A random variable X has the distribution law as given below:X123P(X =

1.	A random variable X has the distribution law as given below:X123P(X = x)0.30.40.3The variance of the distribution is:1. 0.42. 0.63. 0.24. None of these
Answer» Correct Answer - Option 2 : 0.6 Concept: For a random variable X = xi with probabilities P(X = x) = pi: Mean/Expected Value: μ = ∑pixi. Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2. Standard Deviation: σ = \(\rm \sqrt{Var(X)}\). Calculation: We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3. Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3) = 0.3 + 0.8 + 0.9 = 2 And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²) = 0.3 + 1.6 + 2.7 = 4.6 ∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6. The variance of the distribution is Var(X) = 0.6.

A random variable X has the distribution law as given below:X123P(X = x)0.30.40.3The variance of the distribution is:1. 0.42. 0.63. 0.24. None of these

Answer» Correct Answer - Option 2 : 0.6

Concept:

For a random variable X = xi with probabilities P(X = x) = pi:

Mean/Expected Value: μ = ∑pixi.
Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
Standard Deviation: σ = \(\rm \sqrt{Var(X)}\).

Calculation:

We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.

Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)

= 0.3 + 0.8 + 0.9

= 2

And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²)

= 0.3 + 1.6 + 2.7

= 4.6

∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.

The variance of the distribution is Var(X) = 0.6.

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A random variable X has the distribution law as given below:X123P(X = x)0.30.40.3The variance of the distribution is:1. 0.42. 0.63. 0.24. None of these

A random variable X has the distribution law as given below:X123P(X = x)0.30.40.3The variance of the distribution is:1. 0.42. 0.63. 0.24. None of these

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