Theoretical DistributionsPYQ Dec 22 - Series IQuestion 3974 of 230
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For a Poisson variate X, P(X=1)=P(X=2)\displaystyle P(X=1) = P(X=2), what is the mean of X?

Options

A1\displaystyle 1
B3/2\displaystyle 3/2
C2\displaystyle 2
D5/2\displaystyle 5/2
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Correct Answer

✅ Option c — 2\displaystyle 2

All Options:

  • A1\displaystyle 1
  • B3/2\displaystyle 3/2
  • C2\displaystyle 2
  • D5/2\displaystyle 5/2

Detailed Solution & Explanation

**Mean of a Poisson Variate given P(X=1)=P(X=2)\displaystyle P(X=1) = P(X=2)** Let m\displaystyle m be the mean (parameter) of the Poisson distribution. The probability mass function of a Poisson distribution is: P(X=x)=frace−mmxx!P(X = x) = \\frac{e^{-m} m^x}{x!} **Step 1: Write the expressions for P(X=1)\displaystyle P(X=1) and P(X=2)\displaystyle P(X=2)** - For x=1\displaystyle x = 1: P(X=1)=frace−mm11!=e−mmP(X = 1) = \\frac{e^{-m} m^1}{1!} = e^{-m} m - For x=2\displaystyle x = 2: P(X=2)=frace−mm22!=frace−mm22P(X = 2) = \\frac{e^{-m} m^2}{2!} = \\frac{e^{-m} m^2}{2} **Step 2: Equate the two probabilities** According to the problem: P(X=1)=P(X=2)P(X = 1) = P(X = 2) e−mm=frace−mm22e^{-m} m = \\frac{e^{-m} m^2}{2} **Step 3: Solve for m\displaystyle m** Since e−m>0\displaystyle e^{-m} > 0 and the parameter m>0\displaystyle m > 0 for a non-degenerate Poisson distribution, we can divide both sides by e−mm\displaystyle e^{-m} m: 1=fracm21 = \\frac{m}{2} m=2m = 2 Thus, the mean of the Poisson variate X\displaystyle X is 2\displaystyle 2, which corresponds to Option C. 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.

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