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

X

1

2

3

P(X = x)

0.3

0.4

0.3


The variance of the distribution is:

This question was previously asked in
NIMCET 2013 Official Paper
View all NIMCET Papers >
  1. 0.4
  2. 0.6
  3. 0.2
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 0.6
Free
NIMCET 2020 Official Paper
120 Qs. 480 Marks 120 Mins

Detailed Solution

Download Solution PDF

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 = .

 

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.

Latest NIMCET Updates

Last updated on Jun 12, 2025

->The NIMCET 2025 provisional answer key is out now. Candidates can log in to the official website to check their responses and submit objections, if any till June 13, 2025.

-> NIMCET exam was conducted on June 8, 2025.

-> NIMCET 2025 admit card was out on June 3, 2025.

-> NIMCET 2025 results will be declared on June 27, 2025. Candidates are advised to keep their login details ready to check their scrores as soon as the result is out.

-> Check NIMCET 2025 previous year papers to know the exam pattern and improve your preparation.

More Mean and Variance of Random variables Questions

More Probability Questions

Hot Links: teen patti real money app teen patti master teen patti gold new version