Question

In Exercises , we $3.65-3.70$have provided simple data sets for you to practice the basics of finding measures of variation . For each data set determine.

a) Range

b) sample standard deviation

2,5,0,-1

Short answer

$$\begin{gathered} a)Range=6 \\ b)samples\tan darddeviation=2.6458 \end{gathered}$$

Step-by-step solution

Step 1. Given information. 

We have given data :$2,5,0,-1$

Step 2. Range of the data.  

$$\begin{gathered} Range=Max.value-Min.value \\ Range=5-(-1) \\ Range=5+1 \\ Range=6 \end{gathered}$$

Step 3. Mean of the data.   

$$\begin{gathered} \sqrt{\frac{{\displaystyle \sum _{}\{x-\overline{x}{)}^2}}{n-1}} \\ Mean,\overline{x}=\frac{{\displaystyle \sum _{}}x}{n} \\ Mean=\frac{2+5+0+(-1)}{4} \\ Mean=\frac{6}{4} \\ Mean=1.5 \\ sampiesize,n=4 \end{gathered}$$

Step 4. Standard deviation of the data.  

Variable $x$	Deviation from mean $x-\overline{x}$	Squared deviation ${(x-\overline{x})}^2$
$2$	$0.5$	$0.25$
$5$	$3.5$	$12.25$
$0$	$-1.5$	$2.25$
$-1$	$-2.5$	$6.25$
		$21.0$

Thus, the standard deviation is

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \sum _{}}{(x_1-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{21.0}{4-1}} \\ =\sqrt{7} \\ =2.6458 \end{gathered}$$