# What is the standard deviation of the data set 11, 5, 12, 8, 5, 12, 10 rounded to the nearest tenth?

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#### Explanation:

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Aug 3, 2017

$s = 3.1$

#### Explanation:

First, we need to determine whether we are looking for the population or sample standard deviation. It is not clear from the question which it should be, but let's guess that it is the sample version, since this says it's a "data set" implying that there is a larger population that this is just a sample of. The sample standard deviation is given by the formula:

$s = \sqrt{\frac{1}{N - 1} {\sum}_{i = 1}^{N} {\left({x}_{i} - \overline{x}\right)}^{2}}$

where $\overline{x}$ is the sample mean. We can calculate the sample mean as

$\overline{x} = \frac{1}{N} {\sum}_{i = 1}^{N} {x}_{i} = \frac{1}{7} \times \left(11 + 5 + 12 + 8 + 5 + 12 + 10\right) \to$

where N=7, the number of data points in the sample.

$\overline{x} = \frac{11 + 5 + 12 + 8 + 5 + 12 + 10}{7} = 9$

In the same way, we can plug our values into the formula above to get:

$s = \sqrt{\frac{1}{7 - 1} \times \left({\left(11 - 9\right)}^{2} + {\left(5 - 9\right)}^{2} + {\left(12 - 9\right)}^{2} + {\left(8 - 9\right)}^{2} + {\left(5 - 9\right)}^{2} + {\left(12 - 9\right)}^{2} + {\left(10 - 9\right)}^{2}\right)}$

$s = \sqrt{\frac{{\left(11 - 9\right)}^{2} + {\left(5 - 9\right)}^{2} + {\left(12 - 9\right)}^{2} + {\left(8 - 9\right)}^{2} + {\left(5 - 9\right)}^{2} + {\left(12 - 9\right)}^{2} + {\left(10 - 9\right)}^{2}}{6}}$

$s = \sqrt{\frac{4 + 16 + 9 + 1 + 16 + 9 + 1}{6}} = \sqrt{\frac{56}{6}} = \sqrt{9.33}$

$s = 3.1$ rounded to the nearest tenth

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