assume that when human resource managers are randomly selected, 42% say job applicants should follow up…

assume that when human resource managers are randomly selected, 42% say job applicants should follow up within two weeks. if 9 human resource managers are randomly selected, find the probability that at least 3 of them say job applicants should follow up within two weeks.\n\nthe probability is \n(round to four decimal places as needed.)

assume that when human resource managers are randomly selected, 42% say job applicants should follow up within two weeks. if 9 human resource managers are randomly selected, find the probability that at least 3 of them say job applicants should follow up within two weeks.\n\nthe probability is \n(round to four decimal places as needed.)

Answer

Explanation:

Step1: Define binomial parameters

$n=9$, $p=0.42$, $q=1-p=0.58$ We calculate $P(X\geq3) = 1 - P(X=0)-P(X=1)-P(X=2)$

Step2: Calculate $P(X=0)$

Binomial formula: $P(X=k)=\binom{n}{k}p^kq^{n-k}$ $P(X=0)=\binom{9}{0}(0.42)^0(0.58)^9 = 1\times1\times0.58^9\approx0.0073$

Step3: Calculate $P(X=1)$

$P(X=1)=\binom{9}{1}(0.42)^1(0.58)^8 = 9\times0.42\times0.58^8\approx0.0476$

Step4: Calculate $P(X=2)$

$P(X=2)=\binom{9}{2}(0.42)^2(0.58)^7 = 36\times0.42^2\times0.58^7\approx0.1433$

Step5: Compute complementary probability

$P(X\geq3)=1 - 0.0073 - 0.0476 - 0.1433$

Answer:

0.8018