X is a normally-distributed random variable, and
$P[X<20] = 1/10 = P[X>100]$
I am trying to solve for the mean and the variance.
I know that $\mu=60$ by symmetry.
How can I solve for $\sigma^2$?
2026-03-31 18:58:22.1774983502
Finding the variance of a normally-distributed random variable
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The mean, as you say, must be 60. So we know that the 90th percentile is 40 away from the mean. With a standard normal, one with standard deviation 1, the 90th percentile is approximately 1.29 (you can find this in a table of standard normal percentiles). So the standard deviation of your normal must be 40/1.29. From which it follows that the variance is the square, $[40/1.29]^2$