Let X = the time (in 10−1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the m
inimum 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.)
(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.)
1 answer:
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(C)
Step-by-step explanation:
The volume of the conical pile is given by
Taking the derivative of V with respect to time, we get
Since r is always equal to h, we can set
so that our expression for dV/dt becomes
Solving for dh/dt, we get