If, for a Poisson distributed random variable X , the probability for X taking value 2 is 3 times the probability for X taking value 4, then the variance of X is

Option c: 2. Solution: For a Poisson distributed random variable X with parameter m , the probability mass function is: P(X = k) = -m mkk! We are given: P(X = 2) = 3 P(X = 4) Substituting the formulas: -m m22! = 3 × -m m44! -m m22 = 3 × -m m424 Since e-m ≠ 0 and m > 0 for a Poisson variate: m2/2 = m4/8 4 m2 = m4 m2(m2 - 4) = 0 Since m > 0 , we have: m2 = 4 m = 2 For a Poisson distribution, the variance is equal to the parameter m . Therefore, the variance of X is 2 . Hence, Option C is the correct answer.

If, for a Poisson distributed random variable $X$, the probability for $X$ taking value 2 is 3 times the probability for $X$ taking value 4, then the variance of $X$ is

The correct answer is Option c: 2. For a Poisson distributed random variable $X$ with parameter $m$, the probability mass function is: $$P(X = k) = \frac{e^{-m} m^k}{k!}$$ We are given: $$P(X = 2) = 3 P(X = 4)$$ Substituting the formulas: $$\frac{e^{-m} m^2}{2!} = 3 \times \frac{e^{-m} m^4}{4!}$$ $$\frac{e^{-m} m^2}{2} = 3 \times \frac{e^{-m} m^4}{24}$$ Since $e^{-m} \neq 0$ and $m > 0$ for a Poisson variate: $$\frac{m^2}{2} = \frac{m^4}{8}$$ $$4 m^2 = m^4 \implies m^2(m^2 - 4) = 0$$ Since $m > 0$, we have: $$m^2 = 4 \implies m = 2$$ For a Poisson distribution, the variance is equal to the parameter $m$. Therefore, the variance of $X$ is $2$. Hence, **Option C** is the correct answer.

Theoretical DistributionsMCQPYQ Dec 21Question 3495 of 230

All Questions

If, for a Poisson distributed random variable $\displaystyle X$ , the probability for $\displaystyle X$ taking value 2 is 3 times the probability for $\displaystyle X$ taking value 4, then the variance of $\displaystyle X$ is

Options

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Correct Answer

✅ Option c — 2

All Options:

Detailed Solution & Explanation

For a Poisson distributed random variable

\displaystyle X

with parameter

\displaystyle m

, the probability mass function is:

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

We are given:

P(X = 2) = 3 P(X = 4)

Substituting the formulas:

\frac{e^{-m} m^2}{2!} = 3 \times \frac{e^{-m} m^4}{4!}

\frac{e^{-m} m^2}{2} = 3 \times \frac{e^{-m} m^4}{24}

Since

\displaystyle e^{-m} \neq 0

and

\displaystyle m > 0

for a Poisson variate:

\frac{m^2}{2} = \frac{m^4}{8}

4 m^2 = m^4 \implies m^2(m^2 - 4) = 0

Since

\displaystyle m > 0

, we have:

m^2 = 4 \implies m = 2

For a Poisson distribution, the variance is equal to the parameter

\displaystyle m

. Therefore, the variance of

\displaystyle X

\displaystyle 2

. Hence, **Option C** 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

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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, for a Poisson distributed random variable X\displaystyle XX, the probability for X\displaystyle XX taking value 2 is 3 times the probability for X\displaystyle XX taking value 4, then the variance of X\displaystyle XX is