Chapter 11, Problem 6CR

To determine

To find the range and standard deviation of each data set.

Answer to Problem 6CR

Standard deviation and mean for store A are $39.37 and $190 respectively.

Standard deviation and mean for store B is $48.08 and $232.5.

Explanation of Solution

Given information:

Two datasets of tablet price

Formula used:

Mean, $\overline{X}=\frac{{\displaystyle \sum _{}^{}X}}{n}$

Standard deviation, $\sigma =\sqrt{{\displaystyle \sum _{}^{}\frac{{(X_i-\overline{X})}^2}{N}}}$ .

Calculation:

Store A,

N =8

Substituting values,

  $\overline{X}$

  $=\frac{140+200+150+250+180+250+190+160}{8}$

Solving,

  $\overline{X}$

  $=\frac{1250}{8}$

Mean is $232.5

Standard deviation,

Items	$X_i-\overline{X}$	${(X_i-\overline{X})}^2$
140	-7.5	56.25,
200	27.5	756.25
150	-42.5	1806.25
250	-72.5	5256.25
180	77.5	6006.25
250	-42.5	1806.25
190	52.5	2756.25
160	7.5	56.25

  ${\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}$ =18500

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

Substituting values,

  $\sigma =\sqrt{\frac{18500}{8}}$

Hence, standard deviation for store A is $48.08.

Store B,

N =8

Substituting values,

  $\overline{X}$

  $=\frac{225+260+190+160+310+190+285+240}{8}$

Solving,

  $\overline{X}$

  $=\frac{1860}{8}$

Mean is $190.

Standard deviation,

Items	$X_i-\overline{X}$	${(X_i-\overline{X})}^2$
225	-50	2500
260	10	100
190	-40	1600
160	60	3600
310	-10	100
190	60	3600
285	0	0
240	-30	900

  ${\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}$ =12400

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

Substituting values,

  $\sigma =\sqrt{\frac{12400}{8}}$

Hence, standard deviation for store A is $39.37.