Question

Of all customers purchasing automatic garage-door openers, 60% purchase a chain-driven model. Let X = the...

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

Homework Answers

Answer #1

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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