TITLE: How to calculate the shortest interval, for $P ( X ≤ 1 . 645) = 0 . 95$?
QUESTION [5 upvotes]: The problem statement said:

Based on the fact that $\Phi(1 . 645) = 0 . 95$ ﬁnd an interval in which $X$
  will fall with $95\%$ probability.

Therefore:
Since $P ( X ≤ 1 . 645) = 0 . 95, ( -∞ , 1 . 645)$ is a $95\%$ conﬁdence interval for $X$
The question I have problem to understand is:

Among all possible intervals into which $X$ falls with $95\%$ probability,
  ﬁnd the shortest one.

How can I compute or see which is the shortest interval?
Thanks!

REPLY [3 votes]: Because the normal curve has a bell shape, there is more area under it near the average. It appears that we are dealing with a standard normal distribution, so you want to solve for little $x$
Using the CDF,
\begin{align*}
.95 &= P(X\leq x) -P(X\leq -x) \\
&=P(X\leq x)-[1-P(X\leq x)]\tag{1}\\
&=2P(X\leq x)-1\\
&=2\Phi(x)-1
\end{align*}
where in $(1)$ I recognize symmetry.
This gives
$$\Phi(x)=\frac{.95+1}{2}\implies x = \Phi^{-1}\left(0.975\right) = 1.959964$$
So the interval is $(-1.959964,1.959964)$.

Alternatively, using the survival gives
\begin{align*}
.95 &= P(X>-x)-P(X>x)\\
&= P(X\leq x)-[1-P(X\leq x)]\tag 2\\
&=2P(X\leq x) - 1
\end{align*}
where in $(2)$ I recognize symmetry for the value $P(X>-x)$.
This results in the same interval.