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Question
Using the following activity, find the expected value and variance of the r.v.X if its probability distribution is as follows.
| x | 1 | 2 | 3 |
| P(X = x) | `1/5` | `2/5` | `2/5` |
Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`
E(X) = `square + square + square = square`
Var(X) = `"E"("X"^2) - {"E"("X")}^2`
= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`
= `square - square`
= `square`
Solution
µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`
E(X) = `1(1/5)+ 2(2/5) + 3(2/5) = 11/5`
`sumx^2 "P"_"i" = 1(1/5) + 4(2/5) + 9(2/5) = 27/5`
Var(X) = `"E"("X"^2) - {"E"("X")}^2`
= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`
= `27/5 - 121/25`
= `14/25`
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