Chapter B.2, Problem 21E

(a)

To determine

To find: The mean and standard deviation of the sets of data.

Answer to Problem 21E

The mean and standard deviation are given below.

Explanation of Solution

Given:

The line plot of data sets is given as:

  

The data in discrete form is written as:

  $\left\{8,10,10,14,14,16\right\}$

The mean and standard deviation are given as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{N}\left[\begin{array}{l}\text{where are }{x}_i\text{ number of elements of data and }\\ N\text{are the total number of elements ofdata}\end{array}\right]\\ \sigma =\sqrt{\frac{{\left(x_i-\overline{x}\right)}^2}{N}}\end{array}$

The mean and standard deviation are calculated as:

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

Further simplified as:

  $\sigma \approx 2.828$ .

Therefore, the mean and standard deviation are given above.

(b)

To determine

To find: The mean and standard deviation of the sets of data.

Answer to Problem 21E

The mean and standard deviation are given below.

Explanation of Solution

Given:

The line plot of data sets is given as:

  

The data in discrete form is written as:

  $\left\{16,18,18,22,22,24\right\}$

The mean and standard deviation are given as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{N}\left[\begin{array}{l}\text{where are }{x}_i\text{ number of elements of data and }\\ N\text{are the total number of elements ofdata}\end{array}\right]\\ \sigma =\sqrt{\frac{{\left(x_i-\overline{x}\right)}^2}{N}}\end{array}$

The mean and standard deviation are calculated as:

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

Further simplified as:

  $\sigma \approx 2.828$ .

Therefore, the mean and standard deviation are given above.

(c)

To determine

To find: The mean and standard deviation of the sets of data.

Answer to Problem 21E

The mean and standard deviation are given below.

Explanation of Solution

Given:

The line plot of data sets is given as:

  

The data in discrete form is written as:

  $\left\{10,11,11,13,13,14\right\}$

The mean and standard deviation are given as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{N}\left[\begin{array}{l}\text{where are }{x}_i\text{ number of elements of data and }\\ N\text{are the total number of elements ofdata}\end{array}\right]\\ \sigma =\sqrt{\frac{{\left(x_i-\overline{x}\right)}^2}{N}}\end{array}$

The mean and standard deviation are calculated as:

  $\begin{array}{c}\overline{x}=\frac{10+11+11+13+13+14}{6}\\ =12\\ \sigma =\sqrt{\frac{{\left(10-12\right)}^2+{\left(11-12\right)}^2+{\left(11-12\right)}^2+{\left(13-12\right)}^2+{\left(13-12\right)}^2+{\left(14-12\right)}^2}{6}}\\ =\sqrt{2}\end{array}$

Further simplified as:

  $\sigma \approx 1.414$ .

Therefore, the mean and standard deviation are given above.

(d)

To determine

To find: The mean and standard deviation of the sets of data.

Answer to Problem 21E

The mean and standard deviation are given below.

Explanation of Solution

Given:

The line plot of data sets is given as:

  

The data in discrete form is written as:

  $\left\{10,11,11,13,13,14\right\}$

The mean and standard deviation are given as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_i}}{N}\left[\begin{array}{l}\text{where are }{x}_i\text{ number of elements of data and }\\ N\text{are the total number of elements ofdata}\end{array}\right]\\ \sigma =\sqrt{\frac{{\left(x_i-\overline{x}\right)}^2}{N}}\end{array}$

The mean and standard deviation are calculated as:

  $\begin{array}{c}\overline{x}=\frac{10+11+11+13+13+14}{6}\\ =12\\ \sigma =\sqrt{\frac{{\left(10-12\right)}^2+{\left(11-12\right)}^2+{\left(11-12\right)}^2+{\left(13-12\right)}^2+{\left(13-12\right)}^2+{\left(14-12\right)}^2}{6}}\\ =\sqrt{2}\end{array}$

Further simplified as:

  $\sigma \approx 1.414$ .

Therefore, the mean and standard deviation are given above.