I'm taking a probability theory class and I'm stuck on a question.
Here's the question:
A manufacturing plant utilizes 3000 electric light bulbs whose length of life is normal distributed with mean 500 hours and standard deviation 50 hours. To minimize the number of bubs that burn out during operating hours, all the bulbs are replaced after a given period of operation. How often should the bulbs be replaced if we want no more than 1% of the bulbs to burn out between replacement periods?
I understand that to solve this, I need to find the 1st percentile of distribution of light bulb lives. I just don't really get how to approach answering the question. (For reference, the answer is 383.5)
Any help would be greatly appreciated! Thank you!
Best Answer
Let $X$ be the lifetime of a bulb. We want to find $w$ such that $\Pr(X\le w)=0.01$. Now $$\Pr(X\le w)=\Pr\left(Z\le \frac{w-500}{50}\right),$$ where $Z$ is standard normal.
We want to find $z$ such that $\Pr(Z\le z)=0.01$. So we want the lower tail, the part below $z$, to have probability $0.01$.
The lower tail is not dealt with in the usual tables. But we can use symmetry. From a table, we have $\Pr(Z\gt 2.32)\approx 0.01$. So $\Pr(Z\le -2.32)\approx 0.01$.
It follows that we want $$ \frac{w-500}{50}\approx -2.32.$$ That gives $w\approx 384$.
Remark: A more precise evaluation of the point $z$ such that $\Pr(Z\le z)=0.01$ will give a more precise estimate.
In these matters, great precision is usually pointless, since the normality of the length of life is at best an approximation, and the mean and standard deviation are imperfectly known.