Assume that when human resource managers are randomly selected, 45% say job applicants should follow up within 2 weeks if 10 human resource manages are randomly selected find the probability that fewer than 3 of them say job applicants should follow up within 2 weeks

Math Apr 29, 2022

Expert Solution

=0.0995

Step-by-step explanation

let p denote the probability job applicants should follow up within 2 weeks thus p=0.45

if we choose 10 of then and let x denote the number job applicants who follow up within 2 weeks

for 10 trials

x~bin(10, 0.45)

we know for for a binomial distribution

P(X=x) nCx p^x(1-p)^n-x

in our case

P(X=x)=10Cx 0.45^x(1-0.45)^10-x

so P(X<3)

=P(X=0)+P(X=1)+P(X=2)

=10C0 0.45⁰(1-0.45)^10-0 +10C1 0.45¹(1-0.45)^10-1 +10C2 0.45²(1-0.45)^10-2

=0.0025+0.0207+0.0763

=0.0995

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