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 s 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 The required sample size is (Round up to the nearest whole number as needed.)

To determine the required sample size for a \( 98 \% \) confidence level, you can use the formula: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Here, \( Z \) is the Z-score corresponding to the \( 98 \% \) confidence level (which is approximately \( 2.33 \)), \( \sigma \) is the estimated standard deviation (\( 1.25 \)), and \( E \) is the margin of error you are willing to accept. Let's say you want a margin of error of \( 0.5 \). Plugging these values into the formula gives: \[ n = \left( \frac{2.33 \cdot 1.25}{0.5} \right)^2 \approx \left( 5.8125 \right)^2 \approx 33.8 \] Rounding up gives you a required sample size of \( 34 \). This ensures you have a robust sample to make confident inferences about the average grade-point average within the specified range, while capturing enough variability with the standard deviation you've estimated.