The Weibull distribution is widely used in statistical problems relating to agin

Question

The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress. Use this distribution as a model for time (in hours) to failure of solid insulating specimens subjected to AC voltage. The values of the parameters depend on the voltage and temperature; suppose alpha = 2.2 and beta = 220. What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300? (Round your answers to four decimal places.) at most 250 less than 250 more than 300 What is the probability that a specimen's lifetime is between 100 and 250? (Round your answer to four decimal places.) What value is such that exactly 50% of all specimens have lifetimes exceeding that value? () hr

Explanation / Answer

X -speciement lifetime

(a)P(X<=250)= 0.734 ( using ms-excel function =WEIBULL(250,2.2,220,1))

P(X<250)=0.734 (( using ms-excel function =WEIBULL(300,2.2,220,1))

P(X>300)=1-P(X<300)=1-0.862-0.138

P(X<300)=0.862

(b) P(100<X<250)=P(X<250)-P(X<100)=0.734 - 0.162=0.572

(c) we want to find x such that

P(X<=x)=0.5 and corresponding x =187

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