Question 91 markMCQSection A

Let X be a discrete random variable, then the variance of X is :

Probability Distributions
Variance of a Random Variable

Options

(A)E(X2)E(X^2)
(B)E(X2)−[E(X)]2E(X^2) - [E(X)]^2
(C)E(X2)+[E(X)]2E(X^2) + [E(X)]^2
(D)E(X2)−[E(X)]2\sqrt{E(X^2) - [E(X)]^2}

Official answer

The correct option is (B) E(X2)−[E(X)]2E(X^2) - [E(X)]^2.


Reason

  • Variance measures spread about the mean: Var(X)=E[(X−μ)2]\text{Var}(X) = E[(X - \mu)^2].
  • Expanding gives the computational form Var(X)=E(X2)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2.
Answer keywords:variancediscrete random variableexpectationE(X^2)meansecond momentspreadcomputational formula

Marking scheme

  1. 1.1 mark: correct option (B) E(X2)−[E(X)]2E(X^2) - [E(X)]^2.
  2. 2.No marks for (A), (C) or (D); the raw second moment, the additive form, or the square-root (standard deviation) form are not accepted.

Hint

Variance = (mean of squares) minus (square of the mean).

Quick oral answer

Variance of a discrete random variable is the expected squared deviation from the mean, and its computational form is E(X squared) minus the square of E(X).

Common mistakes

  1. 1.Writing E(X2)E(X^2) alone and forgetting to subtract [E(X)]2[E(X)]^2.
  2. 2.Adding instead of subtracting the square of the mean.
  3. 3.Confusing variance with standard deviation by taking the square root, E(X2)−[E(X)]2\sqrt{E(X^2) - [E(X)]^2}.

Analysis and explanation

This is the standard short-cut (computational) formula for variance of a discrete random variable.


Concept

  • By definition Var(X)=E[(X−μ)2]\text{Var}(X) = E[(X - \mu)^2], where μ=E(X)\mu = E(X).
  • Expanding: E[X2−2μX+μ2]=E(X2)−2μE(X)+μ2=E(X2)−[E(X)]2E[X^2 - 2\mu X + \mu^2] = E(X^2) - 2\mu E(X) + \mu^2 = E(X^2) - [E(X)]^2.

Why the key is right

  • Option (B) is exactly this expanded form, so it is correct.

Why the distractors are wrong

  • (A) E(X2)E(X^2) alone is the raw second moment, not the variance; it ignores the mean correction term.
  • (C) uses a plus sign, which over-counts and can never give a spread measure.
  • (D) E(X2)−[E(X)]2\sqrt{E(X^2) - [E(X)]^2} is the standard deviation (the square root of the variance), not the variance itself.

Interesting facts

The identity Var(X)=E(X2)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2 is the discrete analogue of the 'mean of squares minus square of mean' rule used for grouped-data variance in statistics.

Because Var(X)≥0\text{Var}(X) \ge 0 always, this formula proves that E(X2)≥[E(X)]2E(X^2) \ge [E(X)]^2, a special case of the Cauchy-Schwarz inequality.

Question details

Q9 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Probability Distributions
Topic
Variance of a Random Variable
Bloom's level
remember
Syllabus topic
Unit: Probability Distributions — Expectation and Variance of a Discrete Random Variable
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is variance E(X2)−[E(X)]2E(X^2) - [E(X)]^2 and not E(X2)E(X^2)?

E(X2)E(X^2) is only the average of the squared values (raw second moment). Variance measures spread about the mean, so the square of the mean, [E(X)]2[E(X)]^2, must be subtracted to remove the location and leave only the dispersion.

How is variance different from standard deviation here?

Variance is E(X2)−[E(X)]2E(X^2) - [E(X)]^2. Its square root, E(X2)−[E(X)]2\sqrt{E(X^2) - [E(X)]^2}, is the standard deviation — which is exactly option (D), and therefore not the answer to a question asking for the variance.