Chapter 11, Problem 5CR

To determine

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

Answer to Problem 5CR

Standard deviation and mean for player A are 15.56 and 201 respectively.

Standard deviation and mean for player B 30.32 and 201.

Explanation of Solution

Given information:

Two datasets of bowling scores.

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 =10

Substituting values,

  $\overline{X}$ $=\frac{205+185+210+174+194+190+200+219+203+230}{10}$

Solving,

  $\overline{X}$ $=\frac{2010}{10}$

Mean is 201.0

Standard deviation,

Items		$X_i-\overline{X}$
205	4	16
185	-16	256
210	9	81
174	-27	729
194	-7	49
190	-11	121
200	-1	1
219	18	324
203	2	4
230	29	841

  ${\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}$ =2422

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

Substituting values,

  $\sigma =\sqrt{\frac{2422}{10}}$

Hence, standard deviation for store A is 15.56.

Player B,

N =10

Substituting values,

  $\overline{X}$ $=\frac{228+172+154+235+168+205+181+240+235+192}{10}$

Solving,

  $\overline{X}$ $=\frac{2010}{10}$

Mean is 201.0

Standard deviation,

Items	$X_i-\overline{X}$	${(X_i-\overline{X})}^2$
228	27	729
172	29	841
154	47	2209
235	34	1156
168	-33	1089
205	4	16
181	-20	400
240	39	1521
235	34	1156
192	-9	81

  ${\displaystyle \sum _{i=1}^n{(X_i-\overline{X})}^2}$ =9198

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

Substituting values,

  $\sigma =\sqrt{\frac{9198}{10}}$

Hence, standard deviation for player B is 30.32.