Finding the mean is easy and what about maximizing the variance?

Question

The random variable has a normal distribution, and $$ p (- 16 <\beta < - 11 ) = p (-13 <\beta < - 8) $$ Find the variance of $ \beta $ for which the probability $$ p (- 11 <\beta < - 8 ) $$ is maximal. I found the mean, but I do not know how to proceed further.

Answer 1

It is pretty clear from symmetry that the mean is $-12$ (and I presume that's what you got). Now let's think about how to maximize the probability that $-11<\beta< -8.$ Let's de-mean and flip $\beta,$ so this is the probability that a zero-mean gaussian is between $1$ and $4.$ This is given by $$ \Phi(4/\sigma) -\Phi(1/\sigma)$$ where $\Phi$ is the normal cumulative and $\sigma=\operatorname{Var}(\beta).$ We can maximize by differentiating and setting equal to zero, using the fact that $\Phi'(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$ This gives $$ 4e^{-\frac{16}{2\sigma^2}} = e^{-\frac{1}{2\sigma^2}}\implies \sigma^2 = \frac{15}{2\ln(4)} $$