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Last updated at Feb. 15, 2020 by Teachoo
Example 28 Find the variance of the number obtained on a throw of an unbiased die. Let X be number obtained on a throw So, value of X can be 1, 2, 3, 4, 5 or 6 Since die unbiased, Probability of getting of each number is equal P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6 Hence, probability distribution The mean Expectation value is given by E(X) = β2_(π = π)^πβππππ = 1 Γ 1/6+2 Γ 1/6+ 3 Γ 1/6+ 3 Γ 1/6+ 5 Γ 1/6+ 6 Γ 1/6 = 21/6 The variance of x is given by : Var (πΏ)=π¬(πΏ^π )β[π¬(πΏ)]^π So, finding πΈ(π^2 ) E(π^2 )=β2_(π = 1)^πβγγπ₯_πγ^2 ππγ = 12 Γ 1/6+22 Γ 1/6+ 32 Γ 1/6+ 42 Γ 1/6+ 52 Γ 1/6+ 62 Γ 1/6 = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 Now, Var (π)=πΈ(π^2 )β[πΈ(π)]^2 = 91/6β[21/6]^2 = 91/6β441/36 = (546 β 441)/36 = 105/36 = 35/12 Hence, variance is ππ/ππ