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tino4ka555 [31]
3 years ago
13

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.)

Mathematics
1 answer:
Aleks [24]3 years ago
5 0

Answer:

Step-by-step explanation: see attachment for solution

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4 years ago
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Answer: 0.243.

Step-by-step explanation:

The probability in this case is approcimated by the probability distribution formula for random variables. This formula is denoted as:

P(X=r) = nCr * p^r * q^n-r

Where n = selected number to be sampled and in this case, n= 3

r = varied number of sample and in this case, r=1

P = probability of success and in this case p = 0.10

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P(X=1) = 0.243.

3 0
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