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

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

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.

• How about giving a specific numerical example, showing what you have tried, and saying where your difficulty arises. Commented Nov 4, 2019 at 23:30
• @BruceET I just added some actual values from my data. Commented Nov 4, 2019 at 23:47
• I don't know why you resist giving one forthright numerical example in detail, along with your attempted solution and indications what is stopping you from getting a final answer. Commented Nov 5, 2019 at 0:24
• @BillChen: Welcome. Your answer (now deleted) seemed on target, giving a good clue, but not a final answer. Commented Nov 5, 2019 at 0:35
• You need to include the self-study tag. Note that the P(X<-2) =0.4 is equivalent to P([X-$\mu$]/1.5 <[-2-$\mu$]/1.5]. Then you should be able to solve for $\mu$ from the table of the standard normal distribution. Commented Nov 5, 2019 at 1:36

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.