Question

This data is from a sample.

Calculate the mean, standard deviation, and variance.

Suggestion: use technology. Round answers to two decimal places. $x$ $10.1$ $45.6$, $18.9$, $17.9$, $18.1$, $15.7$, $26.7$, $36.9$

Mean =
Standard Deviation =
Variance =
Loops - now you discover that the data was actually from a population! So now you must give the population standard deviation.
Population Standard Deviation =

Answer 1

The given data from the sample are:

$10.1$, $45.6$, $18.9$, $17.9$, $18.1$, $15.7$, $26.7$, $36.9$.

Step 1. Calculation of Mean.

The formula for the calculation of the mean is:

$\overline{x}=\frac{\sum _{}x}{n}$

Substitute the values:

$\begin{array}{rcl}& \Rightarrow & \overline{x}=\frac{10.1+45.6+18.9+17.9+18.1+15.7+26.7+36.9}{8}\\ & \Rightarrow & \overline{x}=\frac{189.9}{8}\\ & \Rightarrow & \overline{x}=23.7375\end{array}$

Step 2. Calculation of Variance.

The formula for the calculation of the variance is:

$S^2=\frac{1}{n-1}\left(\sum {x_i}^2-n{\overline{x}}^2\right)$

Substitute the values:

$\begin{array}{rcl}& \Rightarrow & S^2=\frac{1}{8-1}\left({\left(189.9\right)}^2-8{\left(23.7375\right)}^2\right)\\ & \Rightarrow & S^2=\frac{999.73875}{7}\\ & \Rightarrow & S^2=142.8198\end{array}$

Step 3. Calculation of standard deviation.

The formula for the calculation of the standard deviation is:

${\sigma }^2=\frac{1}{n}\left(\sum _{}{x_i}^2-n{\overline{x}}^2\right)$

Substitute the values:

$\begin{array}{rcl}& \Rightarrow & {\sigma }^2=\frac{1}{8}\left({\left(189.9\right)}^2-8{\left(23.7375\right)}^2\right)\\ & \Rightarrow & {\sigma }^2=\frac{999.73875}{8}\\ & \Rightarrow & {\sigma }^2=124.9673\\ & \Rightarrow & \sigma =11.17888\end{array}$

Hence, the mean, standard deviation, and variance are $\mathbf{23}\mathbf{.}\mathbf{7375}$, $\mathbf{11}\mathbf{.}\mathbf{17888}$ and $\mathbf{142}\mathbf{.}\mathbf{8198}$.