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CUET · MATHS · PYQ PAPER 2025

A random variable X has the following probability distribution :
X0123
P(X)0.10.20.30.4
The variance of the X will be :

  1. A 1
  2. B 2
  3. C 2.3
  4. D 1.2
Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

\(E[X] = \sum x P(x) = (0)(0.1) + (1)(0.2) + (2)(0.3) + (3)(0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0\) \(E[X^2] = \sum x^2 P(x) = (0^2)(0.1) + (1^2)(0.2) + (2^2)(0.3) + (3^2)(0.4) = 0 + 0.2 + 1.2 + 3.6 = 5.0\) \(Var(X) = E[X^2] - (E[X])^2 = 5.0 - (2.0)^2 = 5.0 - 4.0 = 1.0\)