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Suppose that the mean for some data is 75 mm and the standard deviation is 10 mm. The distribution for this data is not necessarily normal.

Is it possible to determine the probability of the distance being greater than 73 mm?

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You can not determine the probability, but using a one sided Chebyshev inequality you can get bounds. I suggest you play around by considering different distributions where your data can only take two values to see how different the answers can be. What is the max/min that you can make it for these limited distributions? –  Aaron Oct 4 '12 at 3:57
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Expanding on my comment, let us consider distributions which take only two values and have mean 75mm and standard deviation 10mm. Every such distribution will be a shifted and resealed Bernoulli distribution.

Let $X_p$ be the random variable that takes the value $1$ with probability $p$ and the value $0$ with probability $q=1-p$. The mean is $p$ and the standard deviation is $\sqrt{pq}$. Therefore, we must scale by $\frac{10}{\sqrt{pq}}$ to correct the standard deviation, and since $\frac{10X_p}{\sqrt{pq}}$ has mean $10\sqrt{p/q}$, our desired distribution is the distribution of $Y_p=\frac{10X_p}{\sqrt{pq}}+75-10\sqrt{p/q}$

What is the probability that $Y_p\geq 73$? Since $(75-73)/10=1/5$, it will be the same as the probability that $X_p\geq p-\frac{\sqrt{pq}}{5}$. This will be $p$ if $p-\frac{\sqrt{pq}}{5}>0$, and $1$ otherwise. Since $p-\frac{\sqrt{pq}}{5}$ has a root at $p=1/26$, we have that the probability for this simple family of distributions lies in $(1/26,1]$.

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