To find the percentage of scores that fall between 58.9 and 75.1 in a normal distribution, we first need to calculate the z-scores for these values using the provided mean (\( \mu \)) and standard deviation (\( \sigma \)).
- Calculate the z-score for 58.9:
\[ z = \frac{X - \mu}{\sigma} \]
\[ z = \frac{58.9 - 64.3}{5.4} = \frac{-5.4}{5.4} = -1 \]
- Calculate the z-score for 75.1:
\[ z = \frac{75.1 - 64.3}{5.4} = \frac{10.8}{5.4} = 2 \]
Now we have the z-scores: \( z = -1 \) for 58.9 and \( z = 2 \) for 75.1.
- Look up the z-scores in the standard normal distribution table:
- The area to the left of \( z = -1 \) is approximately 0.1587 (15.87%).
- The area to the left of \( z = 2 \) is approximately 0.9772 (97.72%).
- Calculate the percentage of scores between these two z-scores:
\[ \text{Percentage} = \text{Area to the left of } z = 2 - \text{Area to the left of } z = -1 \]
\[ \text{Percentage} = 0.9772 - 0.1587 = 0.8185 \quad \text{(or } 81.85% \text{)} \]
Thus, the best approximation of the percentage of scores that fell between 58.9 and 75.1 is approximately 81.5%.
Final answer: 81.5%