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 σ2 , 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 = σx2 = E(X2)-(E(X))2

so to find Variance let us find E(X2) and (E(X))2

= 0 + ½ + 1
= 3/2
= 1.5

and (E(X))2 = 12=1

∴ Var(X)= E(X2)-(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).
Solution:
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(X2)-(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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