Question

The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 42 hours. A single plant operates three galvanizing lines that are assumed to operate independently. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that none of the lines experiences a surface finish problem in 42 hours of operation

Answer 1

0.0498

Explanation:

given,

The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 42 hours.

`\lambda = (1)/(42)`

a) probability that none of the lines experiences a surface finish problem in 42 hours of operation

=
`(1)/(42)(1)/(42)(1)/(42) \int_(42)^(\infty) \int_(42)^(\infty) \int_(42)^(\infty) (e^{(-x)/(42)})\ (e^{(-y)/(42)})\ (e^{(-z)/(42)})dx\ dy\ dz`

=
`((1)/(42))^3\int_(42)^(\infty) \int_(42)^(\infty) \int_(42)^(\infty) (e^{(-x)/(42)})\ (e^{(-y)/(42)})\ [\frac{e^{(-z)/(42)}}{(-1)/(42)}]_(42)^(\infty)dx\ dy\ dz`

=
`((1)/(42))^2\int_(42)^(\infty) \int_(42)^(\infty) (e^{(-x)/(42)})\ (e^{(-y)/(42)})\ e^(-1)dx\ dy`

=
`((1)/(42))^2\ e^(-1)[\frac{e^{(-y)/(42)}}{(-1)/(42)}]_(42)^(\infty)`

=
`((1)/(42)) \ e^(-1)\ e^(-1)[\frac{e^{(-x)/(42)}}{(-1)/(42)}]_(42)^(\infty)`

=
`e^(-1)\ e^(-1)\ e^(-1)`

=
`e^(-3)`

= 0.0498