# How do you find the range, variance, and standard deviation for 1, 2, 3, 4, 5, 6, 7?

Jan 21, 2017

The range is $6$ the variance is $\frac{14}{3}$, and the standard deviation is $\sqrt{\frac{14}{3}}$.

#### Explanation:

For a sample set $S = \left\{1 , 2 , 3 , 4 , 5 , 6 , 7\right\}$

The range is $\max \left(S\right) - \min \left(S\right) = 7 - 1 = 6$

The average $\overline{x} = \frac{1}{n} {\sum}_{k = 1}^{n} k$ where $n = 7$

Then $\overline{x} = \frac{1}{7} {\sum}_{k = 1}^{7} k = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7}{7} = \frac{28}{7} = 4$

This is important because the variance

${s}^{2} = \frac{1}{n - 1} {\sum}_{k = 1}^{n} {\left({x}_{k} - \overline{x}\right)}_{k}^{2}$, and since $n = 7$ and $\overline{x} = 4$

${s}^{2} = \frac{1}{6} {\sum}_{k = 1}^{7} {\left(x - 4\right)}_{k}^{2} = \frac{1}{6} \left({\left(- 3\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} + {0}^{2} + {1}^{2} + {2}^{2} + {3}^{2}\right) = \frac{{\left(- 3\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} + {0}^{2} + {1}^{2} + {2}^{2} + {3}^{2}}{6} = \frac{9 + 4 + 1 + 0 + 1 + 4 + 9}{6} = \frac{2 + 8 + 18}{6} = \frac{28}{6} = \frac{14}{3}$

And since the standard deviation is the square root of the variance $s = \sqrt{{s}^{2}}$

$s = \sqrt{\frac{14}{3}}$