Question

Find the​ z-scores for which 8​% of the​ distribution's area lies between minus−z and z.

Answer 1

$\mathbb P(-z$

Because the distribution (normal) is symmetric, you know that $\mathbb P(Zz)$, so

$\mathbb P(Zz)=2\mathbb P(Z$

Now,

$\mathbb P(-z$
$\implies\mathbb P(Z$

If $F_Z(z)$ is the CDF of the normal distribution, so that $F_Z(z)=\mathbb P(Z$, then the z-score satisfies

$\begin{cases}z={F_Z}^{-1}(0.54)\\z=-{F_Z}^{-1}(0.46)\end{cases}$

so that $z\approx0.1004$.