Find the confidence interval for $\mu$ if: $\bar{x}=81.4, \mathrm{~s}=6.5, \mathrm{n}$ $=37$ and $c=0.95$

Step 1 ：We are given the sample mean \(\bar{x}=81.4\), the standard deviation of the sample \(s=6.5\), the sample size \(n=37\), and the confidence level \(c=0.95\).

Step 2 ：The z-score for a 95% confidence level is approximately 1.96.

Step 3 ：We can calculate the confidence interval for a population mean, given a sample, using the formula: \[\bar{x} \pm z \frac{s}{\sqrt{n}}\]

Step 4 ：Substituting the given values into the formula, we get: \[81.4 \pm 1.96 \frac{6.5}{\sqrt{37}}\]

Step 5 ：Calculating the margin of error, we get approximately 2.094443098270252.

Step 6 ：Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval: \[81.4 - 2.094443098270252 = 79.30555690172976\]

Step 7 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval: \[81.4 + 2.094443098270252 = 83.49444309827025\]

Step 8 ：Rounding to two decimal places, the 95% confidence interval for \(\mu\) is approximately \(\boxed{[79.31, 83.49]}\).