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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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Transcript

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) = βˆ‘2_(π’Š = 𝟏)^π’β–’π’™π’Šπ’‘π’Š = 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 (𝑿)=𝑬(𝑿^𝟐 )βˆ’[𝑬(𝑿)]^𝟐 So, finding 𝐸(𝑋^2 ) E(𝑋^2 )=βˆ‘2_(𝑖 = 1)^𝑛▒〖〖π‘₯_𝑖〗^2 𝑝𝑖〗 = 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 (𝑋)=𝐸(𝑋^2 )βˆ’[𝐸(𝑋)]^2 = 91/6βˆ’[21/6]^2 = 91/6βˆ’441/36 = (546 βˆ’ 441)/36 = 105/36 = 35/12 Hence, variance is πŸ‘πŸ“/𝟏𝟐

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.