Finding mean of distribution which gives specific percentage of results in specified interval

Question

Let's assume that there is a known interval [a, b] and a normal distribution with a known standard deviation. How to find mean xbar of the distribution for which the interval [a, b] contains defined percentage P of results? So the only variable is mean of distribution. I can find its approximation with numeric method, but looking for analytic solution (general formula).

The assumption is that when the distribution is centered in the interval [a, b], there exists percentate greater or equal to defined percentage P.

There will be two solutions, at the same distance from the values ​​a and b.

Answer 1

First, let's clear up some of the terminology. The variable $\bar X$ is the mean of a sample; the mean of the distribution is the fixed parameter $\mu$, for which we are aiming to find a point estimate using only $a$, $b$, $\sigma$, and $P$, the latter of which is the percentage of observations within the interval from a single sample of size $n$.

We can model each observed value as observations from a Bernoulli distribution, with the probability parameter represented by $p=\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)$ (note that little $p$ is the percentage of the distribution, different from $P$ the percentage from the sample). The sum of all of the $n$ observations within the region follows a binomial distribution which we will assign the variable $Y=nP$:

$$f_Y={n \choose y}p^y(1-p)^{n-y}={n \choose y}\left[\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right]^{nP}\left[1-\left(\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right)\right]^{n(1-P)}$$

We will use maximum likelihood estimation to try to find the estimator. The log likelihood function is

$$\ell(\mu\mid n,P,\sigma)=\log L=\\\log{n \choose y}+nP\log\left[\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right]+n(1-P)\log\left[1-\left(\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right)\right]$$

We now take the derivative and set equal to zero to obtain the maximum.

$$\frac{\partial \ell}{\partial\mu}=\frac{nP\left[-\left(\frac{b-\mu}{\sigma}\right)\phi\left(\frac{b-\mu}{\sigma}\right)+\left(\frac{a-\mu}{\sigma}\right)\phi\left(\frac{a-\mu}{\sigma}\right)\right]}{\left[\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right]}+\frac{n(1-P)\left[\left(\frac{b-\mu}{\sigma}\right)\phi\left(\frac{b-\mu}{\sigma}\right)-\left(\frac{a-\mu}{\sigma}\right)\phi\left(\frac{a-\mu}{\sigma}\right)\right]}{1-\left[\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)\right]}:=0$$

Solving, we get $\Phi\left(\frac{b-\hat\mu}{\sigma}\right)-\Phi\left(\frac{a-\hat\mu}{\sigma}\right)=P$. (Perhaps not unexpectedly.) I'm not aware of a function that represents the proportion associated with the width of the standard normal distribution as a function of the distance of that width from the mean; if this function exists and is invertible we could take the inverse and therefore solve analytically for $\hat\mu$. We can observe that this function is a convolution of a uniform distribution with the normal distribution and therefore find its quasi-characteristic function (offset by a constant), but I don't believe this can provide any unique insight into the problem.

Hope this helps to some degree.