Theoretical DistributionsMTP March 22/RTP Sep 24Question 3519 of 230
All Questions

For a Poisson variate X, P(X=2)=3P(X=4)\displaystyle P(X=2) = 3 P(X=4). Then the standard deviation of X is

Options

A2\displaystyle 2
B4\displaystyle 4
C2\displaystyle \sqrt{2}
D3\displaystyle 3
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — 2\displaystyle \sqrt{2}

All Options:

  • A2\displaystyle 2
  • B4\displaystyle 4
  • C2\displaystyle \sqrt{2}
  • D3\displaystyle 3

Detailed Solution & Explanation

Let X\displaystyle X follow a Poisson distribution with parameter m\displaystyle m. The probability mass function is given by: P(X=k)=e−mmkk!P(X = k) = \frac{e^{-m} m^k}{k!} We are given the relation: P(X=2)=3P(X=4)P(X = 2) = 3 P(X = 4) Substituting the PMF values into this relation: e−mm22!=3×e−mm44!\frac{e^{-m} m^2}{2!} = 3 \times \frac{e^{-m} m^4}{4!} e−mm22=3×e−mm424\frac{e^{-m} m^2}{2} = 3 \times \frac{e^{-m} m^4}{24} e−mm22=e−mm48\frac{e^{-m} m^2}{2} = \frac{e^{-m} m^4}{8} Since e−m≠0\displaystyle e^{-m} \neq 0 and m>0\displaystyle m > 0, we can divide both sides by e−mm22\displaystyle \frac{e^{-m} m^2}{2}: 1=m24  ⟹  m2=4  ⟹  m=21 = \frac{m^2}{4} \implies m^2 = 4 \implies m = 2 Thus, the mean parameter is m=2\displaystyle m = 2. For a Poisson distribution, the standard deviation is defined as m\displaystyle \sqrt{m}: SD=m=2\text{SD} = \sqrt{m} = \sqrt{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.

More Questions from Theoretical Distributions

Find the variance of binomial distribution with n=10\displaystyle n=10, p=0.3\displaystyle p=0.3

When p=0.5\displaystyle p=0.5, the binomial distribution is

If mean and standard deviation of a binomial distribution is 10\displaystyle 10 and 2\displaystyle 2 respectively, q\displaystyle q will be

A random variable X\displaystyle X follows Binomial Distribution With E(x)=2\displaystyle E(x)=2 and V(x)=1.2\displaystyle V(x)=1.2, then the value of n\displaystyle n is

When 'p\displaystyle p' is large than 0.5\displaystyle 0.5, the Binomial Distribution is:

For a Poisson distribution,

Ready to Master Theoretical Distributions?

Practice all 230 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free