Question

Find the variance of the distribution:

X	0	1	2	3	4	5
P(X)	`1/6`	`5/18`	`2/9`	`1/6`	`1/9`	`1/18`

Answer 1

We know that,

Variance (X) = E(X²) – [E(X)]²

E(X) = `sum_("i" = 1)^"n" "p"_"i"x_"i"`

= `0 xx 1/6 + 1 xx 5/18 + 2 xx 2/9 + 3 xx 1/6 + 4 xx 1/9 + 5 xx 1/18`

= `0 + 5/18 + 4/9 + 3/6 + 4/9 + 5/118`

= `(5 + 8 + 9 + 8 + 5)/18`

= `35/18`

E(X)² = `0 xx 1/6 + 1 xx 5/18 + 4 xx 2/9 + 9 xx 1/6 + 16 xx 1/9 + 25 xx 1/18`

= `5/18 + 8/9 + 9/6 + 16/9 + 25/18`

= `(5 + 16 + 27 + 32 + 25)/18`

= `105/18`

∴ Var(x) = `105/18 - 35/18 xx 35/18`

= `(1890 - 1225)/324`

= `665/324`

Hence, the required variance is `665/324`.