Answer:
The frequency by which the equipment should receive periodic maintenance is [tex]x= 13.59[/tex]
Step-by-step explanation:
A Wellbull distribution is mathematically represented as
[tex]f(x,\alpha, \beta ) = \left \{ { {\frac{\alpha }{\beta^{\alpha } }x^{\alpha -1} e^{-[\frac{x}{\beta } ]^\alpha} \ \ \ \ \ \ \\ \\ , x \ge0} \atop {0 \ \ \ \ \ \ \ \ Otherwise }} \right.[/tex]
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
[tex]P(x) = \int\limits^x_0 {\frac{\alpha }{\beta^{\alpha } } x^{\alpha -1 } e^{-[\frac{x}{\beta } ]^{\alpha }} } \, dx[/tex]
Substituting 2 for [tex]\alpha[/tex] and 60 for [tex]\beta[/tex] 0.05 for P(x)
[tex]0.05 = \int\limits^x_0 {\frac{2}{60^2} x^{2-1} e^{-[\frac{x}{60}]^2 } \, dx[/tex]
[tex]0.05 = \frac{2}{3600}\int\limits^x_0 {xe^{[-\frac{x^2}{3600} ]}} \, dx[/tex]
Let [tex]v = - \frac{x^2}{3600}[/tex]
=> [tex]dv = \frac{-xdx}{1800}[/tex]
=> [tex]-xdx = 1800\ dv[/tex]
Multiply both sides by minus
=> [tex]xdx = -1800dv[/tex]
Substituting this into the equation
[tex]0.05 = \frac{2}{3600} \int\limits^x_0 {-1800 e^u} \, du[/tex]
Integrating and substituting back for x
[tex]0.05 = -\frac{1800}{1800} [e^{[-\frac{x^2}{3600} ]}]\left {{x} \atop {0}} \right.[/tex]
substituting for the range of x and 0
[tex]0.05 = - [e^{-[\frac{x^2}{3600}] } -1][/tex]
[tex]0.05 = -e^{[-\frac{x^2}{3600} ]} +1[/tex]
[tex]0.05 -1 = -e^{-\frac{x^2}{3600} }[/tex]
[tex]-0.95 = -e^{[-\frac{x^2}{3600} ]}[/tex]
[tex]0.95 = e^{[-\frac{x^2}{3600} ]}[/tex]
Taking ln of both sides
[tex]-0.05129 = - \frac{x^2}{3600}[/tex]
making x the subject of the formula
[tex]x = \sqrt{0.05129 * 3600 }[/tex]
[tex]x= 13.59[/tex]
