Q10. - Algebra 1

Question

Find the mean, variance, and standard deviation to the nearest tenth for each set of data.

$\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{8}\mathbf{,}\mathbf{9}\mathbf{,}\mathbf{10}$

Step-by-Step Solution

Answer

The mean, variance and standard variation of the given data set is $\mathbf{6}\mathbf{.}\mathbf{625}\mathbf{,}\mathbf{3}\mathbf{.}\mathbf{984}$ and 2 respectively.

Step 1. Find the mean.

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

$\begin{array}{l}\overline{x}=\frac{4+5+5+6+6+8+9+10}{8}\\ =\frac{53}{8}\\ =6.625\end{array}$

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.

$\begin{array}{l}{\sigma }^2=\frac{{(4-6.63)}^2+{(5-6.63)}^2+{(5-6.63)}^2+{(6-6.63)}^2+{(6-6.63)}^2+{(8-6.63)}^2+{(9-6.63)}^2+{(10-6.63)}^2}{8}\\ =\frac{{(-2.625)}^2+{(-1.625)}^2+{(-1.625)}^2+{(-0.625)}^2+{(-0.625)}^2+{\left(1.375\right)}^2+{\left(2.375\right)}^2+{\left(3.375\right)}^2}{8}\\ =\frac{6.890625+2.640625+2.640625+0.390625+0.390625+1.890625+5.640625+11.390625}{8}\\ =\frac{31.875}{8}\\ =3.984\end{array}$

Step 3. Find the standard deviation.

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

$\begin{array}{c}{\sigma }^2=3.984\\ \sigma =\sqrt{3.984}\\ =1.996\approx 2\end{array}$

Therefore, the mean, variance and standard variation of the given data set is $6.625,3.984$ and 2 respectively.