Questions: Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of four points. How many scores would the professor need so that 99% confidence level would have a margin of error of 0.75 points?

Solution Steps

To find the Z-score corresponding to a \(99\%\) confidence level, we use the formula:

\[ Z = \text{PPF}\left(1 - \frac{1 - 0.99}{2}\right) = \text{PPF}(0.995) = 2.5758 \]

The sample size \(n\) required to achieve the desired margin of error can be calculated using the formula:

\[ n = \left(\frac{Z \cdot \sigma}{\text{Margin of Error}}\right)^2 \]

Substituting the known values:

\[ n = \left(\frac{2.5758 \cdot 4}{0.75}\right)^2 \]

Calculating this gives:

\[ n = (2.5758 \cdot 5.3333)^2 \approx (13.7478)^2 \approx 188.7259 \]

Rounding this to the nearest whole number, we find:

\[ n \approx 189 \]

Final Answer