Answer
See explanation
Work Step by Step
### **Step 1: Estimate the Larger Standard Deviation**
#### Dataset a: 5, 4, 3, 2, 1, 0
* Range: $5 - 0 = 5$
* Fairly spread out, symmetrical
#### Dataset b: 5, 5, 5, 0, 0, 0
* Range: $5 - 0 = 5$
* Split into two clusters (0s and 5s), with no middle ground
**Estimate**: Distribution **b** is likely to have **larger** standard deviation due to the extreme clustering at both ends—values are farther from the mean than in a.
---
### **Step 2: Use the Raw-Score Formula**
$$
S = \sqrt{\frac{\sum X^2 - \frac{(\sum X)^2}{N}}{N}}
$$
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### **Compute for a: 5, 4, 3, 2, 1, 0**
* $N = 6$
* $\sum X = 5 + 4 + 3 + 2 + 1 + 0 = 15$
* $\sum X^2 = 5^2 + 4^2 + 3^2 + 2^2 + 1^2 + 0^2 = 25 + 16 + 9 + 4 + 1 + 0 = 55$
$$
S_a = \sqrt{ \frac{55 - \frac{15^2}{6}}{6} } = \sqrt{ \frac{55 - 37.5}{6} } = \sqrt{ \frac{17.5}{6} } = \sqrt{2.9167} \approx 1.71
$$
---
### **Compute for b: 5, 5, 5, 0, 0, 0**
* $N = 6$
* $\sum X = 5 + 5 + 5 + 0 + 0 + 0 = 15$
* $\sum X^2 = 25 + 25 + 25 + 0 + 0 + 0 = 75$
$$
S_b = \sqrt{ \frac{75 - \frac{15^2}{6}}{6} } = \sqrt{ \frac{75 - 37.5}{6} } = \sqrt{ \frac{37.5}{6} } = \sqrt{6.25} = 2.5
$$
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### **Results**
* $S_a \approx 1.71$
* $S_b = 2.5$
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### **Conclusion**
* **Distribution b has the larger standard deviation**, confirming the visual estimate.
* Both distributions have the **same range (5)**, but the values in b are clustered at the extremes, inflating the standard deviation.
* **Raw-score formula and estimation align**.
