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CA Foundation · Quantitative Aptitude · Theoretical Distributions

A random variable X has the probability distribution: P(X=0)=0.2, P(X=1)=0.5, P(X=2)=0.3. What is the variance of X?

The variance is 0.49. The mean E(X) is 1.1 and E(X²) is 1.7. Variance equals E(X²) minus the square of the mean, which is 1.7 minus 1.21, giving 0.49. Quoting 1.21 would confuse the squared mean with the variance.

  1. A0.61
  2. B0.49Correct
  3. C1.10
  4. D1.21

Explanation

E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1. E(X²) = 0 + 0.5 + 4(0.3) = 1.7. Variance = 1.7 − 1.21 = 0.49. The value 1.21 is (E(X))², and 1.10 is the mean itself.

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