Chapter 13.2, Problem 23E

(a)

To determine

To find: The mean and standard deviation of given data set.

Answer to Problem 23E

The mean is $12$ and the standard deviation is $2\sqrt{2}$ .

Explanation of Solution

Given:

The given line plot represents the data is given below:

  

Concept used:

Mean is given as:

  $\overline{X}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Standard deviation is given by:

  ${\sigma }_X=\left(\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}{n}}\right)$

Here $n$ is the total number of terms.

Calculation:

  $\left\{8,10,10,14,14,16\right\}$ The given data in discrete form is given as,

The mean and standard deviation are calculated as below:

Mean of given data is calculated as:

  $\begin{array}{c}\overline{X}=\frac{8+10+10+14+14+16}{6}\\ =\frac{72}{6}\\ =12\end{array}$

Standard deviation is calculated as:

  $\begin{array}{c}{\sigma }_X=\sqrt{\frac{{\left(8-12\right)}^2+{\left(10-12\right)}^2+{\left(10-12\right)}^2+{\left(12-12\right)}^2+{\left(14-12\right)}^2+{\left(16-12\right)}^2}{6}}\\ =\sqrt{\frac{48}{6}}\\ =2\sqrt{2}\end{array}$

  $$

Therefore, mean is $12$ and the standard deviation is $2\sqrt{2}$ .

(b)

To determine

To find: The mean and standard deviation of given data set.

Answer to Problem 23E

The mean is $20$ and the standard deviation is $2\sqrt{2}$ .

Explanation of Solution

Given:

The given line plot represents the data is given below:

  

Concept used:

Mean is given as:

  $\overline{X}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Standard deviation is given by:

  ${\sigma }_X=\left(\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}{n}}\right)$

  $$Calculation:

The given data in discrete form is given as:

  $\left\{16,18,18,22,22,24\right\}$

The mean and standard deviation are calculated as below:

Mean of given data is calculated as:

  $\begin{array}{c}\overline{X}=\frac{16+18+18+22+22+24}{6}\\ =\frac{120}{6}\\ =20\end{array}$

Standard deviation is calculated as:

  $\begin{array}{c}{\sigma }_X=\sqrt{\frac{{\left(16-20\right)}^2+{\left(18-20\right)}^2+{\left(18-20\right)}^2+{\left(22-20\right)}^2+{\left(22-20\right)}^2+{\left(24-20\right)}^2}{6}}\\ =\sqrt{\frac{48}{6}}\\ =2\sqrt{2}\end{array}$

  $$

Therefore, mean is $20$ and the standard deviation is $2\sqrt{2}$ .