Answer:
a) [tex]\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}[/tex]
b) [tex]\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}[/tex]
Step-by-step explanation:
Given:
The lifetimes of the individual items are independent exponential random variables.
Mean = 200 hours.
Assume, Ti be the time between ( [tex]i-1[/tex] )st and the [tex]ith[/tex] failures. Then, the [tex]T_{i}[/tex] are independent with [tex]\mathrm{T}_{\mathrm{i}}[/tex] being exponential with rate [tex]\frac{(101-i)}{200} .[/tex]
Therefore,
a) [tex]E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right][/tex]
[tex]=\sum_{i=1}^{5} \frac{200}{101-i}[/tex]
[tex]\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}[/tex]
[tex]b)[/tex]
The variance is given by, [tex]\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T][/tex]
[tex]\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}[/tex]
