Answer
Even though both were the fastest in their events, **Heslope’s performance** was slightly more **exceptional relative to her peers**, as reflected by a more negative z-score.
Work Step by Step
Here is the full **step-by-step breakdown** of the z-score calculations for **Harriett** and **Heslope**:
---
## **Harriett – 10k Runners**
### 1. Raw Times (in minutes):
* Harriett: 37
* Dott: 39
* Liz: 40
* Marette: 42
### 2. Mean ($\mu$):
$$
\mu = \frac{37 + 39 + 40 + 42}{4} = \frac{158}{4} = 39.5
$$
### 3. Population Standard Deviation ($\sigma$):
$$
\sigma = \sqrt{\frac{(37 - 39.5)^2 + (39 - 39.5)^2 + (40 - 39.5)^2 + (42 - 39.5)^2}{4}}
$$
$$
= \sqrt{\frac{6.25 + 0.25 + 0.25 + 6.25}{4}} = \sqrt{\frac{13}{4}} = \sqrt{3.25} \approx 1.80
$$
### 4. Z-Score for Harriett:
$$
z = \frac{X - \mu}{\sigma} = \frac{37 - 39.5}{1.80} = \frac{-2.5}{1.80} \approx -1.39
$$
---
## **Heslope – 50m Swimmers**
### 1. Raw Times (in seconds):
* Heslope: 24
* Ta-Li: 26
* Deb: 27
* Betty: 28
### 2. Mean ($\mu$):
$$
\mu = \frac{24 + 26 + 27 + 28}{4} = \frac{105}{4} = 26.25
$$
### 3. Population Standard Deviation ($\sigma$):
$$
\sigma = \sqrt{\frac{(24 - 26.25)^2 + (26 - 26.25)^2 + (27 - 26.25)^2 + (28 - 26.25)^2}{4}}
$$
$$
= \sqrt{\frac{5.0625 + 0.0625 + 0.5625 + 3.0625}{4}} = \sqrt{\frac{8.75}{4}} = \sqrt{2.1875} \approx 1.48
$$
### 4. Z-Score for Heslope:
$$
z = \frac{24 - 26.25}{1.48} = \frac{-2.25}{1.48} \approx -1.52
$$
---
## Final Comparison
| Athlete | Mean | Std Dev | Time | Z-Score |
| -------- | ----- | ------- | ---- | --------- |
| Harriett | 39.5 | 1.80 | 37 | **-1.39** |
| Heslope | 26.25 | 1.48 | 24 | **-1.52** |
