By Dan Kernler [CC BY-SA 4.0 ], from Wikimedia Commons
Consider the following test data. Each form was normally distributed.
| Student | Raw score | Form | Mean | Standard deviation |
|---|---|---|---|---|
| Jonce | 90 | A | 80 | 15 |
| Tashika | 85 | B | 75 | 5 |
1. For each test form, what are the test scores that contain the middle 68% of the data?
A: [65, 95], B: [70, 80]
2. For each test form, what score would a student need to be at the 84th percentile?
A: 95, B: 80
3. What are the standard scores for each student? (see next section)
\(z_{\text{jonce}} = \dfrac{90 - 80}{15} = \dfrac{2}{3}; z_{\text{tashika}} = \dfrac{85 - 75}{5} = 2\)
4. Which student scored better than other test takers? Why?
Although Jonce had the higher raw score, their z-score is 2/3 (~75 percentile), whereas Taskika's z-score is 2 (~97.5 percentile). Tashika's raw score is lower, but they did better than 97.5% of their classmates.
A standard score (or z-score) measures how many standard deviations a score is from the mean. It is computed by…
\( z = \dfrac{x - \mu}{\sigma} \)
… where \(x\) is a data point, \(\mu\) is the mean, \(\sigma\) is the standard deviation. In the normal distribution above, \(\mu\) has \(z=0\), \(\mu + \sigma\) has \(z = 1\), and so on.
An exam was normally distributed with μ = 75 and σ = 17.6. The professor wants to base grades on standard deviations. Please complete the following:
| Grade | Z-score | Exam score needed | Percentile |
|---|---|---|---|
| A | 2 | 110.2 | 97.5th |
| B | 1 | 92.6 | 84th |
| C | 0 | 75 | 50th |
| D | -1 | 57.4 | 16th |
Note: We use the 68-95-99.7 rule because more accurate results require conversion tables or calculus. Also see Percentile calculator or what about national percentile ranks?