# Find variance of a normal distribution given 2 probabilites are equal

Sorry if the question is phrased awkwardly, I'm not sure what to properly call the question.

Here's my question:

$X$ is a normal distribution ~$N(0,\sigma^2)$

$P( X < A ) = P( A < X < B )$

A and B are 2 (negative) numbers that are given.

Basically, it's that for a given range $A$ to $B$, the normal distribution has the cumulative probability of the left-tail below $A$ equal to the cumulative probability of the range $A$ to $B$. How do I calculate the variance, $\sigma^2$?

For what it's worth, I've been able to get:

$Z$ ~ $N(0,1)$

$P( Z < -A/\sigma ) = x$

$P( Z < -B/\sigma ) = 2x$

I don't know if I'm even on the right track though.

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You can carry on from where you are. As you say, we standardise the distribution to get $Z\tilde~N(0,1)$. We have $2\Phi\left(\frac A\sigma\right)=\Phi\left(\frac B\sigma\right)$ (note, no negative signs). Since $A$ and $B$ are known, we can solve for $\sigma$ by using a graphic calculator or computer (e.g. by solving $2\Phi\left(\frac A\sigma\right)-\Phi\left(\frac B\sigma\right)=0$). As far as I am aware, there is no closed-form expression and the answer cannot be computed manually.
@Eric May I know which calculator you're using? You just need the function $\Phi(x)$ and the ability to plot a graph. Just plot the graph of $y=2\Phi\left(\frac Ax\right)-\Phi\left(\frac Bx\right)$ and find where it cuts the 'x-axis'. On TI calculators, $\Phi(x)$ is normcdf. – Sarastro Dec 6 '12 at 4:42