Question

The probability distribution of a random variable X is p (x). Variable X can assume the values x₁ = – 2, x₂ = – 1, x₃ = 1 and x₄ = 2 and if 4p(x₁) = 2p (x₂) = 3p (x₃) = 4p (x₄), then obtain mean and variance of this probability distribution.

Answer 1

p(x) is the probability distribution of a random variable X.

X = x₁; x₂, x₃, x₄ and x₁ = – 2, x₂ = – 1, x₃ = 1, x₄ = 2.

Σp(x) = 1

∴ p (x₁) + p(x₂) + p(x₃) + p(x₄) = 1 ……………… (1)

Now, 4p(x₁) = 2p(x₂) = 3p(x₃) = 4p(x₄)

∴ 4p (x₁) = 4p(x₄), 2p(x₂) = 4p (x₄) and 3p(x₃) = 4p (x₄)

∴ p(x₁) = p(x₄), p(x₂) = 2p (x₄) and

x₁ = – 2, x₂ = – 1, x₃ = 1, and x₄ = 2, the probability distribution of random variable X is written as follows:

To find the mean and variance, the table is prepared as follows:

Mean of the distribution:

µ = E (X) = Σx ∙ p(x) = –\(\frac{1}{8}\)

Variance of the distribution:

Hence, the mean and variance of the distribution obtained are \(\frac{1}{8}\) and \(\frac{135}{64}\) respectively.