Chapter 4.6, Problem 7EE

a.

To determine

Calculate the average, the deviations from average, and the SD for given lists.

Answer to Problem 7EE

(i):

The average of the given list is 2.

The deviations from average are 3, –6, 1, –3, and 5.

The SD of the given list is 4.

(ii):

The average of the given list is –2.

The deviations from average are –3, 6, –1, 3, and –5.

The SD of the given list is 4.

Explanation of Solution

Calculation:

(i):

The average of the given list is obtained as follows:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Number of data values}}\\ =\frac{5-4+3-1+7}{5}\\ =2\end{array}$

Thus, the average of the given list is 2.

The deviations from average is calculated as follows:

$\text{Deviation from average = }\left(\text{Entry}-\text{ Average}\right)$

$\begin{array}{c}\text{Deviation from average of 5}=5-2\\ =3\end{array}$

$\begin{array}{c}\text{Deviation from average of }\left(-4\right)=-4-2\\ =-6\end{array}$

$\begin{array}{c}\text{Deviation from average of 3}=3-2\\ =1\end{array}$

$\begin{array}{c}\text{Deviation from average of }\left(-1\right)=-1-2\\ =-3\end{array}$

$\begin{array}{c}\text{Deviation from average of 7}=7-2\\ =5\end{array}$

Thus, deviations from average are 3, –6, 1, –3, and 5.

The SD is obtained as follows:

$\begin{array}{c}\text{SD}=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Number of data values }}}\\ =\sqrt{\frac{{\left(5-2\right)}^2+{\left(-4-2\right)}^2+{\left(3-2\right)}^2+{\left(-1-2\right)}^2+{\left(7-2\right)}^2}{5}}\\ =\sqrt{\frac{80}{5}}\\ =4\end{array}$

Thus, the SD of the given list is 4.

(ii):

The average of the given list is obtained as follows:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Number of data values}}\\ =\frac{-5+4-3+1-7}{5}\\ =-2\end{array}$

Thus, the average of the given list is –2.

The deviations from average are calculated as follows:

$\text{Deviation from average = }\left(\text{Entry}-\text{ Average}\right)$

$\begin{array}{c}\text{Deviation from average of }\left(-\text{5}\right)=-5+2\\ =-3\end{array}$

$\begin{array}{c}\text{Deviation from average of }4=4+2\\ =6\end{array}$

$\begin{array}{c}\text{Deviation from average of }\left(-3\right)=-3+2\\ =-1\end{array}$

$\begin{array}{c}\text{Deviation from average of }1=1+2\\ =3\end{array}$

$\begin{array}{c}\text{Deviation from average of }\left(-\text{7}\right)=-7+2\\ =-5\end{array}$

Thus, deviations from average are –3, 6, –1, 3, and –5.

The SD is obtained as follows:

$\begin{array}{c}\text{SD}=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Number of data values }}}\\ =\sqrt{\frac{{\left(-5+2\right)}^2+{\left(4+2\right)}^2+{\left(-3+2\right)}^2+{\left(1+2\right)}^2+{\left(-7+2\right)}^2}{5}}\\ =\sqrt{\frac{80}{5}}\\ =4\end{array}$

Thus, the SD of the given list is 4.

b.

To determine

Explain how the list (ii) related to list (i).

In what way the relationship carries over to the average, deviations from the average, and the SD.

Explanation of Solution

From the given lists (i) & (ii), list (ii) is obtained by changing the sign of each entry of list (i).

Hence, average of list (ii) becomes negative. The signs of deviations from the average are also changes.

Since there is no change in the sum of squares of the deviations, SD of list (ii) remains same.