Mathematics, 07.04.2020 18:20 alanflores40
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?
Answers: 2
![f(x,\alpha, \beta ) = \left \{ { {\frac{\alpha }{\beta^{\alpha } }x^{\alpha -1} e^{-[\frac{x}{\beta } ]^\alpha} \ \ \ \ \ \ \\ \\ , x \ge0} \atop {0 \ \ \ \ \ \ \ \ Otherwise }} \right.](/tpl/images/0586/6829/cd32b.png)
![P(x) = \int\limits^x_0 {\frac{\alpha }{\beta^{\alpha } } x^{\alpha -1 } e^{-[\frac{x}{\beta } ]^{\alpha }} } \, dx](/tpl/images/0586/6829/4050d.png)
and 60 for 
0.05 for P(x)![0.05 = \int\limits^x_0 {\frac{2}{60^2} x^{2-1} e^{-[\frac{x}{60}]^2 } \, dx](/tpl/images/0586/6829/6eb9b.png)
![0.05 = \frac{2}{3600}\int\limits^x_0 {xe^{[-\frac{x^2}{3600} ]}} \, dx](/tpl/images/0586/6829/5ebd3.png)
![0.05 = -\frac{1800}{1800} [e^{[-\frac{x^2}{3600} ]}]\left {{x} \atop {0}} \right.](/tpl/images/0586/6829/11d8c.png)
![0.05 = - [e^{-[\frac{x^2}{3600}] } -1]](/tpl/images/0586/6829/95fe5.png)
![0.05 = -e^{[-\frac{x^2}{3600} ]} +1](/tpl/images/0586/6829/bbb33.png)
![-0.95 = -e^{[-\frac{x^2}{3600} ]}](/tpl/images/0586/6829/08aac.png)
![0.95 = e^{[-\frac{x^2}{3600} ]}](/tpl/images/0586/6829/e6b06.png)
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7. The length of time (in months after maintenance) until failure of a bank’s surveillance televisio...
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