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 20.3% 20.3% 79.7% 79.7% 70.3% 70.3% 29.7%
Answers
GPT-4o mini
To find the percentage of students who scored less than 75 on the history test, we can use the z-score formula, which is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the score we are interested in (75),
- \(\mu\) is the mean (80),
- \(\sigma\) is the standard deviation (6).
Substituting the values:
\[
z = \frac{(75 - 80)}{6} = \frac{-5}{6} \approx -0.8333
\]
Next, we can look up the z-score of approximately -0.83 in the z-score table to find the corresponding percentage.
The z-score table shows that a z-score of -0.83 corresponds roughly to a percentile of around 0.2033, or 20.33%.
Rounding to the nearest tenth gives us 20.3%.
Therefore, the percentage of students who earned less than 75 is **20.3%**.
\[
z = \frac{(X - \mu)}{\sigma}
\]
where:
- \(X\) is the score we are interested in (75),
- \(\mu\) is the mean (80),
- \(\sigma\) is the standard deviation (6).
Substituting the values:
\[
z = \frac{(75 - 80)}{6} = \frac{-5}{6} \approx -0.8333
\]
Next, we can look up the z-score of approximately -0.83 in the z-score table to find the corresponding percentage.
The z-score table shows that a z-score of -0.83 corresponds roughly to a percentile of around 0.2033, or 20.33%.
Rounding to the nearest tenth gives us 20.3%.
Therefore, the percentage of students who earned less than 75 is **20.3%**.