Example 28 - Find variance of number obtained on throw of die

Example 28 - Chapter 13 Class 12 Probability - Part 2
Example 28 - Chapter 13 Class 12 Probability - Part 3

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Question 7 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) = โˆ‘2_(๐’Š = ๐Ÿ)^๐’โ–’๐’™๐’Š๐’‘๐’Š = 1 ร— 1/6+2 ร— 1/6+ 3 ร— 1/6+ 3 ร— 1/6+ 5 ร— 1/6+ 6 ร— 1/6 = 21/6 The variance of x is given by : Var (๐‘ฟ)=๐‘ฌ(๐‘ฟ^๐Ÿ )โˆ’[๐‘ฌ(๐‘ฟ)]^๐Ÿ So, finding ๐ธ(๐‘‹^2 ) E(๐‘‹^2 )=โˆ‘2_(๐‘– = 1)^๐‘›โ–’ใ€–ใ€–๐‘ฅ_๐‘–ใ€—^2 ๐‘๐‘–ใ€— = 12 ร— 1/6+22 ร— 1/6+ 32 ร— 1/6+ 42 ร— 1/6+ 52 ร— 1/6+ 62 ร— 1/6 = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 Now, Var (๐‘‹)=๐ธ(๐‘‹^2 )โˆ’[๐ธ(๐‘‹)]^2 = 91/6โˆ’[21/6]^2 = 91/6โˆ’441/36 = (546 โˆ’ 441)/36 = 105/36 = 35/12 Hence, variance is ๐Ÿ‘๐Ÿ“/๐Ÿ๐Ÿ

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