Question

GRIDDED RESPONSE Find the standard deviation of the set of data below to the nearest tenth.

Short answer

The standard variation of the given data set is $\mathbf{3}\mathbf{.}\mathbf{14}$.

Step-by-step solution

Step 1. Find the mean.

To find the mean, add the numbers and then divide by the number of values in the data set.

$\begin{array}{l}\overline{x}=\frac{14+11+9+6+10+16+15+13+9+12+19+10}{12}\\ =\frac{144}{12}\\ =12\end{array}$

Step 2. Find the variance.

To find the variance, square the difference between each number and the mean. Then add the squares, and divide by the number of values.

$\begin{array}{l}{\sigma }^2=\frac{{(14-12)}^2+{(11-12)}^2+{(9-12)}^2+{(6-12)}^2+{(10-12)}^2+{(16-12)}^2+{(15-12)}^2+{(13-12)}^2+{(9-12)}^2+{(12-12)}^2+{(19-12)}^2+{(10-12)}^2}{12}\\ =\frac{{\left(2\right)}^2+{(-1)}^2+{(-3)}^2+{(-6)}^2+{(-2)}^2+{\left(4\right)}^2+{\left(3\right)}^2+{\left(1\right)}^2+{(-3)}^2+{\left(0\right)}^2+{\left(7\right)}^2+{(-2)}^2}{12}\\ =\frac{4+1+9+12+4+16+9+1+9+49+4}{12}\\ =\frac{118}{12}\\ =9.83\end{array}$

Step 3. Find the standard deviation.

To find the standard deviation, take square root of the variance.

$\begin{array}{l}{\sigma }^2=9.83\\ \sigma =\sqrt{9.83}\\ \approx 3.14\end{array}$

Therefore, the standard variation of the given data set is $3.14$.