Chapter 13.3, Problem 63PPSE

Solution

To find: The mean and standard deviation for the set of data $12,13,15,9,16,5,18,16,12,11,15$ .

The mean of the data is $12.9$ and standard deviation is $3.53$ .

Given:

The set of data is written as $12,13,15,9,16,5,18,16,12,11,15$ .

Concept used:

The mean $\left(\overline{x}\right)$ for the set of data $x_1,x_2,x_3\mathrm{......}{x}_{n-2},x_{n-1},x_n$ is written as $\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\mathrm{......}+x_{n-2}+x_{n-1}+x_n}{n}$ where $\left(n\right)$ is the number of terms in the given data.

And, standard deviation $\left(\sigma \right)$ for the set of data is evaluated as:

  $\left(\sigma \right)=\sqrt{{\displaystyle \sum _{i=1}^n\frac{{\left(x_i-\overline{x}\right)}^2}{n}}}$ where $\left(\overline{x}\right)$ is the mean of the set of data.

Calculation:

Apply the formula $\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\mathrm{......}+x_{n-2}+x_{n-1}+x_n}{n}$ for the set of data $12,13,15,9,16,5,18,16,12,11,15$ to find the mean $\left(\overline{x}\right)$ where the number of terms $\left(n\right)=11$ .

  $\begin{array}{c}\text{Mean }\left(\overline{x}\right)=\frac{x_1+x_2+x_3+\mathrm{......}+x_{n-2}+x_{n-1}+x_n}{n}\\ =\frac{12+13+15+9+16+5+18+16+12+11+15}{11}\\ \approx \text{12}\text{.9}\end{array}$

Now, in order to find the standard deviation $\left(\sigma \right)$ apply the formula for standard deviation.

  $\begin{array}{c}\left(\sigma \right)=\sqrt{{\displaystyle \sum _{i=1}^n\frac{{\left(x_i-\overline{x}\right)}^2}{n}}}\left(12,13,15,9,16,5,18,16,12,11,15\right)\\ =\sqrt{\frac{\begin{array}{l}{\left(12-12.9\right)}^2+{\left(13-12.9\right)}^2+{\left(15-12.9\right)}^2\\ +{\left(9-12.9\right)}^2+{\left(16-12.9\right)}^2+{\left(5-12.9\right)}^2\\ +{\left(18-12.9\right)}^2+{\left(16-12.9\right)}^2+{\left(12-12.9\right)}^2\\ +{\left(11-12.9\right)}^2+{\left(15-12.9\right)}^2\end{array}}{11}}\\ =\sqrt{\frac{0.81+0.01+4.41+15.21+9.61+62.41+26.01+9.61+0.81+3.61+4.41}{11}}\\ \approx 3.53\end{array}$

Conclusion:

The mean of the data is $12.9$ and standard deviation is $3.53$ .