Answer
* A **single low outlier (2)** in distribution **a** drastically increases the standard deviation.
* Distribution **b** has tighter clustering near the mean, so the deviation is much smaller.
* This perfectly illustrates how **just one score** can **inflate the variability** significantly.
Work Step by Step
$$
\sigma = \sqrt{\frac{\sum X^2 - \frac{(\sum X)^2}{N}}{N}}
$$
---
### **Dataset a: 9, 8, 8, 7, 2**
* $N = 5$
* $\sum X = 9 + 8 + 8 + 7 + 2 = 34$
* $\sum X^2 = 81 + 64 + 64 + 49 + 4 = 262$
$$
\sigma_a = \sqrt{\frac{262 - \frac{34^2}{5}}{5}} = \sqrt{\frac{262 - 231.2}{5}} = \sqrt{\frac{30.8}{5}} = \sqrt{6.16} \approx 2.48
$$
---
### **Dataset b: 9, 8, 8, 7, 6**
* $N = 5$
* $\sum X = 9 + 8 + 8 + 7 + 6 = 38$
* $\sum X^2 = 81 + 64 + 64 + 49 + 36 = 294$
$$
\sigma_b = \sqrt{\frac{294 - \frac{38^2}{5}}{5}} = \sqrt{\frac{294 - 288.8}{5}} = \sqrt{\frac{5.2}{5}} = \sqrt{1.04} \approx 1.02
$$
---
### **Results**
* $\sigma_a \approx 2.48$
* $\sigma_b \approx 1.02$
