Converting and comparing with z-scores
Converting a z-score back to a raw score, comparing performances across different tests using z-scores, and finding percentages from whole-number z-scores.
Mia got $72$ in English and $80$ in Maths. Her parents congratulate her on Maths. But the Maths test was easy (mean $74$) and the English test was hard (mean $60$). Raw marks from different tests can't be compared directly. z-scores put them on the same scale.
Comparing results
To compare scores from different datasets, convert each one to a z-score. The higher z-score is the better result relative to the group.
Mia's results are below. In which subject did she do better compared with her class?
| Subject | Mia's mark | Class mean | Standard deviation |
|---|---|---|---|
| English | 72 | 60 | 8 |
| Maths | 80 | 74 | 6 |
For negative z-scores the same rule holds: $-0.5$ is a better result than $-1.2$, because it's closer to (less far below) the mean. Careful with times in a race, though. There a lower time is better, so the lower z-score is the better performance.
Converting a z-score to a raw score
Rearrange the z-score formula to go backwards:
This just says: start at the mean, then move $z$ standard deviations. A negative $z$ moves down.
An exam has mean $65$ and $s = 12$. Students with a z-score of $2$ or more get a merit certificate. What mark is needed?
z-scores and percentages
For whole-number z-scores, the 68–95–99.7 rule gives percentages directly. About $68\%$ of scores have $z$ between $-1$ and $1$, $95\%$ between $-2$ and $2$, and $99.7\%$ between $-3$ and $3$. The table from section 8B works the same way with z-values marked on the curve instead of raw scores.
Each exam has its own mean and standard deviation. Set a mark for each and see both z-scores side by side on matching bell curves.
Find two marks where the raw scores say one thing but the z-scores say the other. Then press New challenge and decide which result is better before the answer shows.
Your turn
Jai ran $100$ m in $12.4$ s (his age group: mean $13.0$ s, $s = 0.4$ s) and threw a discus $38$ m (mean $32$ m, $s = 4$ m). Which was his better performance?
Show the answer
Sprint: $z = \dfrac{12.4 - 13.0}{0.4} = \dfrac{-0.6}{0.4} = -1.5$. A faster time is better, so being $1.5$ standard deviations below the mean time is a strong result.
Discus: $z = \dfrac{38 - 32}{4} = 1.5$, so $1.5$ standard deviations above the mean distance.
Both are $1.5$ standard deviations better than average, so the two performances are equally good relative to the group.
A test has mean $60$ and $s = 7$. Find the score with a z-score of $2$.
$x = \bar{x} + zs = 60 + 2 \times 7 = 60 + 14 = 74$.
A test has mean $54$ and $s = 6$. Find the score with a z-score of $-1.5$.
$x = 54 + (-1.5)(6) = 54 - 9 = 45$.
Sam's z-score in Biology is $0.8$ and in Chemistry is $1.3$. Relative to the class, Sam did better in:
The higher z-score is the better relative result: $1.3 \gt 0.8$, so Chemistry.
Which z-score is the better result in a test?
All are below the mean. The best is the one closest to (least far below) the mean: $-0.3$.
Ella scored $68$ in Test A (mean $60$, $s = 4$) and $75$ in Test B (mean $65$, $s = 10$). Which is true?
Test A: $z = \dfrac{68 - 60}{4} = 2$. Test B: $z = \dfrac{75 - 65}{10} = 1$. Test A has the higher z-score.
What percentage of normally distributed scores have a z-score between $-2$ and $2$?
By the 68–95–99.7 rule, $z$ between $-2$ and $2$ covers about $95\%$.
What percentage of normally distributed scores have a z-score less than $1$?
Below $z = 0$ is $50\%$. Between $0$ and $1$ is $68 \div 2 = 34\%$. Total: $50 + 34 = 84\%$.
In a swimming race, lower times are better. Which z-score for a swimmer's time shows the best performance?
A negative z-score means a time below the mean, which is faster. The most negative, $-1.2$, is the fastest relative to the group.
Exam style: In a Year 12 English exam, the mean was $66$ and the standard deviation $6$. Priya's z-score was $-1.5$. What was her mark?
$x = \bar{x} + zs = 66 + (-1.5)(6) = 66 - 9 = 57$.
Exam style: Liam scored $81$ in Legal Studies (mean $72$, $s = 6$) and $84$ in Business Studies (mean $78$, $s = 3$). In which subject did he perform better relative to his classes?
Legal: $z = \dfrac{81 - 72}{6} = \dfrac{9}{6} = 1.5$. Business: $z = \dfrac{84 - 78}{3} = \dfrac{6}{3} = 2$. Business Studies has the higher z-score, so it is the better relative result.
Coaching Centre