Chapter 3.2, Problem 13P

a.

To determine

Find the range.

Answer to Problem 13P

The range is 15.

Explanation of Solution

The range is obtained below:

$\begin{array}{c}\text{Range}=\text{Largest value}-\text{Smallest value}\\ =30-15\\ =15\end{array}$

Thus, the range is 15.

b.

To determine

Verify that ${\displaystyle \sum x}=110$ and ${\displaystyle \sum x^2}=2568$ using calculator.

Explanation of Solution

Step-by-step procedure to verify ${\displaystyle \sum x}=110$ and ${\displaystyle \sum x^2}=2568$ using Ti83 calculator:

• Press STAT.
• Select Edit.
• Enter the values in L1.
• Press STAT and Choose CALC.
• Select 1-Var Stats.
• To select the variable L1, Press 2-nd and then press 1.
• Press Enter.

Output obtained using the Ti83 calculator is given below:

From the output, the values of ${\displaystyle \sum x}=110$ and ${\displaystyle \sum x^2}=2,568$.

c.

To determine

Find the sample variance and the sample standard deviation using the computation formula.

Answer to Problem 13P

The sample variance using the computation formula is 37.

The sample standard deviation using the computation formula is 6.08.

Explanation of Solution

The sample variance using the computation formula is given below:

$s^2=\frac{{\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}}{n-1}$

Where $\overline{x}$ is the mean, n is the number of values in the data, and x is the value in the data.

The sample standard deviation using the computation formula is given below:

$s=\sqrt{\frac{{\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}}{n-1}}$

Where $\overline{x}$ is the mean, n is the number of values in the data, and x is the value in the data.

The sample variance using the computation formula is obtained below:

$\begin{array}{c}{s}^2=\frac{2,568-\frac{{\left(110\right)}^2}{5}}{5-1}\\ =37\end{array}$

Thus, the sample variance using the computation formula is 37.

The sample standard deviation using the computation formula is obtained below:

$\begin{array}{c}s=\sqrt{\frac{2,568-\frac{{\left(110\right)}^2}{5}}{5-1}}\\ =\sqrt{37}\\ =6.08\end{array}$

Thus, the sample standard deviation using the computation formula is 6.08.

d.

To determine

Find the sample variance using the defining formula.

Find the sample standard deviation using the defining formula.

Answer to Problem 13P

The sample variance using the defining formula is 37.

The sample standard deviation using the defining formula is 6.08.

Explanation of Solution

The sample variance using the defining formula is given below:

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

Where $\overline{x}$ is the mean, n is the number of values in the data, and x is the value in the data.

The sample standard deviation using the defining formula is given below:

$s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}$

Where $\overline{x}$ is the mean, n is the number of values in the data, and x is the value in the data.

The value of $\overline{x}$ is obtained below:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{23+17+15+30+25}{5}\\ =22\end{array}$

The value of ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ is obtained below:

x	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
23	1	1
17	–5	25
15	–7	49
30	8	64
25	3	9
		${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=148$

The sample variance using the defining formula is given below:

$\begin{array}{c}{s}^2=\frac{148}{5-1}\\ =\frac{148}{4}\\ =37\end{array}$

Thus, the sample variance using the defining formula is 37.

The sample standard deviation using the defining formula is given below:

$\begin{array}{c}s=\sqrt{\frac{148}{5-1}}\\ =\sqrt{\frac{148}{4}}\\ =6.08\end{array}$

Thus, the sample standard deviation using the defining formula is 6.08.

e.

To determine

Find the population variance using the defining formula.

Find the population standard deviation using the defining formula.

Answer to Problem 13P

The population variance using the defining formula is 29.6.

The population standard deviation using the defining formula is 5.44.

Explanation of Solution

The population variance using the defining formula is given below:

${\sigma }^2=\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{n}$

Where $\mu$ is the population mean, N is the number of values in the population, and x is the value in the data.

The population standard deviation using the defining formula is given below:

$\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{N}}$

Where $\mu$ is the population mean, N is the number of values in the population, and x is the value in the data.

The value of $\mu$ is obtained below:

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x}}{N}\\ =\frac{23+17+15+30+25}{5}\\ =22\end{array}$

The value of ${\displaystyle \sum {\left(x-\mu \right)}^2}$ is obtained below:

x	$x-\mu$	${\left(x-\mu \right)}^2$
23	1	1
17	–5	25
15	–7	49
30	8	64
25	3	9
		${\displaystyle \sum {\left(x-\mu \right)}^2}=148$

The population variance using the defining formula is given below:

$\begin{array}{c}{\sigma }^2=\frac{148}{5}\\ =29.6\end{array}$

Thus, the population variance using the defining formula is 29.6.

The population standard deviation using the defining formula is given below:

$\begin{array}{c}\sigma =\sqrt{\frac{148}{5}}\\ =5.44\end{array}$

Thus, the population standard deviation using the defining formula is 5.44.