Weibull distribution The Weibull distribution is widely used in statistical prob

Question

Weibull distribution

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 = 2.8 and = 220. (a) 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 (b) What is the probability that a specimen's lifetime is between 100 and 250? (Round your answer to four decimal places.) c what value is such that exactly 50% of all specimens have lifetimes exceeding that value? Round your answer to three decimal places. hr

Explanation / Answer

Here,

= 2.8

= 220

(a) Here CDF of weibull distribution is

F(x) = 1 - e-(x/)^

At most 250 :

F(x < 250) = 1 - exp [-(250/220)^ 2.8]

F(x =< 250) = 1- 0.2392 = 0.7608

Less than 250

F(x < 250) = F(x =< 250) - F(X = 250) = 0.7608 - 0.0038 = 0.7570

More than 300 :

F(x > 300) = 1 - [ 1- exp [-(300/220)^ 2.8] = 0.0923

(b) Pr(100 < X < 250) = F (250) - F(100) = 1 - exp [-(250/220)^ 2.8] - 1 + exp [-(100/220)^ 2.8]

= 0.7608 - 0.1041 = 0.6567

(c) Here let say 50% of all specimen exceeds that life. That means we want to know what is median here. Let say that value is X0 .

P(x > X0) = 0.5

so, 1 - [1 - e-(x/)^] = 0.5

1 - e-(x/)^ = 0.5

e-(x/)^ = 0.5

- (x0/)^ = ln (2)

(x0/220)2.8 = ln (2)

2.8 ln (x0/220) = ln ln (2)

X0 = 193 hours

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