d. Between what two values of $Z$ (symmetrically distributed around the mean) will $92.82 \%$ of all possible $Z$ values be contained? The two values of $Z$ for which $92.82 \%$ of all possible $Z$ values are contained between are $\square$ and $\square$. (Use ascending order. Round to two decimal places as needed.)
Solution
Step 1 ：Convert the percentage to a decimal: 92.82% = 0.9282
Step 2 ：Since the distribution is symmetric around the mean, half of 0.9282 will be on the left side of the mean and half will be on the right side. So, we need to find the z-score for \(0.9282/2 = 0.4641\) on each side of the mean.
Step 3 ：Look up 0.4641 in the z-table (or use a z-score calculator). The z-score that corresponds to 0.4641 is approximately 1.81.
Step 4 ：Since the distribution is symmetric, the z-scores that contain 92.82% of the data are -1.81 and 1.81.
Step 5 ：To check the answer, you can look up the z-scores in the z-table or use a z-score calculator. The area between -1.81 and 1.81 should be approximately 0.9282 or 92.82%, which confirms the answer.
Step 6 ：So, the two values of Z for which 92.82% of all possible Z values are contained between are \(\boxed{-1.81}\) and \(\boxed{1.81}\).
