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

step-by-step explanation:

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00