Class 12 Maths Probability | Mean and Variance of a Random variable |

**Mean and Variance of a Random variable**:

Let X be a random variable assuming values x1, x2,...., xn with probabilities p1, p2, ..., pn, respectively such that pi ≥ 0,

Mean of X, denoted by μ [or expected value of X denoted by E (X)] is defined as:

and variance denoted by σ^{2} , is defined as

Standard deviation of the random variable X is defined as

Let us take the above table

Now mean is

= X1P1+X2P2+X3P3

= (0X¼ ) + (1 X ½)+ (2X ¼)

= 0 + ½ + ½

=1

Now Variance = σ_{x}^{2} = E(X^{2})-(E(X))^{2}

so to find Variance let us find E(X^{2}) and (E(X))^{2}

= 0 + ½ + 1

= 3/2

= 1.5

and (E(X))^{2} = 1^{2}=1

∴ Var(X)= E(X^{2})-(E(X))^{2} = 1.5-1=0.5

now standard deviation of X = √(variance (X ))

= √0.5

≈ 0.707

* Example*: In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed,

and X = 1 if he is in favour. Find E(X) and Var(X).

It is given that P(X = 0) = 30% = 30/100 = 0.3

P(X=1)=70%= 70/100 = 0.7

Therefore, the probability distribution is as follows.

x | 0 | 1 |

P(x) | 0.3 | 0.7 |

Then, Mean= E(X) =

= 0×0.3+1×0.7

=0.7

= 02×0.3+12×0.7 =0.7

Now,

∴ Var(X)= E(X^{2})-(E(X))^{2}

= 0.7−(0.7)^{2}

= 0.7 − 0.49

= 0.21

* Example*:The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is (A) 1 (B) 2 (C) 5 (D) 8/3

* Solution*:

Let X be the random variable representing a number on the dice.

The total number of observations is six.

∴P(X=1)= 3/6 = 1/2

P(X=2)= 2/6 = 1/3

P(X=5)= 1/6

∴ The probability distribution is as follows.

X | 1 | 2 | 5 |

P(X) | 1/2 | 1/3 | 1/6 |

Now, Mean=E(X)

=1×1/2 +2×1/3 +5×1/6

= 1/2 +2/3 +5/6

= 12/6

= 2

∴ The correct option is B.

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