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.

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 \( 100% - 82% = 18% \) 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:

\[ -1.34 \quad \text{and} \quad 1.34 \]

Rounded to two decimal places, the answer is:

\[ \boxed{-1.34} \text{ and } \boxed{1.34} \]