Percentage of the normal area - How to find the area under the normal distribution?

Question

First this is not my homework is an example for learn and solve this little problem.

(Reference online pdf here.)

I have problems only with this section in exercise text:

The middle $60\%$ means that $30\%$ of the area will be on either side of the mean. The corresponding $z$ score is found using $\rm{area} = 0.20$. The $z$ scores are $\pm 0.84$.

How did they find the area is $0.2$ they skipped this part?

Answer 1

I have problems only with this section in exercise text:

The middle $60\%$ means that $30\%$ of the area will be on either side of the mean. The corresponding $z$ score is found using $\rm{area} = 0.20$. The $z$ scores are $\pm 0.84$.

There is an evident typo....it is not 30% but 40% (total) that is on both tails, 20% on the left one and 20% on the right one. Obviously $20\%=0.2$

thus you will claculate the two quantiles corresponding to

$$P(Z<z)=0.2$$

that is

$$\Phi^{-1}(0.2)=-0.84$$

and

$$P(Z>z)=0.2$$

that is

$$\Phi^{-1}(0.8)=0.84$$

Answer 2

Hint: How big is the not shaded area?