Answer to Question #120908 in Statistics and Probability for Abdul Rafay

Question

Recent studies have shown that 30% of employees in the insurance industry telecommute four days a week. If a random sample of 15 insurance industry employees is taken, what is the probability that
(a) Less than 6 telecommute four days a week?
(b) At least 9 telecommute four days a week?

Answer 1

This is a binomial distribution with $n=15,\;p=0.3.$

(a) $P(X<6)=\Sigma_{i=0}^5C_{15}^i0.3^i(1-0.3)^{15-i}=0.7216.$

(b) $P(X\ge9)=\Sigma_{i=9}^{15}C_{15}^i0.3^i(1-0.3)^{15-i}=0.0152.$