Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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.

share|cite|improve this question
up vote 2 down vote accepted

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.

We did not use the Central Limit Theorem, nor the Markov Inequality, since we were asked to assume normality. The Poisson distribution has no connection with the problem.

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.