4)+P(X = 5)$.
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate $P(X = 4)$

$C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=\frac{5\times4!}{4!×1}=5$.
$P(X = 4)=C(5,4)\times p^{4}\times q^{5 - 4}=5\times(0.53)^{4}\times(0.47)^{1}=5\times0.07890481\times0.47 = 0.184925$.

Step3: Calculate $P(X = 5)$

$C(5,5)=\frac{5!}{5!(5 - 5)!}=1$.
$P(X = 5)=C(5,5)\times p^{5}\times q^{5 - 5}=(0.53)^{5}=0.0418195493$.

Step4: Calculate $P(X\geq4)$

$P(X\geq4)=P(X = 4)+P(X = 5)=0.184925+0.0418195493\approx0.2267$.

Answer:

$0.2267$