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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`

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#### 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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