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Let the p.m.f. ( probability mass function ) of random variable x be `P(x) = (4/x)(5/9)^(x)(4/x)^(4-x), x = 0, 1, 2, 3, 4` = 0 , otherwise Find E(x) and Var.(x) |
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Answer» `P(x) = (4/x)(5/9)^(x)(4/x)^(4-x)` ` x = 0, 1, 2, 3, 4` = 0 otherwise Comparing this with `P(x) = .^(n)C_(x) p^(x)q^(n-x)` we get ` n = 4, p = 5/9, q = 4/9` ` E(x) = np = 4 (5/9) = 20/9` ` = 2*22` Var (x) = npq = ` 4 xx 5/9 xx 4/9` `= 80/81 ` `= 0.9876 ` Hence ` E(x) = 2*22 and Var(x) = 0.9876`. |
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