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 ? = 2.7 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

Answer 1

Explanation:

Given that

`\alpha =2.5` ,
`\beta =220`

The weibull distribution with parameters
`\alpha \ \ and \ \ \beta`

where
`\alpha =0 \ \ , \beta =0`

`F(x,\alpha ,\beta) \left|\begin{array}{cc}(\alpha )/(\beta ) x^{\alpha-1e^-(x/\beta)^\alpha &x\geq 0\\0&x<0\end{array}\right`

Then,

`F(x,\alpha ,\beta )=\left|\begin{array}{cc}0&x<0\\1-e^-^((x/\beta))^\alpha &x\geq 0\end{array}\right`

A) The probability that a specimen's lifetime is at most 250 is

`P(X\leq 250)=F(250,2.7,220)\\\\=1-e^-^(250/220)^(2.7)\\\\=1-0.2436\\\\=0.7564`

The probability that the specimen's life time is more than 300 is

`P(X>300)=1-P(X\leq 300)\\\\=1-F(300;2.7,220)\\\\=1-(1-e^-^{(300/220)^(2.7)`

`=e^-^{(300/220)^(2.7)`

= 0.0992

b)The probability of the specimen's lifetime is between 100 and 250

`P(100<X<250)=P(X<250)-P(X<100)\\\\=F(250;2.7,220)-F(100;2.7,220)\\\\=(1-e^-^((250/220)^2^.^7))-(1-e^-^((100/220)^2^.^7))\\\\=(1-0.2436)-1(1-0.8878)\\\\=0.6442`

c) The value such that exactly 50% of all specimens have lifetimes exceeding that value is

`P(X>x)=0.50\\\\1-P(X<x)=0.50\\\\1-(1-e^-^{(x/220)^(2.7)}=0.50\\\\e^-^{(x/220)^(2.7)}=0.50\\\\-(x/220)^(2.7)=In(0.50)\\\\(x/220)^(2.7)=-In(0.50)\\\\(x/220)=[-In(0.50)]^{((1)/(2.7))} \\\\x=220[-In(0.50)]^{((1)/(2.7))}`

x = 192.07