Notes On Variance of a Random Variable - CBSE Class 12 Maths
In statistics, variance is a measure of the spread or scatter in data. Variance can also be a measure of the spread of the values of a random variable in a probability distribution. Var(X) = σx2 = ∑i=1n(xi - μ)2p(xi) x1,x2,x3,.....,xn are the possible values of X and they occur with the probabilities p(x1), p(x2),p(x3), ...p(xn), respectively. μ = E(X) = Mean of X σx2 = E(X - μ)2 The square root of the variance of random variable X is called the standard deviation of X. σx = √Var(X) = √(∑i=1n(xi - μ)2p(xi)) Var(X) = σx2 = ∑i=1n(xi - μ)2p(xi) = ∑i=1n(xi2 + μ2 - 2μxi)p(xi) = ∑i=1n xi2 p(xi) + ∑i=1n μ2 p(xi) - ∑i=1n 2μxi p(xi) = ∑i=1n xi2 p(xi) + ∑i=1n μ2 p(xi) - 2μ ∑i=1n xi p(xi) = ∑i=1n xi2 p(xi) + μ2 - 2μ2 ['.' ∑i=1n p(xi) = 1 and μ = ∑i=1n xi p(xi) ] = ∑i=1n xi2 p(xi) - μ2 = ∑i=1n xi2 p(xi) - [∑i=1n xi p(xi)]2 ⇒ Var(X) = E(X2) - [E(X)]2 Ex: The probability distributions of two experiments are   X  0   1   2   3    Y   ‑ 1   0   3   5   6  P(X) 1/8 3/8 3/8 1/8   P(Y)  2/8  3/8 1/8 1/8 1/8 Calculate the mean of the random variables in both the cases. Mean of X = E(X) = ∑i=14 xi pi = x1 p1 + x2 p2 + x3 p3 + x4 p4 = (0 x 1/8) + (1 x 3/8) + (2 x 3/8) + (3 x 1/8) = 0 + 3/8 + 6/8 + 3/8 = (3+6+3)/8 = 12/8 = 1.5 Mean of Y = E(Y) = ∑i=15 xi pi = x1 p1 + x2 p2 + x3 p3 + x4 p4 + x5 p5 = (-1 x 2/8) + (0 x 3/8) + (3 x 1/8) + (5 x 1/8) + (6 x 1/8) = -2/8 + 0 + 3/8 + 5/8 + 6/8 = (-2+3+5+6)/8 = 12/8 = 1.5 Observe that both the distributions have the same mean. It does not give us any information about the variability in the values of the random variables. To understand the variability in the variables, draw two graphs using the distributions ... by taking the values of the random variables along the X-axis and their probability along the Y-axis. Var(X) = σx2 = ∑i=14(xi - μ)2p(xi) = (0-1.5)2 x 1/8 + (1 - 1.5)2 x 3/8 + (2 - 1.5)2 x 3/8 + (3 - 1.5)2x 1/8 = (-1.5)2 x 1/8 + (- 0.5)2 x 3/8 + (0.5)2 x 3/8 + (1.5)2x 1/8 = 2.25 x 1/8 + 0.25 x 3/8 + 0.25 x 3/8 + 2.25 x 1/8 = 0.281 + 0.094 + 0.094 + 0.281 = 0.469 Var(Y) = σy2 = ∑i=14(yi - μ)2p(yi) = (-1 -1.5)2 x 2/8 + (0 - 1.5)2 x 3/8 + (3 - 1.5)2 x 1/8 + (5 - 1.5)2x 1/8 + (6 - 1.5)2x 1/8 = (-2.5)2 x 2/8 + (- 1.5)2 x 3/8 + (1.5)2 x 1/8 + (3.5)2x 1/8 + (4.5)2x 1/8 = 6.25 x 2/8 + 2.25 x 3/8 + 2.25 x 1/8 + 12.25 x 1/8 + 20.25 x 1/8 = 1.562 + 0.844 + 0.781 + 2.531 + 3.781 = 6.749 ⇒ Var(Y) > Var(X) The spread of values is more in Y than in X.

#### Summary

In statistics, variance is a measure of the spread or scatter in data. Variance can also be a measure of the spread of the values of a random variable in a probability distribution. Var(X) = σx2 = ∑i=1n(xi - μ)2p(xi) x1,x2,x3,.....,xn are the possible values of X and they occur with the probabilities p(x1), p(x2),p(x3), ...p(xn), respectively. μ = E(X) = Mean of X σx2 = E(X - μ)2 The square root of the variance of random variable X is called the standard deviation of X. σx = √Var(X) = √(∑i=1n(xi - μ)2p(xi)) Var(X) = σx2 = ∑i=1n(xi - μ)2p(xi) = ∑i=1n(xi2 + μ2 - 2μxi)p(xi) = ∑i=1n xi2 p(xi) + ∑i=1n μ2 p(xi) - ∑i=1n 2μxi p(xi) = ∑i=1n xi2 p(xi) + ∑i=1n μ2 p(xi) - 2μ ∑i=1n xi p(xi) = ∑i=1n xi2 p(xi) + μ2 - 2μ2 ['.' ∑i=1n p(xi) = 1 and μ = ∑i=1n xi p(xi) ] = ∑i=1n xi2 p(xi) - μ2 = ∑i=1n xi2 p(xi) - [∑i=1n xi p(xi)]2 ⇒ Var(X) = E(X2) - [E(X)]2 Ex: The probability distributions of two experiments are   X  0   1   2   3    Y   ‑ 1   0   3   5   6  P(X) 1/8 3/8 3/8 1/8   P(Y)  2/8  3/8 1/8 1/8 1/8 Calculate the mean of the random variables in both the cases. Mean of X = E(X) = ∑i=14 xi pi = x1 p1 + x2 p2 + x3 p3 + x4 p4 = (0 x 1/8) + (1 x 3/8) + (2 x 3/8) + (3 x 1/8) = 0 + 3/8 + 6/8 + 3/8 = (3+6+3)/8 = 12/8 = 1.5 Mean of Y = E(Y) = ∑i=15 xi pi = x1 p1 + x2 p2 + x3 p3 + x4 p4 + x5 p5 = (-1 x 2/8) + (0 x 3/8) + (3 x 1/8) + (5 x 1/8) + (6 x 1/8) = -2/8 + 0 + 3/8 + 5/8 + 6/8 = (-2+3+5+6)/8 = 12/8 = 1.5 Observe that both the distributions have the same mean. It does not give us any information about the variability in the values of the random variables. To understand the variability in the variables, draw two graphs using the distributions ... by taking the values of the random variables along the X-axis and their probability along the Y-axis. Var(X) = σx2 = ∑i=14(xi - μ)2p(xi) = (0-1.5)2 x 1/8 + (1 - 1.5)2 x 3/8 + (2 - 1.5)2 x 3/8 + (3 - 1.5)2x 1/8 = (-1.5)2 x 1/8 + (- 0.5)2 x 3/8 + (0.5)2 x 3/8 + (1.5)2x 1/8 = 2.25 x 1/8 + 0.25 x 3/8 + 0.25 x 3/8 + 2.25 x 1/8 = 0.281 + 0.094 + 0.094 + 0.281 = 0.469 Var(Y) = σy2 = ∑i=14(yi - μ)2p(yi) = (-1 -1.5)2 x 2/8 + (0 - 1.5)2 x 3/8 + (3 - 1.5)2 x 1/8 + (5 - 1.5)2x 1/8 + (6 - 1.5)2x 1/8 = (-2.5)2 x 2/8 + (- 1.5)2 x 3/8 + (1.5)2 x 1/8 + (3.5)2x 1/8 + (4.5)2x 1/8 = 6.25 x 2/8 + 2.25 x 3/8 + 2.25 x 1/8 + 12.25 x 1/8 + 20.25 x 1/8 = 1.562 + 0.844 + 0.781 + 2.531 + 3.781 = 6.749 ⇒ Var(Y) > Var(X) The spread of values is more in Y than in X.

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