Answer :
To determine how well Teresa and Raina performed on their respective subject-matter tests, we can use the concept of Z-scores, which indicate how many standard deviations a particular score is away from the mean.
Step-by-step Calculation of Z-scores:
Calculate Teresa's Z-score for the English Test:
- Teresa's score: [tex]X = 580[/tex]
- Mean score for the English test: [tex]\mu = 639[/tex]
- Standard deviation for the English test: [tex]\sigma = 22[/tex]
- Z-score formula: [tex]Z = \frac{X - \mu}{\sigma}[/tex]
- Z-score calculation: [tex]Z = \frac{580 - 639}{22} = \frac{-59}{22} \approx -2.68[/tex]
- This Z-score means that Teresa scored about 2.68 standard deviations below the mean.
Calculate Raina’s Z-score for the Chemistry Test:
- Raina's score: [tex]X = 495[/tex]
- Mean score for the chemistry test: [tex]\mu = 533[/tex]
- Standard deviation for the chemistry test: [tex]\sigma = 18[/tex]
- Z-score calculation: [tex]Z = \frac{495 - 533}{18} = \frac{-38}{18} \approx -2.11[/tex]
- This Z-score means that Raina scored about 2.11 standard deviations below the mean.
Interpretation of Results:
- Both Teresa and Raina scored below the average for their respective tests. Teresa's score is further below the average than Raina's in terms of standard deviations from their respective means. This suggests that compared to the average test-takers in each category, Teresa had a more anomalous low score than Raina did.
Understanding Z-scores helps us see how a particular score stands relative to the typical performance in that subject area, making it a valuable tool for comparing scores from tests with different means and standard deviations.