find the -scores that bound the middle 82% of the area under the standard normal curve. Enter the answers in ascending order and round to two decimal places.

Answer 1

To find the $z$-scores that bound the middle 82% of the area under the standard normal curve, we need to determine the critical values that leave $9$ in the lower tail and $9$ in the upper tail, since $100$ and half of that is $9$.

Here are the steps to find the $z$-scores:

1. Find the area in the lower tail: This will be $0.09$.

2. Find the area in the upper tail: Because the middle area is $82$, the area in the upper tail will also be $0.09$.

3. Find the $z$-score corresponding to the lower tail: We look up the cumulative area of $0.09$ in the standard normal distribution table (or use a calculator). The $z$-score corresponding to $0.09$ is approximately $-1.34$.

4. Find the $z$-score corresponding to the area of $0.91$ (which is $1-0.09$): For the upper tail, the cumulative area of $0.91$ (which corresponds to finding the critical value for the upper end) corresponds to a $z$-score of approximately $1.34$.

Thus, the $z$-scores that bound the middle 82% of the area under the standard normal curve are approximately:

$$-1.34\text{and}1.34$$

Rounded to two decimal places, the answer is:

$$\boxed{{\displaystyle -1.34}}\text{ and }\boxed{{\displaystyle 1.34}}$$