# How to calculate minimal member of set having mean and standard deviation?

Assume that we have given standard deviation and mean, and we know that our data follow normal distribution. ~ N(100, 16).

What is the technique to calculate the minimal value that can appear in the set?

We know that tennis players ball speed is distributed with a mean of 100 and standard deviation of 16. Find the minimal speed needed to be member of top players list (2% of all players).

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There is no minimum possible value (other than that imposed by reality: e.g., ball speed can’t less than $0$). Given a large enough data set, the minimum can be arbitrarily small. In your example it’s unlikely that you’ll find anyone below $3$ standard deviations below the mean. – Brian M. Scott Jan 19 '13 at 17:51

We have to give an explicit criterion for "top players list." There is no universally applicable criterion. Perhaps we can use $99$-th percentile.

Edit: The question was changed, and now defines top player as $98$-th percentile. So we use that from here on.

Let $X$ be the ball speed of a randomly chosen player. We may want to find the speed $x$ such that $\Pr(X\le x)=0.98$. For any $x$, we have $$\Pr(X\le x)=\Pr\left(\frac{X-\mu}{\sigma } \right)\le \frac{x-\mu}{\sigma},$$ where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.

But $\frac{X-\mu}{\sigma}$ has standard normal distribution. So let $Z$ be standard normal. We want $$\Pr\left(Z\le \frac{x-100}{16} \right)=0.98.$$ From tables of the standard normal, we have $\Pr(Z\le 2.05)\approx 0.98$.

So the cutoff point $x$ is given approximately by $$\frac{x-100}{16}\approx 2.05.$$

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As I read the question, $\mu=100$ and $\sigma=16$ apply to the list of top players, not to the list of all players. – Brian M. Scott Jan 19 '13 at 17:53
Ok I see that I skipped meaningfull part of question. So simply now I have assume eg. 2% and take 0.98 quantile. Am I right? Also, why did u squareroot standard deviation? In formula you given there's just to put standard deviation as is. – mickula Jan 19 '13 at 17:56
@mickula: I usually specify normal by giving mean, variance. But it looks as if your question says standard deviation is $16$. Have changed answer to reflect that. – André Nicolas Jan 19 '13 at 18:04
Seems like it was really easy. Thanks to @BrianM.Scott he pointed me that it goes about whole population of tennis players, not only about top, and ofc to you, Andre – mickula Jan 19 '13 at 18:06