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?

a-

b- 0.4612

c- 0.1475

d- µ = 11.25 ; = 15*0.75*0.25 = 2.8125

e- 0.3093

Step-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?

where  

are combinations of 15 elements taken k at a time

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

µ = 15*0.75 = 11.25

= 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

instagram is better than go there instead lol

step-by-step explanation: