# 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 y = 5.5 and that the excess X - 5.5 over the minimum has a Weibull distribution with parameters a = 2 and B = 1.5.(a) What is the cdf of X?x< 5.5F(x) = {x2 5.5(b) What are the expected return time and variance of return time? [Hint: First obtain E(X - 5.5) and V(X - 5.5).] (Round your answers to three decimal places.)E(X) =V(X) =|10-1 weeks(10-1 weeks)?(c) Compute P(X > 7). (Round your answer to four decimal places.)(d) Compute P(7 SXS 8.5). (Round your answer to four decimal places.)

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Step 1

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Step 2

(a)

The random variable (X ­– 5.5) is a 2-parameter Weibull distribution, with parameters α = 2, and β = 1.5. Thus, the random variable X is a 3-parameter Weibull distribution, with parameters α = 2, β = 1.5, and γ = 5.5.

The probability distribution function (pdf) and cumulative distribution function (cdf) of this distribution are given below:

Step 3

(b)

The expectation and variance for the 2-parameter Weibull distribution are:

Calculation:

Thus, the expectation and variance for the 3-parameter Weibull distribution are:

Calculation:

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