A researcher want to determine an interval estimate for the average weight of adult gorillas in pounds. She wants to be $90\mathrm{\%}$ certain that she is within 5.8 pounds of the true average. From past studies, it is known that the standard deviation of the weights of adult gorillas is 19.5 pounds.
a. What is the critical value that corresponds to the given level of confidence?
Round your answer to two decimal places.
b. What is the minimum sample size necessary to estimate the population mean?
Round your answer up to the next integer.

Step 1 ：The problem is asking for two things: the critical value for a 90% confidence interval and the minimum sample size necessary to estimate the population mean with a margin of error of 5.8 pounds.

Step 2 ：The critical value corresponds to the z-score for a 90% confidence interval. We can find this value using a standard normal distribution table or a z-score calculator.

Step 3 ：The minimum sample size necessary to estimate the population mean can be found using the formula for the margin of error in a confidence interval, which is $E=z\ast (\sigma /\surd n)$, where $E$ is the margin of error, $z$ is the z-score, $\sigma$ is the standard deviation, and $n$ is the sample size. We can rearrange this formula to solve for $n$: $n=(z\ast \sigma /E{)}^2$.

Step 4 ：Given that the confidence level is 0.9, the standard deviation is 19.5, and the margin of error is 5.8, we can calculate the critical value and the sample size.

Step 5 ：The critical value that corresponds to the given level of confidence is approximately $\boxed{{\displaystyle 1.64}}$.

Step 6 ：The minimum sample size necessary to estimate the population mean is $\boxed{{\displaystyle 31}}$.