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Find expected value and variance of X for the following p.m.f.

x |
-2 | -1 | 0 | 1 | 2 |

P(X) |
0.2 | 0.3 | 0.1 | 0.15 | 0.25 |

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#### Solution 1

We construct the following table to calculate E (X) and V (X) :

X = x_{i} |
_{pi }=P [X = x_{i}] |
x_{i} · p_{i} |
x_{i} ^{2}·p_{i} = x_{i} × x_{i}·p_{i} |

-2 | 0.2 | -0.4 | 0.8 |

-1 | 0.3 | -0.3 | 0.3 |

0 | 0.1 | 0 | 0 |

1 | 0.15 | 0.15 | 0.15 |

2 | 0.25 | 0.5 | 1 |

Total | 1 | -0.05 | 2.25 |

From the table, Σx_{i} · p_{i} = -0.05 and Σx_{i}^{2} · p_{i} = 2.25

∴E (X) = Σx_{i} · p_{i}= -0.05

and V (X) = Σx_{i}^{ 2 }·p_{i }- ( Σx_{i} · p_{i})^{2}

= 2.25 - (-0.05)^{2}

= 2.25 - 0.0025 = 2.2475

Hence, E (X)= -0.05 and V (X) = 2.2475.

#### Solution 2

Expected value of X 5

= E(X) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]

= (–2) x (0.2) + (–1) x (0.3) + 0 x (0.1) + 1 x (0.15) + 2 x (0.25)

= – 0.4 – 0.3 + 0 + 0.15 + 0.5

= – 0.05

E(X^{2}) = \[\sum\limits_{i=1}^{5} x_i.\text{P}_i(x_i)\]

= (–2)^{2} x (0.2) + (–1)2 x (0.3) + 02 x (0.1) + 12 x (0.15) + 22 x (0.25)

= 0.8 + 0.3 + 0 + 0.15 + 1

= 2.25

∴ Variance of X

= Var(X)

= E(X^{2}) – [E(X)]^{2 }

= 2.25 – (– 0.05)^{2 }

= 2.2475.

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