Question

7. The length of time (in months after maintenance) until failure of a bank’s surveillance television equipment follows a Weibull distribution with α=2 and β=60. If the bank wants the probability of a breakdown before the next scheduled maintenance to be .05, how frequently should the equipment receive periodic maintenance?

Answer 1

The frequency by which the equipment should receive periodic maintenance is
`x= 13.59`

Explanation:

A Wellbull distribution is mathematically represented as

`f(x,\alpha, \beta ) = \left \{ { {(\alpha )/(\beta^(\alpha ) )x^(\alpha -1) e^{-[(x)/(\beta ) ]^\alpha} \ \ \ \ \ \ \\ \\ , x \ge0} \atop {0 \ \ \ \ \ \ \ \ Otherwise }} \right.`

Where x is the frequency that the equipment should receive periodic maintenance

probability of a breakdown before the next scheduled maintenance is mathematically represented as

`P(x) = \int\limits^x_0 {(\alpha )/(\beta^(\alpha ) ) x^(\alpha -1 ) e^{-[(x)/(\beta ) ]^(\alpha )} } \, dx`

Substituting 2 for
`\alpha` and 60 for
`\beta` 0.05 for P(x)

`0.05 = \int\limits^x_0 {(2)/(60^2) x^(2-1) e^{-[(x)/(60)]^2 } \, dx`

`0.05 = (2)/(3600)\int\limits^x_0 {xe^{[-(x^2)/(3600) ]}} \, dx`

Let
`v = - (x^2)/(3600)`

=>
`dv = (-xdx)/(1800)`

=>
`-xdx = 1800\ dv`

Multiply both sides by minus

=>
`xdx = -1800dv`

Substituting this into the equation

`0.05 = (2)/(3600) \int\limits^x_0 {-1800 e^u} \, du`

Integrating and substituting back for x

`0.05 = -(1800)/(1800) [e^{[-(x^2)/(3600) ]}]\left {{x} \atop {0}} \right.`

substituting for the range of x and 0

`0.05 = - [e^{-[(x^2)/(3600)] } -1]`

`0.05 = -e^{[-(x^2)/(3600) ]} +1`

`0.05 -1 = -e^{-(x^2)/(3600) }`

`-0.95 = -e^{[-(x^2)/(3600) ]}`

`0.95 = e^{[-(x^2)/(3600) ]}`

Taking ln of both sides

`-0.05129 = - (x^2)/(3600)`

making x the subject of the formula

`x = √(0.05129 * 3600 )`

`x= 13.59`