Question
Download Solution PDFA 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:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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.
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.