Question

Scores on a history test have an average of 80 with a standard deviation of 6. What percentage of students who took this test earned less than 75? Round to the nearest tenth, and use the z-score table to determine the percentage.

(1 point)
Responses

70.3%
70.3%

79.7%
79.7%

20.3%
20.3%

29.7%
29.7%
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Answer 1

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%**