Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$. Assume that the population has a normal distribution.
$n=30, \bar{x}=84.6, s=10.5,90 \%$ confidence
A. $81.36< \mu< 87.84$
B. $81.34< \mu< 87.86$
C. $79.32< \mu< 89.88$
D. $80.68< \mu< 88.52$

Step 1 ：Given the sample size \(n=30\), the sample mean \(\bar{x}=84.6\), the sample standard deviation \(s=10.5\), and a 90% degree of confidence.

Step 2 ：The z-score for a 90% confidence interval is approximately 1.645. This value can be found in a standard z-table or using a calculator.

Step 3 ：The formula for a confidence interval is \(\bar{x} \pm z*\frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired degree of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 4 ：Substitute the given values into the formula to find the confidence interval: \(84.6 \pm 1.645*\frac{10.5}{\sqrt{30}}\).

Step 5 ：Calculate the lower and upper bounds of the confidence interval to get approximately 81.45 and 87.75 respectively.

Step 6 ：The calculated confidence interval is approximately \(81.45<\mu<87.75\). This interval does not exactly match any of the given options, but it is closest to option A.

Step 7 ：\(\boxed{\text{A. } 81.36<\mu<87.84}}\) is the best choice based on the calculated confidence interval.