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How do I calculate the mean of a portion of a normal distribution. In other words, say I have a normal distribution of the heights of adult males. The mean is 70" and the standard deviation is 4". What is the average height of males above the 95th percentile? What is the average height of all males below the 95th percentile? How do I calculate this?

Someone asked a similar question here which was never answered clearly:

Mean value from part of normal distribution

This question has a practical application for my work where I am running a power plant at 10.3 MW mean operating point with a standard deviation of 0.2 MW. I would like to know the average power when I am above 10 MW. Or the mean of all points above 10 MW.

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4 Answers 4

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Suppose $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $Z=\frac{X-\mu}{\sigma}$ is normal with mean $0$ and standard deviation $1$, and $X=\sigma Z + \mu$. Then

$$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ \frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma} \right ] \right. \right ] \\ = \mu + \sigma \frac{E \left [ Z 1_{[\frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma}]}\right ]}{P \left ( Z \in \left [\frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma} \right ] \right )} \\ = \mu + \sigma \frac{\int_{\frac{a-\mu}{\sigma}}^{\frac{b-\mu}{\sigma}} x e^{-\frac{x^2}{2}} dx}{\int_{\frac{a-\mu}{\sigma}}^{\frac{b-\mu}{\sigma}} e^{-\frac{x^2}{2}} dx}.$$

The numerator can be calculated in terms of elementary functions using the FTC, since $-x e^{-x^2/2}$ is the derivative of $e^{-x^2/2}$.

The denominator is an integral of the standard normal density, which cannot be calculated in terms of elementary functions, but can be easily evaluated using software, using functions generally called either "erf" (for "error function") or "normcdf".

$$= \mu + \sigma \frac{ e^{-\frac{\left ( a-\mu\right ) ^2}{2\sigma^2}} - e^{-\frac{\left ( b-\mu\right ) ^2}{2\sigma^2}} }{ \sqrt{\frac{\pi}{2}}\left ( \text{erf}\left ( \frac{b-\mu}{\sqrt{2}\sigma}\right ) - \text{erf}\left ( \frac{a-\mu}{\sqrt{2}\sigma}\right ) \right ) }$$

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  • $\begingroup$ Sorry about that. I realize it's an indicator now! @Ian $\endgroup$
    – aaiezza
    Jun 22, 2017 at 16:05
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I think you just need to take the values/points (which are above 10MW), sum them up and divide the sum by their count. That's all, no? Whether the distribution is normal or not, that's not relevant here, I think.

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  • $\begingroup$ That only applies if you have discrete points, otherwise use integral $\endgroup$
    – qwr
    Nov 25, 2015 at 16:58
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The "mean" of a continuous probability distribution, P(x), is, by definition, the integral of xP(x). To restrict a normal distribution, $y= Ae^{\frac{(x- \mu)^2}{\sigma^2}}$, between x= a and x= b, with a< b, we have to divide by the probability x is between a and b, the integral of P(x) between a and b. Here, mean is $\frac{\int_a^b xe^{\frac{(x- \mu)^2}{\sigma^2}}dx}{\int_a^b e^\frac{(x- \mu)^2}{\sigma^2}dx}$

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Suppose for simplicity that you have a standard normal $X$ with pdf $f$. One of the main properties of $f$ is that it satisfies $f'=-xf$ which implies $\int_a^b xf\,dx=-\int_a^b df=f(a)-f(b)$.

It follows that $$E(X|X\in[a,b])=\frac {f(a)-f(b)}{\Phi(b)-\Phi(a)}$$

Set $b=\infty$ ($\Phi(\infty)=1, f(\infty)=0$) if you want a one-sided bound.

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  • $\begingroup$ What if there is a sign change prohibiting the substitution? (Your formula actually still works out, but you haven't explained why.) Also, you should have $-\int_{f(a)}^{f(b)} df$. $\endgroup$
    – Ian
    Nov 25, 2015 at 17:02
  • $\begingroup$ @Ian What sign change and how can it prohibit a substitution? $\endgroup$
    – A.S.
    Nov 25, 2015 at 17:06
  • $\begingroup$ You may only make a substitution if the substitution function is monotonic over the entire interval in question. This is part of the hypotheses of the change of variables theorem. $\endgroup$
    – Ian
    Nov 25, 2015 at 17:08
  • $\begingroup$ @Ian $\int_a^b f'dx=\int_a^b df=f(b)-f(a)$ is a FTC. No change of variables., no need for monotonicity. $\endgroup$
    – A.S.
    Nov 25, 2015 at 17:10
  • $\begingroup$ Ah, I see, carry on. You still have the error in the limits, though. $\endgroup$
    – Ian
    Nov 25, 2015 at 17:11

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