A company manufactures a brand of light bulb with a lifetime in months
that is normally distributed with mean 3 and variance 1. A consumer
buys a number of these bulbs with the intention of replacing them
successively as they burn out. The light bulbs have independent
lifetimes.What is the smallest number of bulbs to be purchased so that the
succession of light bulbs for at least 40 months with probability at
least 0.9772?
The chance that we only use one light bulb is $P(Z>\frac{40-3}{1})$. I don't know how to go about finding the probability that we use more than one light bulb. I don't know how to distribute the lifetime of each of the light bulb so that they add up more than 40. For example, if we use two light bulbs, we want the sum of their life to be more than 40 months. But we could have the first light bulb lives for 10 months and the second lives for more than 30 months or the first lives for more than 25 months and the second lives for 15 months. There are an infinite number of cases.
Best Answer
Let i.i.d. lifetimes $T_i \sim N(3,1)$. Then, $S_n= T_1+T_2+...+T_n \sim N(3n,n)$, where n is the variance of $S_n$.
We want $P(S_n > 40 ) > 0.9772$.
We have, $P(S_n>40)=P(Z=\frac{S_n-3n}{\sqrt{n}} > q) > 0.9772$ implies $q=2=\frac{ 40-3n}{\sqrt{n}}$. Thus, $n \ge 40$.