9. Of all customers purchasing automatic garage door openers, 75% purchase chain-driven model. Let X = the number among the next 15 purchasers who select the chain-driven model. a. What is the frequency function (pmf) of X? b. Compute P(X > 11). c. Compute P( 6 ≤ X < 10). d. Compute µ and σ2. e. If the store currently has in stock 10 chain-driven models and 8 shaft-driven models, what is the probability that at least 7 out of the 15 customers select a chain-driven model from this stock?

Answer:a- [tex]\bf P(X=k)=\binom{15}{k}0.75^k(1-0.75)^{15-k}=\binom{15}{k}0.75^k(0.25)^{15-k}\;(0\leq k\leq 15)[/tex]b- 0.4612c- 0.1475d- µ = 11.25 ; [tex]\bf \sigma^2[/tex] = 15*0.75*0.25 = 2.8125e- 0.3093Step-by-step explanation:This is a binomial distribution with probability of “success” (probability of purchasing chain-driven model) is 75%=0.75 a. What is the frequency function (pmf) of X? [tex]\bf P(X=k)=\binom{15}{k}0.75^k(1-0.75)^{15-k}=\binom{15}{k}0.75^k(0.25)^{15-k}\;(0\leq k\leq 15)[/tex] where  [tex]\bf \binom{15}{k}[/tex] are combinations of 15 elements taken k at a time [tex]\bf \binom{15}{k}=\frac{15!}{k!(15-k)!}[/tex] b. Compute P(X > 11) P(X > 11) = P(X = 12)+P(X = 13)+P(X = 14)+P(X = 15) = 0.4612 c. Compute P( 6 ≤ X < 10). P( 6 ≤ X < 10) = P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9) = 0.1475 d. Compute µ and [tex]\bf \sigma^2[/tex] µ = 15*0.75 = 11.25 [tex]\bf \sigma^2[/tex] = 15*0.75*0.25 = 2.8125 e. If the store currently has in stock 10 chain-driven models and 8 shaft-driven models, what is the probability that at least 7 out of the 15 customers select a chain-driven model from this stock? We want P( 7≤ X ≤ 10) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.3093