The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3. Compute the following.

math Statistics-And-Probability

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Answer 1

Answer #1

a)

mean=E(X)=μ =

β*τ(1+1/α)=

2.659

varianceVar(x)=σ²=

β[τ(1+2/α)-(τ(1+1/α))²]=

1.931

b)

P(X<5)=(1-exp(-(5/3)^2))=

0.938

c)

P(1.50.717

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Answer 2

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3

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