Chapter 10.3, Problem 3GP

To determine

To calculate standard deviation

Answer to Problem 3GP

Standard deviation=73.124

Explanation of Solution

Given information:

Data points are 1950, 2000, 2100, 1900, 2100, 2000

Formula used:

Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

  ${\displaystyle \sum _{i=1}^n{X}_i}$ = sum of all terms Standard deviation, ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$

Where $x_i^{}$ = each value from data

  $\mu$ =mean of data

  $N$ = size of data

Calculation:

For mean,

Substituting values,

Mean= $\frac{1950+2000+2100+1900+2100+2000}{6}$ = $\frac{12,050}{6}=2008.3$

Values	Difference= $x_{}^{}-\mu$	${(x_{}^{}-\mu )}^2$
1950	-58.3	3402.7
2000	-8.3	69.4
2100	91.6	8402.7
1900	-108.3	11736.1
2100	91.6	8402.7
2000	-8.3	69.4
		${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =32083.3

Standard deviation, ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$ = $\sqrt{\frac{32083.3}{6}}$ = 73.124