Standard Deviation Math Example 4

Solution

1. 1
   Both sets have mean $4$.
2. 2
   For $A$, every value equals the mean, so every deviation is $0$. Therefore $\sigma_A = 0$.
3. 3
   For $B$, the squared deviations are $(2-4)^2 = 4$, $(4-4)^2 = 0$, $(4-4)^2 = 0$, and $(6-4)^2 = 4$. Their sum is $8$.
4. 4
   Population variance for $B$ is $\frac{8}{4} = 2$, so $\sigma_B = \sqrt{2} \approx 1.41$.