Question

assume that when human resource managers are randomly selected, 53% say job applicants should follow up within two weeks. if 6 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 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, $n = 6$, $p=0.53$, and $1 - p = 0.47$. We want to find $P(X\geq4)=P(X = 4)+P(X = 5)+P(X = 6)$.

Step2: Calculate $P(X = 4)$

$C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6!}{4!2!}=\frac{6\times5}{2\times1}=15$.
$P(X = 4)=C(6,4)\times(0.53)^{4}\times(0.47)^{2}=15\times0.53^{4}\times0.47^{2}=15\times0.07890481\times0.2209 = 15\times0.017430072529\approx0.2615$.

Step3: Calculate $P(X = 5)$

$C(6,5)=\frac{6!}{5!(6 - 5)!}=6$.
$P(X = 5)=C(6,5)\times(0.53)^{5}\times(0.47)^{1}=6\times0.0418195493\times0.47=6\times0.019655188171\approx0.1179$.

Step4: Calculate $P(X = 6)$

$C(6,6)=\frac{6!}{6!(6 - 6)!}=1$.
$P(X = 6)=C(6,6)\times(0.53)^{6}\times(0.47)^{0}=(0.53)^{6}=0.0221$.

Step5: Calculate $P(X\geq4)$

$P(X\geq4)=P(X = 4)+P(X = 5)+P(X = 6)=0.2615 + 0.1179+0.0221=0.4015$.

Answer:

$0.4015$