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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.

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1 Answer 1

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.

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