Question

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

Answer 1

`\mathbb P(-z<Z<z)=0.08\implies \mathbb P(Z<-z)+\mathbb P(Z>z)=0.92`

Because the distribution (normal) is symmetric, you know that
`\mathbb P(Z<-z)=\mathbb P(Z>z)`, so


`\mathbb P(Z<-z)+\mathbb P(Z>z)=2\mathbb P(Z<-z)=0.92\implies\mathbb P(Z<-z)=\mathbb P(Z>z)=0.46`

Now,


`\mathbb P(-z<Z<z)=\mathbb P(Z<z)-\mathbb P(Z<-z)=0.08`

`\implies\mathbb P(Z<z)=0.08+0.46=0.54`

If
`F_Z(z)` is the CDF of the normal distribution, so that
`F_Z(z)=\mathbb P(Z<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`.