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. (round your answers to three decimal places.) (a) e(x) and v(x) e(x) = 2.659 v(x) = 1.931 (b) p(x ≤ 3) .6321 (c) p(1 ≤ x ≤ 3) .5269

I'm assuming is the shape parameter and is the scale parameter. Then the PDF is

a. The expectation is

To compute this integral, recall the definition of the Gamma function,

For this particular integral, first integrate by parts, taking

Substitute , so that :

The variance is

The second moment is

Integrate by parts, taking

Substitute again to get

Then the variance is

b. The probability that is

which can be handled with the same substitution used in part (a). We get

c. Same procedure as in (b). We have

and

Then

D:1/2

Step-by-step explanation: 3/6 simplifies to 1/2, making them equal. Landon got his answer by simplifying the problem incorrectly. 1/3 is equal to 2/6