Question

Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is γ = 3.5 and that the excess X − 3.5 over the minimum has a Weibull distribution with parameters α = 2 and β = 1.5.
(a) What is the cdf of X?
F(x) = 0 x < 3.5
1−e^−((x−3.5)2.5​)2 x ≥ 3.5
(b) What are the expected return time and variance of return time? [Hint: First obtain
E(X − 3.5)
and
V(X − 3.5).]
(Round your answers to three decimal places.)

E(X) = 10^−1 weeks
V(X) = (10^−1 weeks)2

(c) Compute
P(X > 6).
(Round your answer to four decimal places.)

Answer 1

Answer:

Step-by-step explanation: see attachment for solution

Answer 2

If X is the time in 10 to 1 week of the shipment of a defective product until the customer returns the product.  

• Now suppose the mini returns y = 3.5 and the excess X is 3.5 over the mini has a Weibull distribution with parameters Alfa 2 and Beta 1.5.

The shipment of the defective product will be  

• A. 1-e x(25)2.

• B. 5/4 .

Expected value is E (X) = 5.7125.

• C. P(X>5) = 0.3679.

Learn more about the shipment of a defective product

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