# What is the variance of the following set of numbers?: {40, 56, 59, 60, 60, 62, 65, 69, 75, 84}

May 15, 2018

$123.8$

#### Explanation:

First of all, find the average, defined as the sum of all items divided by the number of items:

$\setminus \mu = \frac{1}{N} \setminus {\sum}_{i = 1}^{N} {x}_{i}$

$\setminus \mu = \setminus \frac{40 + 56 + 59 + 60 + 60 + 62 + 65 + 69 + 75 + 84}{10} = \setminus \frac{630}{10} = 63$

Then, you must compute the squared distance of each item from the average:

$40 \setminus \to {\left(40 - 63\right)}^{2} = 529$

$56 \setminus \to {\left(56 - 63\right)}^{2} = 49$

$59 \setminus \to {\left(59 - 63\right)}^{2} = 16$

$60 \setminus \to {\left(60 - 63\right)}^{2} = 9$

$60 \setminus \to {\left(60 - 63\right)}^{2} = 9$

$62 \setminus \to {\left(62 - 63\right)}^{2} = 1$

$65 \setminus \to {\left(65 - 63\right)}^{2} = 4$

$69 \setminus \to {\left(69 - 63\right)}^{2} = 36$

$75 \setminus \to {\left(75 - 63\right)}^{2} = 144$

$84 \setminus \to {\left(84 - 63\right)}^{2} = 441$

The variance is defined as the sum of the squared distances, divided by the number of items:

${\sigma}^{2} = \setminus \frac{529 + 49 + 16 + 9 + 9 + 1 + 4 + 36 + 144 + 441}{10} = \setminus \frac{1238}{10} = 123.8$