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CA Foundation · Quantitative Aptitude · Theoretical Distributions

A discrete random variable X takes the values 1, 2, 3 and 4 with probabilities k, 2k, 3k and 4k respectively. What is the value of E(X)?

E(X) equals 3.0. Since the probabilities k, 2k, 3k and 4k must total 1, k is 0.1. The mean is the sum of each value times its probability: 0.1 + 0.4 + 0.9 + 1.6 = 3.0. A simple average of values would wrongly give 2.5.

  1. A2.5
  2. B3.0Correct
  3. C3.5
  4. D4.0

Explanation

The probabilities must sum to 1, so 10k = 1 and k = 0.1. E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0. The value 2.5 is the simple average of the values, which ignores the unequal probabilities.

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