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**Solve the following problem :**

Let the p. m. f. of the r. v. X be

`"P"(x) = {((3 - x)/(10)", ","for" x = -1", "0", "1", "2.),(0,"otherwise".):}`

Calculate E(X) and Var(X).

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#### Solution 1

P(X)=`(3-x) /10`

X takes values -1, 0, 1, 2

P(X = -1)= P(-1) = `(3+1) /10 =4/ 10`

P(X = 0)= P(0) = `(3-0) /10 =3/ 10`

P(X = 1)= P(1) = `(3-1) /10 =2/ 10`

P(X = 2)= P(2) = `(3-2) /10 =1/ 10`

We construct the following table to calculate the mean and variance of X :

x_{i} |
P (x_{i}) |
x_{i }P (x_{i}) |
x_{i}^{2} P (x_{i}) |

-1 | `4/10` | -`4/10` | `4/10` |

0 | `3/10` | 0 | 0 |

1 | `2/ 10` | `2/10` | `2/10` |

2 | `1/10` | `2/10` | `4/10` |

Total |
1 |
0 |
1 |

From the table

∑x_{i} P(x_{i})0 and ∑x_{i}^{2 }·P(x_{i})= 1

E(X) = x_{i} P(x_{i}) = 0

Var (X) = ∑x_{i}^{2} · P(x_{i}) - [E(X)]^{2}

= 1 - 0 = 1

Hence, E(X) = 0, Var (X) = 1.

#### Solution 2

E(X) = `sum_("i" = 1)^4x_"i""P"(x_"i")`

= `-1 xx ((3 - (-1))/10) + 0 xx ((3 - 0)/10) + 1 xx ((3 - 1)/10) + 2 xx ((3 - 2)/10)`

= `(-4 + 0 + 2 + 2)/10`

= 0

E(X^{2}) = `sum_("i" = 1)^4x_"i"^2"P"(x_"i")`

= `(-1)^2 xx 4/10 + (0)^2 xx 3/10 + (1)^2 xx 2/10 + (2)^2 xx 1/10`

= `(4 + 0 + 2 + 4)/10`

= 1

Var(X) = E(X^{2}) – [E(X)]^{2}

= 1 – (0)^{2 }

= 1

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