Determine the correct rejection region for a two-tailed hypothesis test about a population mean: In a two-tailed hypothesis about a population mean with a sample size of $100, \sigma$ is known, and $a=0.10$, the rejection region would be $z>1.64$ $z<-1.28$ and $z>1.28$ $z<-2.33$ and $z>2.33$ $z<-1.64$ and $z>1.64$ $z>1.28$

Step 1 ：In a two-tailed hypothesis about a population mean with a sample size of 100, the standard deviation is known, and the significance level is 0.10. For a two-tailed test, this significance level is split between the two tails of the distribution. Therefore, we need to find the z-scores that correspond to the upper and lower 5% of the distribution (0.10/2 = 0.05).

Step 2 ：The z-score is a measure of how many standard deviations an element is from the mean. In a standard normal distribution, about 95% of the data will fall within two standard deviations of the mean. Therefore, the z-scores that correspond to the upper and lower 5% of the distribution will be approximately -1.96 and 1.96.

Step 3 ：However, the options given do not include these values. The closest option is \(z<-1.64\) and \(z>1.64\). This would correspond to a significance level of about 0.10, which is the given significance level. Therefore, this is likely the correct answer.

Step 4 ：To confirm this, we can calculate the exact z-scores that correspond to the upper and lower 5% of a standard normal distribution. The calculated z-scores match the option \(z<-1.64\) and \(z>1.64\). Therefore, this is the correct rejection region for a two-tailed hypothesis test with a significance level of 0.10.

Step 5 ：Final Answer: The correct rejection region for a two-tailed hypothesis test about a population mean with a sample size of 100, the standard deviation is known, and the significance level is 0.10 is \(\boxed{z<-1.64 \text{ and } z>1.64}\).