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 2 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

$n = 7$, $p=0.48$, $q = 1 - p=0.52$

Step2: Find probability of 0 and 1 success

$P(X = 0)=\binom{7}{0}p^{0}q^{7}=\binom{7}{0}(0.48)^{0}(0.52)^{7}=1\times1\times0.014697273344 = 0.0147$
$P(X = 1)=\binom{7}{1}p^{1}q^{6}=\frac{7!}{1!(7 - 1)!}\times0.48\times(0.52)^{6}=7\times0.48\times0.0282639872 = 0.0947$

Step3: Calculate probability of at least 2 successes

$P(X\geq2)=1-(P(X = 0)+P(X = 1))=1-(0.0147 + 0.0947)=1 - 0.1094=0.8666$

Answer:

0.8666