Of all customers purchasing automatic garage-door openers, 60% 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, 0.4)
nb(x; 15, 0.6)
h(x; 9, 15, 60)
b(x; 15, 0.6)
h(x; 6, 15, 60)
nb(x; 15, 0.4)
(b)
Compute
P(X > 11).
(Round your answer to three decimal places.)
P(X > 11) =
(c)
Compute
P(7 ≤ X ≤ 11).
(Round your answer to three decimal places.)
P(7 ≤ X ≤ 11) =
(d)
Compute μ and σ2.
μ = σ2 =
(e)
If the store currently has in stock 11 chain-driven models and 6 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.)
Statistics
a) b(x; 15, 0.6)
b) P(X = x) = nCx * px * (1 - p)n - x
P(X > 11)
= P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
= 15C12 * (0.6)^12 * (0.4)^3 + 15C13 * (0.6)^13 * (0.4)^2 + 15C14 * (0.6)^14 * (0.4)^1 + 15C15 * (0.6)^15 * (0.4)^0 = 0.091
c) P(7 < X < 11)
= P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)
= 15C7 * (0.6)^7 * (0.4)^8 + 15C8 * (0.6)^8 * (0.4)^7 + 15C9 * (0.6)^9 * (0.4)^6 + 15C10 * (0.6)^10 * (0.4)^5 + 15C11 * (0.6)^11 * (0.4)^4 = 0.814
d) = np = 15 * 0.6 = 9
= np(1 - p)
= 15 * 0.6 * (1 - 0.6)
= 3.6
e) P(9 < X < 11)
= 15C9 * (0.6)^9 * (0.4)^6 + 15C10 * (0.6)^10 * (0.4)^5 + 15C11 * (0.6)^11 * (0.4)^4
= 0.519
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