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 α = 2.6 and β = 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

Answer 1

a

P(X \le 250 ) = 0.7564 [/tex] ,
`P(X <  250 )  =  0.7564` ,

`P(X <  300 )  =  0.09922`

b

`P(100 <  X  < 250 ) =0.644`

c

`x  = 192.1`

Explanation:

From the question we are told that

The value for
`\alpha  =  2.6`

The value for
`\beta = 220`

Generally the Weibull distribution function is mathematically represented as

`F( x , \alpha ,  \beta ) =  \left \{  0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  x < 0} \atop { 1- e^{-((x)/(\beta ) )^(\alpha ) }}\ \ \ \ \ \ x \ge 0} \right`

Generally the probability that a specimen's lifetime is at most 250 is mathematically represented as

`P(X \le  250 ) =  F(250, 2.7 , 220 )`

`P(X \le 250 )=1 - e^{- ((250)/(220) )^(2.7)}`

`P(X \le 250 ) =  1 - 0.2436`

`P(X \le 250 ) =  0.7564`

Generally the probability that a specimen's lifetime is less than 250

`P(X <  250 ) =  F(250, 2.7 , 220 )`

[texP(X < 250 ) =1 - e^{- (\frac{250}{220} )^{2.7}}[/tex]

`P(X <  250 )  =  1 - 0.2436`

`P(X <  250 )  =  0.7564`

Generally the probability that a specimen's lifetime is more than 300

`P(X >  300 ) = 1- p(X \le 300 )`

`P(X >  300 ) = 1-  F(300, 2.7 , 220 )`

[texP(X < 300) =1- [1 - e^{- (\frac{300}{220} )^{2.7}}][/tex]

`P(X <  300 )  =  0.09922`

Generally the probability that a specimen's lifetime is between 100 and 250 is

`P(100 <  X  < 250 ) =  P(X < 250) - P(X < 100)`

=>
`P(100 <  X  < 250 ) =F(250 , 2.7 , 220 ) - F(100 , 2.7 , 220 )`

=>
`P(100 <  X  < 250 ) =(1 - e^{-((250)/(220))^(2.7)}) - (1 - e^{-((100)/(220))^(2.7)})`

=>
`P(100 <  X  < 250 ) = (1 - 0.244 ) - (1- 0.888)`

=>
`P(100 <  X  < 250 ) =0.644`

Generally the value such that exactly 50% of all specimens

`P(X > x) = 1-P(X <  x) = 0.50`

=>
`P(X > x) = 1- (1 - e^{- ((x)/(220)) ^(2.7)}) = 0.50`

=>
`P(X> x ) = e^(- (x)/(20))^(2.7)  = 0.50`

=>
`P(X> x ) = (- (x)/(20))^(2.7)  = ln0.50`

=>
`P(X> x ) =  (x)/(20)  =[ -ln0.50 ] ^{frac{1}{2.7}}`

=>
`x  = 220[ -ln0.50 ] ^{frac{1}{2.7}}`

=>
`x  = 192.1`