Suppose that lifetime of bulb of type n is exponentially distributed with mean value n years, where $n = 1, 2, 3, 4, 5$.
A bulb is equally likely be chosen from these five types.
(a) Find the probability that the selected bulb can survive for at least 3 years.
(b) Find the variance of the lifetime of the selected bulb.
For (a) I split the cases into 5 cases and calculated the $1-P(X<2)$ which equated to $e^{2/n}$ which led to $\sum_{i=1}^{5}0.2 * e^{2/i}$
For (b) I'm not sure how to start, am i supposed to use $Var(X) = 0.2Var(X_1) + 0.2Var(X_2) + … + 0.2Var(X_5)$ since they have independent lifetimes.
Best Answer
If you do not want to use the law of total variance, which would give $\frac{1+4+9+16+25}{5} + 2$, then an alternative approach here is:
The mean of $X_n$ is $n$ and the variance of $X_n$ is $n^2$, so $E[X_n^2]=2n^2$.
You can average the means for your mixture distribution so $E[X]=\frac{1+2+3+4+5}5=3$.
You cannot simply average the variances but you can average the $E[X_n^2]$ to get $E[X^2]=\frac{2+8+18+32+50}{5}=22$.
That gives $\text{Var}(X)=E[X^2]-(E[X])^2=22-9=13$ for your mixture distribution.