Question

One hundred items are simultaneously put on a life test. Suppose the lifetimes

Answer 1

Answer:

a) $\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

b) $\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

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 ( $i-1$ )st and the $ith$ failures. Then, the $T_{i}$ are independent with $\mathrm{T}_{\mathrm{i}}$ being exponential with rate $\frac{(101-i)}{200} .$

Therefore,

a) $E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]$

$=\sum_{i=1}^{5} \frac{200}{101-i}$

$\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

$b)$

The variance is given by, $\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]$

$\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$