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Homework answers / question archive / 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
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
=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 px(1-p)n-x
in our case
P(X=x)=10Cx 0.45x(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
