- #1
orangeIV
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Hi, I'm currently trying to work through a problem about calculating the most likely time for a hard disk to fail:
Hard disks fail with a probability per unit time: [itex]\alpha (t) = \alpha _0 t [/itex] where [itex]\alpha_0 = 0.5[/itex] years.
I know that the answer is [itex]t_{modal} = \frac{1}{\sqrt{\alpha_0}}[/itex], but am having problems deriving this. Here is what I've done so far:
The probability distribution can be calculated as follows:
[itex]f(x) = \alpha (t) e^{-\int \alpha (t) dt} = \alpha (t) e^{-\frac{1}{2} \alpha_0 t^2} [/itex]
The most likely time for the disk to fail will be when [itex]\frac{df}{dt} = 0 [/itex]. So when
[itex] 0 = -{\alpha_0}^2 t^2 e^{-\frac{1}{2} \alpha_0 t^2} [/itex]
This is where I get stuck. Is this the correct approach? Any ideas about how how I might proceed :)
Thanks
Hard disks fail with a probability per unit time: [itex]\alpha (t) = \alpha _0 t [/itex] where [itex]\alpha_0 = 0.5[/itex] years.
I know that the answer is [itex]t_{modal} = \frac{1}{\sqrt{\alpha_0}}[/itex], but am having problems deriving this. Here is what I've done so far:
The probability distribution can be calculated as follows:
[itex]f(x) = \alpha (t) e^{-\int \alpha (t) dt} = \alpha (t) e^{-\frac{1}{2} \alpha_0 t^2} [/itex]
The most likely time for the disk to fail will be when [itex]\frac{df}{dt} = 0 [/itex]. So when
[itex] 0 = -{\alpha_0}^2 t^2 e^{-\frac{1}{2} \alpha_0 t^2} [/itex]
This is where I get stuck. Is this the correct approach? Any ideas about how how I might proceed :)
Thanks