Question

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*(

* nb*(

* h*(

* b*(

* h*(

* nb*(

(b)

Compute

* P*(

(Round your answer to three decimal places.)

* P*(

(c)

Compute

* P*(7 ≤

(Round your answer to three decimal places.)

* P*(7 ≤

(d)

Compute *μ* and *σ*^{2}.

*μ* = *σ*^{2} =

(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**

Answer #1

a) b(x; 15, 0.6)

b) P(X = x) = nCx * p^{x} * (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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