What is the variance of {-7, 12, 14, 8, -10, 0, 14}?

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What is the variance of {-7, 12, 14, 8, -10, 0, 14}?

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Answer 1 · 2016-12-29T14:08:10

$2.55 (3 s . f .)$ $2.55 (3s.f.)$

Explanation:

${- 7, 12, 14, 8, - 10, 0, 14}$ ${-7, 12, 14, 8, -10, 0, 14}$

mean:

$\frac{- 7 + 12 + 14 + 8 + - 10 + 0 + 14}{7}$ $(-7+ 12+ 14+ 8+ -10 + 0+ 14)/7$

$= \frac{31}{7}$ $=31/7$

find deviations of each number ( $n$ $n$ -mean):

$- 7 - \frac{31}{7} = - \frac{49}{7} - \frac{31}{7} = \frac{80}{7}$ $-7 - 31/7 =- 49/7 - 31/7 = 80/7$

$12 - \frac{31}{7} = \frac{84}{7} - \frac{31}{7} = \frac{53}{7}$ $12 - 31/7 = 84/7 - 31/7 = 53/7$

$14 - \frac{31}{7} = \frac{98}{7} - \frac{31}{7} = \frac{67}{7}$ $14 - 31/7 = 98/7 - 31/7 = 67/7$

$8 - \frac{31}{7} = \frac{56}{7} - \frac{31}{7} = \frac{25}{7}$ $8 - 31/7 = 56/7 - 31/7 = 25/7$

$- 10 - \frac{31}{7} = - \frac{70}{7} - \frac{31}{7} = - \frac{101}{7}$ $-10 - 31/7 = -70/7 - 31/7 = -101/7$

$0 - \frac{31}{7} = - \frac{31}{7}$ $0 - 31/7 = -31/7$

$14 - \frac{31}{7} = \frac{98}{7} - \frac{67}{7} = \frac{32}{7}$ $14 - 31/7 = 98/7 - 67/7 = 32/7$

variance = mean of deviations:

$\frac{\frac{80}{7} + \frac{53}{7} + \frac{67}{7} + \frac{25}{7} - \frac{101}{7} - \frac{31}{7} + \frac{32}{7}}{7}$ $(80/7 +53/7+ 67/7+ 25/7 -101/7 -31/7 +32/7)/7$

$= \frac{125}{49}$ $=125/49$

$= 2.55 (3 s . f .)$ $=2.55 (3s.f.)$

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What is the variance of {-7, 12, 14, 8, -10, 0, 14}?

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