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I have a normal distribution curve figure. It is vertically chopped into equal-width segments based on the distance from the mean value my (in the middle of the curve, where it tips), the distance expressed as multiples of the standard deviation sigma. In each such segment, there is a percentage digit. Also, the number of observations is known. So, if you'd like to know how many observations are between two vertical bars, you'd simply add the percentages, divide by 100, and multiply the number of observations. (Correct me if I'm wrong.) But, how do I do if I have the interval start and endpoint not precisely at any two bars, but at, say, my - 2.3sigma, and my + 0.8sigma. Then obviously I cannot use the figure; I must use some formula or definition. Although I suspect this is not complicated at all, I encourage you to explain it in as much detail as you see fit.

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If $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, then the random variable $\dfrac{X-\mu}{\sigma}$ has standard normal distribution. We have $$\Pr(X\le x)=\Pr\left(\frac{X-\mu}{\sigma}\le \frac{x-\mu}{\sigma}\right)=\Pr\left(Z\le \frac{x-\mu}{\sigma}\right),$$ where $Z$ is standard normal.

In particular, $\Pr(X\le \mu +k\sigma)=\Pr(Z\le k)$.

For a useful range of $z$, the probability that $Z\le z$ is available from tables. You can find such a table online. There is also software that will compute it.

If you go to the table linked to above, you will find that $\Pr(Z\le 2.3)\approx 0.9893$. It follows that $\Pr(X\le \mu +2.3\sigma)\approx 0.9893$.

Tables of the standard normal generally only give $\Pr(Z\le z)$ for non-negative $z$. But the probabilities for negative $z$ are then easy to figure out, by the symmetry of the standard normal about the origin.

For example, $\Pr(Z\le -1.3)=\Pr(Z\ge 1.3)=1-\Pr(Z\le 1.3)$.

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Excellent, thanks. – Emanuel Berg Feb 9 '13 at 23:08

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