Calculate mean and standard deviation of a normal distribution given probabilities at three points only I was looking for a method to calculate mean and standard deviation of a normal distribution where P(X=x) is given at three points that is P(X=10111) = 0.3, P(X=10840) = 0.5, and P(X=10948) = 0.6 
 A: Assuming you meant probabilities of the sort $P(X\leq x)$, here is a numerical result using R
optim(par=c(1e4,1e2),
      fn=function(x){
        (qnorm(0.3,x[1],x[2])-10111)^2+
        (qnorm(0.5,x[1],x[2])-10840)^2+
        (qnorm(0.6,x[1],x[2])-10948)^2
      },method="L-BFGS-B",lower=c(1,1),upper=c(1e5,1e5))$par

[1] 10734.506  1123.457

which results in the following probabilities
> pnorm(10111,10734.506,1123.457)
[1] 0.2894512
> pnorm(10840,10734.506,1123.457)
[1] 0.5374062
> pnorm(10948,10734.506,1123.457)
[1] 0.5753584

A: If it is a continuous variable, then the statement $P(X=x)$ is not correct. It should be $P(X\leq x)$. This is also called the CDF or cumulative distribution function of a normal distribution. 
If that is the case then, you are looking at $P(X\leq 10111) = 0.3$ and $P(X\leq 10840) = 0.5$ etc. My take on this is as follows
Use the CDF formula (you can easily find it using a simple google search, e.g. here). Put the values of $x=1011, x=10840$ and you will have two algebraic equations. By mathematical manipulations, you may easily find the values of mean and standard deviation.
