Question

Of all customers purchasing automatic garage-door openers, 70% purchase a chain-driven model. Let X = the number among the next 15 purchasers who select the chain-driven model. (a) What is the pmf of X? b(x; 15, 0.7) nb(x; 15, 0.3) h(x; 11, 15, 70) b(x; 15, 0.3) h(x; 4, 15, 70) nb(x; 15, 0.7) Correct: Your answer is correct. (b) Compute P(X > 10). (Round your answer to three decimal places.) P(X > 10) = .515 Correct: Your answer is correct. (c) Compute P(6 ≤ X ≤ 10). (Round your answer to three decimal places.) P(6 ≤ X ≤ 10) = .481 Correct: Your answer is correct. (d) Compute μ and σ2. μ = 10.5 Correct: Your answer is correct. σ2 = 3.15 Correct: Your answer is correct. (e) If the store currently has in stock 10 chain-driven models and 7 shaft-driven models, what is the probability that the requests of these 15 customers can all be met from existing stock? (Round your answer to three decimal places.)

Answer 1

Answer:

P(X > 10), n = 15, p = 0.7

P(X > 10) =P(10 < X ≤ 15) = P(11 ≤ X ≤ 15) = P(X = 11, 12, 13, 14, 15)

=P(X = 11) + P(X =12) + P(X = 13) + P(X =14) + P(X = 15) (because these are disjoint events)

Step-by-step explanation:

See attached image for detailed explanation