Question

find the variance of 101, 102, 100, 100, 109, 110, 109, 108, 109

Answer 1

Σ(x) = 101 + 102 + 100 + 100 + 110 + 109 + 109 + 108 + 109 = 948

Σ(x²) = 101² + 102² + 100² + 100² + 110² + 109² + 109² + 108² + 109²

= 10201 + 10404 + 10000 + 10000 + 12100 + 11881 + 11881 + 11664 + 11881

= 100012

n = 9

Therefore Variance = 100012/9 - (948/9)² = 52/3

Standard deviation = √(σ²) = √(52/3) = 4.16(3sf)

Answer 2

oyvey!

id subtract like 105 from all of them and work the smaller set

Answer 3

{101, 102, 100, 100, 109, 110, 109, 108, 109}- 105

{-4, -3, -5, -5, 4, 5, 4, 3, 4}

thats almost mean centered already .... and alot less scary to work with

Answer 4

sum {-4, -3, -5, -5, 4, 5, 4, 3, 4} = 3

mean = 3/9 = 1/3 ; the original sets mean is therefore 105 + 1/3

but the variance of each set is the same

Answer 5

{-4, -3, -5, -5, 4, 5, 4, 3, 4} - 1/3

[ {-12, -9, -15, -15, 12, 15, 12, 9, 12} - 1]/3

[{-13, -10, -16, -16, 11, 14, 11, 8, 11}/3]^2

{169+100+256+256+121+196+121+64+121}/9 / 8

{169+100+256+256+121+196+121+64+121}/72

1404/72 = 39/2

Answer 6

and to verify

http://www.wolframalpha.com/input/?i=variance+%7B101%2C+102%2C+100%2C+100%2C+109%2C+110%2C+109%2C+108%2C+109%7D

Answer 7

52/3 for a population :)