Question

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

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