# Central Limit Theorem/Markov's inequality

Here is the question:

Chip dies are manufactured in a facility where it was observed that the width of the die is normally distributed with mean 5mm and standard deviation $\sigma$. The manufacturer wants to guarantee that no more than 1 out of 100 dies fall outside the range of (5mm +/- 0.5mm). What should be the maximal standard deviation $\sigma$ of this manufacturing process?

My attempt at a solution:

I figured I could use the central limit theorem and Markov's inequality for this one:

thus-

Pr{die will be in range} = 99/100

I assumed that this should be a normal R.V. (because using a Poisson R.V. to solve this would be tedious)

And now I'm horribly stuck. Any advice as to where I went wrong?

Thank you.

Assume, without much justification except that we were told to do so, that the width $X$ of the die has normal distribution with mean $5$ and variance $\sigma^2$.
The probability that we are within $k\sigma$ of the mean $5$ (formally, $P(5-k\sigma\le X \le 5+k\sigma)$) is equal to the probability that $|Z|\le k$, where $Z$ has standard normal distribution. We want this probability to be $0.99$.
If we look at a table for the standard normal, we find that $k\approx 2.57$.
We want $k\sigma=0.5$ to just meet the specification. Solve for $\sigma$. We get $\sigma\approx 0.19455$, so a standard deviation of about $0.195$ or less will do the job.
Remark: The table that we used shows that the probability that $Z\le 2.57$ is about $0.995$. It follows that $P(Z>2.57)\approx 0.005$, we have $1/2$ of $1$ percent in the right tail. We also have by symmetry $1/2$ of $1$ percent in the left tail, for a total of $1$ percent, as desired.