What is the standard deviation of 1, 2, 3, 4, and 5?

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What is the standard deviation of 1, 2, 3, 4, and 5?

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Answer 1 · 2017-01-24T21:43:27

$\frac{\sqrt{10}}{2}$ $sqrt(10)/2$

Explanation:

The formula for standard deviation is

$s = \sqrt{\frac{\sum {(x - ¯ x)}^{2}}{n - 1}}$ $s=sqrt((sum(x-barx)^2)/(n-1))$

In this case $n = 5$ $n=5$

and $¯ x = \frac{1}{n} n \sum k = 1 x_{k} = \frac{1}{5} 5 \sum k = 1 k = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$ $barx=1/nsum_(k=1)^nx_k=1/5sum_(k=1)^5k=(1+2+3+4+5)/5=15/5=3$

then the sum of squares $n \sum k = 1 {(x - ¯ x)}_{k}^{2}$ $sum_(k=1)^n(x-barx)_k^2$ is

$5 \sum k = 1 {(x - 3)}_{k}^{2} = {(- 2)}^{2} + {(- 1)}^{2} + 0^{2} + 1^{2} + 2^{2}$ $sum_(k=1)^5(x-3)_k^2=(-2)^2+(-1)^2+0^2+1^2+2^2$

$= 4 + 1 + 0 + 1 + 4 = 10$ $=4+1+0+1+4=10$

Then we plug in

$s = \sqrt{\frac{\sum {(x - ¯ x)}^{2}}{n - 1}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$ $s=sqrt((sum(x-barx)^2)/(n-1))=sqrt(10/4)=sqrt(10)/2$

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What is the standard deviation of 1, 2, 3, 4, and 5?

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