To find the percentage of students who earned less than 75 on the history test, we need to calculate the z-score for a score of 75. The z-score is calculated using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where:
- \( X \) is the score we are interested in (75),
- \( \mu \) is the mean (80),
- \( \sigma \) is the standard deviation (6).
Substituting the values:
\[ z = \frac{(75 - 80)}{6} = \frac{-5}{6} \approx -0.8333 \]
Now, we use the z-score table to find the percentage corresponding to a z-score of approximately -0.83.
Looking up -0.83 in the z-table, we find that it corresponds to a percentage of approximately 20.3%. This percentage represents the area under the normal curve to the left of the z-score.
Therefore, the percentage of students who earned less than 75 is approximately 20.3%.
The correct option is:
20.3%