How to compute the mean of a normal distribution, given the standard deviation and the 5% percentile?

I apologize if this is a trivial question, but my Google-fu is failing me. All that my queries are returning are variations of this question.

Let $q_0$ be the 5th percentile (or the 95th, for that matter), and $\sigma$ be the standard deviation of a Normal distribution. How do I find the mean $\mu$?

FIrst, locate the $z$-score for the 5th percentile: $z_5$.

Second, use the formula for the $z$-score with this number, the standard deviation, and $q_0$: $$z_5 = \frac{q_0-\mu}{\sigma}$$ Plug in the three values you know and solve for $\mu$.

Mean and standarddeviation are completely independend. Thus it is not possible to estimate one from the other. In principal all combinations of mean and std are possible.

If you have both the 5% and 95% percentile (p5 and p95). Here the percentile is the x-value for which the integral over the propability distribution reaches 5% thus

$\int_{-\infty}^{p5} p(x') dx = 0.05$

and the 95% percentile respectively is

$\int_{-\infty}^{p95} p(x) dx = 0.95$

Then an easy estimate of the mean is just

$m = \frac{1}{2}(p5 + p95)$

since normal distribution is symmetric (all odd moments are 0).

• The information in this answer is correct, however, it doesn't apply to the scenario above (which is a very common homework style question for intro stats classes). To clarify, the type of questions like above specifically only provide one of the quantiles. – Gregg H Apr 12 '18 at 12:59