a school buys 60% of its light bulbs from supplier A and 40% from supplier B. the light bulbs from both suppliers look identical but light bulbs from supplier A have exponentially distributed lifetimes with a mean of 600 hours....the lifetimes of all bulbs are independent of each other.
c) find approximately the probability that 100 light bulbs from supplier A last more than 50,000 hours in total.
in the mark scheme Xt = N(100*600, 100*600*600) but i thought the parameters are supposed to be N(np,npq) where n = 100, p is a probability and q = (1-p)
Number the bulbs. For $i=1$ to $100$, let $X_i$ be the lifetime of bulb $i$. Let $X=X_1+\cdots +X_{100}$. We want $\Pr(X\gt 50000)$.
Let exponentially distributed random variable $T$ have parameter $\lambda$, that is, density function $\lambda e^{-\lambda t}$ for $t\gt 0$. Then $T$ has mean $\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^2}$. (This is a standard fact that can be verified by integration,)
So all the $X_i$ have variance $(600)^2$. The $X_i$ are independent, so $X$ has variance $(100)(600)^2$.
Since $X$ is a sum of a fairly large number of reasonably nice identically distributed independent random variables, the distribution of $X$ is reasonably close to the normal with the same mean and variance.
Remark: When we are approximating a binomially distributed random variable $Y$with parameters $p$ and $n$, then since the variance of $Y$ is $npq$, that is the appropriate variance for the approximating normal.