Example 28
Last updated at March 11, 2017 by Teachoo
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) = 16 Hence, probability distribution The mean Expectation value is given by E(X) = 𝒊 = 𝟏𝒏𝒙𝒊𝒑𝒊 = 1 × 16+2 × 16+ 3 × 16+ 3 × 16+ 5 × 16+ 6 × 16 = 216 The variance of x is given by : Var 𝑿=𝑬 𝑿𝟐− 𝑬 𝑿𝟐 So, finding 𝐸 𝑋2 E 𝑋2= 𝑖 = 1𝑛 𝑥𝑖2𝑝𝑖 = 12 × 16+22 × 16+ 32 × 16+ 42 × 16+ 52 × 16+ 62 × 16 = 1 + 4 + 9 + 16 + 25 + 366 = 916 Now, Var 𝑋=𝐸 𝑋2− 𝐸 𝑋2 = 916− 2162 = 916− 44136 = 546 − 44136 = 10536 = 3512 Hence, variance is 𝟑𝟓𝟏𝟐
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Chapter 13 Class 12 Probability
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About the Author
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
