Determine the minimum required sample size if you want to be $80 \%$ confident that the sample mean is within 25 units of the population mean given $\sigma=128.5$. Assume the population is normally distributed. What is the critical value that corresponds to a level of confidence of $80 \%$ ? Round your answer to two decimals. What is the minimum sample size required for this problem? Round your answer up to the next integer.

Step 1 ：First, find the critical value that corresponds to a level of confidence of $80 \%$. This value is also known as the z-score. The z-score is a measure of how many standard deviations an element is from the mean. In this case, we want to find the z-score that corresponds to the middle $80 \%$ of the data.

Step 2 ：Since the normal distribution is symmetric, the middle $80 \%$ is equivalent to finding the z-score that leaves $10 \%$ in each tail of the distribution (because $100 \% - 80 \% = 20 \%$, and this $20 \%$ is split evenly between the two tails).

Step 3 ：Looking up $10 \%$ in the z-table gives a z-score of approximately $1.28$.

Step 4 ：Next, we need to find the minimum sample size required for this problem. The formula for the sample size is: \(n = \left(\frac{Z*\sigma}{E}\right)^2\) where: \(n\) is the sample size, \(Z\) is the z-score, \(\sigma\) is the standard deviation of the population, and \(E\) is the margin of error.

Step 5 ：Substituting the given values into the formula gives: \(n = \left(\frac{1.28*128.5}{25}\right)^2 \approx 33.57\)

Step 6 ：Since we can't have a fraction of a sample, we round up to the next integer. Therefore, the minimum sample size required is \(\boxed{34}\).