If a Poisson distribution is such that P(X=2) = P(X=3) then the variance of the distribution

Option b: 3. Solution: Let the Poisson parameter be m . The probability mass function is: P(X = k) = -m mkk! Given that: P(X = 2) = P(X = 3) Substituting the formulas: -m m22! = -m m33! -m m22 = -m m36 Since e-m ≠ 0 and m > 0 : 1/2 = m/6 m = 3 Since the variance of a Poisson distribution is equal to its parameter m , the variance is 3 . Hence, Option B is the correct answer.

If a Poisson distribution is such that $P(X=2) = P(X=3)$ then the variance of the distribution

The correct answer is Option b: 3. Let the Poisson parameter be $m$. The probability mass function is: $$P(X = k) = \frac{e^{-m} m^k}{k!}$$ Given that: $$P(X = 2) = P(X = 3)$$ Substituting the formulas: $$\frac{e^{-m} m^2}{2!} = \frac{e^{-m} m^3}{3!}$$ $$\frac{e^{-m} m^2}{2} = \frac{e^{-m} m^3}{6}$$ Since $e^{-m} \neq 0$ and $m > 0$: $$\frac{1}{2} = \frac{m}{6} \implies m = 3$$ Since the variance of a Poisson distribution is equal to its parameter $m$, the variance is $3$. Hence, **Option B** is the correct answer.

Theoretical DistributionsMCQPYQ Dec 22Question 3499 of 230

All Questions

If a Poisson distribution is such that $\displaystyle P(X=2) = P(X=3)$ then the variance of the distribution

Options

\displaystyle \sqrt{3}

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Correct Answer

✅ Option b — 3

All Options:

A $\displaystyle \sqrt{3}$
B3
C6
D9

Detailed Solution & Explanation

Let the Poisson parameter be

\displaystyle m

. The probability mass function is:

P(X = k) = \frac{e^{-m} m^k}{k!}

Given that:

P(X = 2) = P(X = 3)

Substituting the formulas:

\frac{e^{-m} m^2}{2!} = \frac{e^{-m} m^3}{3!}

\frac{e^{-m} m^2}{2} = \frac{e^{-m} m^3}{6}

Since

\displaystyle e^{-m} \neq 0

and

\displaystyle m > 0

\frac{1}{2} = \frac{m}{6} \implies m = 3

Since the variance of a Poisson distribution is equal to its parameter

\displaystyle m

, the variance is

\displaystyle 3

. Hence, **Option B** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Variance

A measure of dispersion equal to the average of the squared deviations of each data point from the mean. Formula: σ² = Σ(xᵢ − x̄)² / n. Standard Deviation = √Variance.

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If a Poisson distribution is such that P(X=2)=P(X=3)\displaystyle P(X=2) = P(X=3)P(X=2)=P(X=3) then the variance of the distribution