Assume that a sample is used to estimate a population mean $\mu$. Find the $99.5 \%$ confidence interval for a sample of size 256 with a mean of 24.2 and a standard deviation of 15.3. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
\[
\square< \mu< \square
\]

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded

Step 1 ：To find the 99.5% confidence interval, we first need to find the critical value (z-score) for this confidence level. The z-score for a 99.5% confidence interval is approximately 2.807 (from the standard normal distribution table). The formula for a confidence interval is: \[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \] where: - $\bar{x}$ is the sample mean, - $z$ is the z-score, - $\sigma$ is the standard deviation of the population, and - $n$ is the sample size. Substituting the given values into the formula, we get: \[ 24.2 \pm 2.807 \frac{15.3}{\sqrt{256}} \] Solving the above expression, we get: \[ 24.2 \pm 2.807 \times \frac{15.3}{16} \] \[ 24.2 \pm 2.807 \times 0.95625 \] \[ 24.2 \pm 2.681 \] So, the 99.5% confidence interval for the population mean $\mu$ is from 21.5 to 26.9 (rounded to one decimal place). Therefore, the answer is: \[ 21.5 < \mu < 26.9 \]