Data set A is 6, 10, 10, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 14, 14, 18 and data set B is 6, 7, 8, 9, 10, 11, 11, 12, 12, 12, 13, 13, 14, 15, 16, 17, 18. Both have mean 12. Which set has greater variability?

Step-by-step solution

1. 1

   Notice what the two sets have in common. Both have $n = 17$ values summing to $204$, so both have mean exactly $12$, and both run from $6$ to $18$ — identical mean and identical range. Neither of those measures can separate them, which is precisely why standard deviation is needed.

2. 2

   Sum the squared deviations for set A. Deviations from $12$ are mostly $0$ or $\pm 1$, with two outliers at $\pm 6$:

   $\sum (x-12)^2 = 36 + 2(4) + 3(1) + 0 + 3(1) + 2(4) + 36 = 94$

3. 3

   Sum the squared deviations for set B. Here the values march evenly from $6$ to $18$, so the deviations are $-6, -5, -4, -3, -2, -1, -1, 0, 0, 0, 1, 1, 2, 3, 4, 5, 6$:

   $\sum (x-12)^2 = 2(36 + 25 + 16 + 9 + 4 + 1) + 1 + 1 = 184$

4. 4

   Convert both to population standard deviations. Same divisor $n = 17$ for both:

   $\sigma_A = \sqrt{\frac{94}{17}} = \sqrt{5.5294} \approx 2.35, \qquad \sigma_B = \sqrt{\frac{184}{17}} = \sqrt{10.8235} \approx 3.29$

5. 5

   Compare and explain the difference. Since $3.29 > 2.35$, data set B has greater variability. Structurally, A piles most of its mass on $11$–$13$ with two lone extremes, while B spreads its values evenly across the entire range — so a typical B value sits much further from $12$ than a typical A value, even though both sets share the same mean, size and range.