Data set A is 6, 10, 10, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 14, 14, 18 and its mean is 12. Find the population standard deviation.

Step-by-step solution

1. 1

   Confirm the stated mean before trusting it. The $17$ values total

   $\sum x = 204, \qquad \bar{x} = \frac{204}{17} = 12$

   So the mean really is exactly $12$. This matters: an integer mean makes every deviation an integer, so the whole calculation can be done without rounding.

2. 2

   Tabulate the squared deviations. Grouping equal values:

   $6 \rightarrow (-6)^2 = 36, \qquad 10 \ (\times 2) \rightarrow 4 \text{ each}, \qquad 11 \ (\times 3) \rightarrow 1 \text{ each}$

   $12 \ (\times 5) \rightarrow 0, \qquad 13 \ (\times 3) \rightarrow 1, \qquad 14 \ (\times 2) \rightarrow 4, \qquad 18 \rightarrow 36$

3. 3

   Add them up.

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

   The two extreme values $6$ and $18$ contribute $72$ of the $94$ — over three quarters of the total spread comes from just two of the seventeen readings.

4. 4

   Divide by n for the population variance.

   $\sigma^2 = \frac{94}{17} \approx 5.5294$

   Use $n = 17$, not $n - 1$, because the question describes the full data set rather than a sample from a larger group.

5. 5

   Take the square root.

   $\sigma = \sqrt{5.5294} \approx 2.35$

   For comparison, the sample standard deviation would be $s = \sqrt{94/16} \approx 2.42$. The population figure $2.35$ is the one that matches the wording here.