Compute the critical value $\mathrm{z}_{\alpha / 2}$ that corresponds to a $93 \%$ level of confidence.
Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).
\[
\mathrm{z}_{\alpha / 2}=
\]
(Round to two decimal places as needed.)

Step 1 ：The critical value $z_{\alpha / 2}$ is the z-score such that the area under the standard normal curve between $-z_{\alpha / 2}$ and $z_{\alpha / 2}$ is the confidence level (in this case, 93%). The area outside of this range, $\alpha$, is split equally between the two tails of the distribution.

Step 2 ：So, we need to find the z-score such that the area to the left of it is $1 - \frac{\alpha}{2} = 1 - \frac{1 - 0.93}{2} = 0.965$.

Step 3 ：We can use the standard normal distribution table to find the z-score corresponding to a given percentile. This table takes a percentile (as a decimal) and returns the corresponding z-score.

Step 4 ：From the table, we find that the z-score corresponding to the 96.5th percentile is approximately 1.81.

Step 5 ：Final Answer: The critical value $z_{\alpha / 2}$ that corresponds to a $93\%$ level of confidence is \(\boxed{1.81}\).