# How do I find the mean of a normal distribution if I know a percentile and the variance?

I am trying to find a way to calculate the mean of my normal distribution if I know a certain percentile. Sounds simple, but I'm really stuck.

The purpose is to calculate a limit of detection (LD) based on Lloyd Currie's work on the issue and his proposal for a IUPAC definition. First, I calculate a critical value (LC) above which the chance of finding noise is less than e.g. 5%, so that I can control my false positives. Now, in order to control my false negatives, I define the LOD as the point above which a true signal will have a chance of less than e.g. 5% that an estimator falls below the LC.

Critical value LC, limit of detection LD and percentages alpha/beta for false positives/negatives http://www.intechopen.com/source/html/46058/media/image9.png

So I have a value where a certain percentage has to be reached, and in order to achieve that, I need to set my mean (LD) accordingly. (The variance is fixed.)

My question is, can I do this mathematically somehow? In Excel and IDL, the only functions I've found were either for a standard normal distribution, or they needed the mean as an input and couldn't output it. And since the integral of the normal distribution is the error function, I can't simply solve the equation for µ.

Any ideas how to tackle this problem?

Let $\Phi^{-1}(p)$ denote the $p$-quantile of the standard normal distribution. The normal being a location scale family, its quantiles $F^{-1}(p)$ are related to that of the standard normal by $$F^{-1}(p)=\mu+\sigma\Phi^{-1}(p)$$ By assumption, $F^{-1}(p)$ and $\sigma$ are known, and you may compute $\Phi^{-1}(p)$ corresponding to the known $p$-quantile. Then, solve for $\mu$.

Recall that the cumulative distribution function (CDF) $F$ for the normal distribution with mean $\mu$ and standard deviation $\sigma$ can be written in terms of the standard normal CDF $\Phi$ as follows:

$$F(x; \mu, \sigma) = \Phi\left( \frac{x - \mu}{\sigma} \right)$$

So you know $\sigma$, $p = F(x; \mu, \sigma)$ and want to solve for $\mu$.

$$p = \Phi\left( \frac{x - \mu}{\sigma} \right)$$

Hence:

$$\mu = x - \sigma \Phi^{-1}(p)$$

• Thank you! Quick question, are you missing an 1/sigma in your first formula? The result seems to be correct, but the normalisation factor is missing in the first formula (maybe just a typo or I'm an idiot)? – PoorYorick Oct 21 '16 at 12:28
• @Fearabbit What 'normalization' factor are you thinking should be there? – Glen_b -Reinstate Monica Oct 21 '16 at 16:16
• Do you maybe confuse the distribution function and the density? – Christoph Hanck Oct 21 '16 at 17:07
• Yeah, as I suspected, I was confused by the integral. You can transform the CDF into the standard CDF easily by substitution, but then you get a term for the derivative, and that cancels out the $1/\sigma$. – PoorYorick Oct 25 '16 at 9:15