Estimate the mean of a normal distribution, given the standard deviation and the probability a value is below a quantity

Question

I am interested in knowing the mean of a normally distributed random variable for a population. However, in the data I have been given, I am only told the percentage of the population below a given (arbitrary) value. If I also know the standard deviation of that random variable, how can I derive the mean?

So, to try to put it in formal terms, I have random variable $X$ and I am trying to estimate the mean $\mu$ of $X$. I know the probability $p_i$ that $X$ is below a given value $x_i$ and I also know the standard deviation $\sigma$ of $X$. So how can I derive $\mu$ given $\sigma$, $p_i$, and $x_i$, where $p_i = P(X<x_i)$

Thank you for the help.

Edit: Based on a comment, here is a numerical example. I actually have a number of sub-populations I am trying to estimate the mean for, but a representative value is that 40% of the sub-population has a value less than -2. So $P(X<-2)=0.4$. I also know that $\sigma = 1.5$. I was looking through the formulas here, which dont exactly fit my application as it is a guide to estimating $\mu$ and $\sigma$ from two values, $x_1$ and $x_2$ which correspond to two percentiles, $p_1$ and $p_2$. So I am still searching for an example or formula that fits my particular problem.

Answer 1

If you mean that $P(X < -2) = 0.04,$ for a normal random variable $X$ with $SD(X) = \sigma = 1.5,$ then

$$P(X < -2) = P\left(Z = \frac{X-\mu}{\sigma} < \frac{-2-\mu}{1.5}\right) = 0.4,$$ where $Z$ is standard normal. Hence, from printed normal CDF tables $\frac{-2-\mu}{1.5} = -.2533,$ which you can solve for $\mu = -1.62.$ Then finally, $X \sim \mathsf{Norm}(\mu=-1.62, \sigma=1.5).$

In R, where pnorm is a normal CDF and qnorm is a quantile function (inverse CDF):

qnorm(.4)
[1] -0.2533471
pnorm(-2, -1.62, 1.5)
[1] 0.4000053

If you mean something else, please clarify.