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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.

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Hmm, so there's no way to get a answer without using a computer to run iterations to solve it huh? Might you have any idea on which programs are able to handle this type of equations? My graphing calculator apparently doesn't seem to have such a function for solving such equations. – Eric Dec 6 '12 at 4:22
@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
I use a Casio fx-9860G Slim. – Eric Dec 6 '12 at 9:08

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