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.

  • $\begingroup$ How about giving a specific numerical example, showing what you have tried, and saying where your difficulty arises. $\endgroup$ – BruceET Nov 4 '19 at 23:30
  • $\begingroup$ @BruceET I just added some actual values from my data. $\endgroup$ – Amadou Kone Nov 4 '19 at 23:47
  • $\begingroup$ 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. $\endgroup$ – BruceET Nov 5 '19 at 0:24
  • $\begingroup$ @BillChen: Welcome. Your answer (now deleted) seemed on target, giving a good clue, but not a final answer. $\endgroup$ – BruceET Nov 5 '19 at 0:35
  • $\begingroup$ 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. $\endgroup$ – Michael R. Chernick Nov 5 '19 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):

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

If you mean something else, please clarify.


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