Assume that all grade-point averages are to be standardized on a scale between 0 and 5 . How many grade-point averages must be obtained so that the sample mean is within 0005 of the Assume that a \( 98 \% \) confidence level is desired. If using the range rule of thumb, \( \sigma \) can be estimated as \( \frac{\text { range }}{4}=\frac{5-0}{4}=1.25 \). Does the sample size seem practical? The required sample size is \( \square \). (Round up to the nearest whole number as needed.)

To determine the required sample size for a sample mean of grade-point averages within \(0.0005\) of the true population mean at a \(98\%\) confidence level, we can use the formula for the sample size \(n\): \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Where: - \(Z\) is the Z-value corresponding to the desired confidence level (for \(98\%\), \(Z \approx 2.33\)). - \(\sigma\) is the estimated standard deviation (\(1.25\) from the range rule of thumb). - \(E\) is the margin of error (\(0.0005\)). Now, plug in the values and calculate \(n\): \[ n = \left( \frac{2.33 \cdot 1.25}{0.0005} \right)^2 \] \[ = \left( \frac{2.9125}{0.0005} \right)^2 \] \[ = (5825)^2 \] \[ = 33819225 \] Rounding up gives us \(n\), which is \(33819225\). So, the required sample size is \(33819225\). Considering practicality, such a large sample size is quite unrealistic, as it would be cumbersome and resource-intensive to collect and analyze that many grade-point averages. In real-world applications, you typically aim for smaller, feasible samples while balancing confidence and precision.