Let $X$ be a normal distribution with mean and variance $2$ and $4$ respectively. Then, $g(a)=P(a\leq X \leq a+2)$. Find the value of $a$ that maximizes, $g(a)$. I have solved the problem by traditional way and got my answer. But my question is is there a quick way to do the problem since the traditional way took me some time to solve.
How I solved using traditional way?
The given inequality can be written as:
$$g(a)=P(\frac{a-2}{2}\leq Z \leq \frac{a}{2})$$
This can be written as:
$$g(a)=\int_{\frac{a-2}{2}}^{\frac{a}{2}}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}\,dx$$
The above then can be solved using the Leibnitz rule and value of $a$ will turn out to be $1$ after solving. But this involves a lot of work. My question is, is there a quick way to go about solving this problem?.
Thanks
The standard normal distribution is symmetric about $0$ and they take the highest value around $0$.
hence, we would want $$\frac{a}{2} = - \left( \frac{a-2}{2}\right)$$
$$a=2-a$$
and we obtain $a=1$.