Example 16 - Variance of 20 observations is 5. If each - Indirect questions - Multiplication of observation

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  1. Chapter 15 Class 11 Statistics
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Example 16 The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations. Let the observations be 𝑥﷮1﷯, 𝑥﷮2﷯, 𝑥﷮3﷯, ..., 𝑥﷮20﷯ and 𝑥﷯ be their mean. Given that Variance = 5 and n = 20. We know that Variance = 1﷮n﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 5 = 1﷮20﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 5 × 20 = ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ 100 = ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ = 100 If each observation is multiplied by 2 , we get new observations, Let the new observations be 𝑦﷮1﷯, 𝑦﷮2﷯, 𝑦﷮3﷯, ..., 𝑦﷮20﷯ where 𝑦﷮𝑖﷯ = 2 ( 𝑥﷮𝑖﷯) We need to find variance of the new observations i.e. New Variance = 1﷮n﷯ ﷮﷮( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ Now, We know 𝑦﷮𝑖﷯ = 2 ( 𝑥﷮𝑖﷯) Calculating 𝑦﷯ in terms of 𝑥﷯ 𝑦﷯ = 1﷮𝑛﷯ ﷮﷮ 𝑦﷮𝑖﷯﷯ 𝑦﷯ = 1﷮20﷯ ﷮﷮ 2𝑥﷮𝑖﷯﷯ 𝑦﷯ = 2 ( 1﷮20﷯ ﷮﷮ 𝑥﷮𝑖﷯﷯) 𝑦﷯ = 2 𝑥﷯ From (1) ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ = 100 ﷮﷮( 1﷮2﷯ 𝑦﷮𝑖﷯﷯− 1﷮2﷯ 𝑦﷯)﷮2﷯ = 100 ﷮﷮( 1﷮2﷯ (𝑦﷮𝑖﷯﷯− 𝑦﷯))﷮2﷯ = 100 ﷮﷮ 1﷮2﷯﷯﷮2﷯( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 100 1﷮2﷯﷯﷮2﷯ ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 100 ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 100 × 4 ﷮﷮ ( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 400 Now, New Variance = 1﷮n﷯ ﷮﷮( 𝑦﷮𝑖﷯﷯− 𝑦﷯)﷮2﷯ = 1﷮20﷯ × 400 = 20

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