Maximum and minimum value of $P(-4<Y<6)$, where Y has normal distribution with standard deviation 2 and the mean unknown.

Question

Let Y be a random variable has normal distribution with standard deviation 2 and mean is unknown. Find the maximum and minimum value of $P(-4<Y<6)$.

Answer 1

Hint... The maximum occurs when the mean is $\frac {-4+6}{2}$ by symmetry. The lower bound for $P$ is zero, so there is no strict minimum.

Answer 2

There is no minimum, for if we take the mean very small or very large, the probability can be made arbitrarily close to $0$.

For the maximum, intuition says that we should pick mean $1$, in which case computation of the maximum value is straightforward.

To show that the maximum is reached when $\mu=1$, note that our probability is equal to $$\Pr\left(Z\le \frac{6-\mu}{2}\right)-\Pr\left(Z\le \frac{-4-\mu}{2}\right),$$ where $Z$ is standard normal. Or, to put it another way, our probability is $$\Phi\left(\frac{6-\mu}{2}\right)-\Phi\left(\frac{-4-\mu}{2}\right),$$ where $\Phi$ is the cdf of the standard normal. To maximize, differentiate and set the derivative equal to $0$. The density function of the standard normal has shape a constant times $e^{-z^2/2}$. So differentiating and setting the result equal to $0$ gives, after some simplification, $$\exp(-(6-\mu)^2/8)=\exp(-(-4-\mu)^2)/8).$$ Take the logarithm. We get $(6-\mu)^2=(-4-\mu)^2$. Solve. We get $\mu=1$.