Theoretical DistributionsMCQPYQ Nov. 20Question 3487 of 230
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If the parameter of Poisson distribution is m\displaystyle m and (Mean+S.D.)=625\displaystyle (Mean + S.D.) = \frac{6}{25} then find m\displaystyle m:

Options

A425\displaystyle \frac{4}{25}
B125\displaystyle \frac{1}{25}
C425\displaystyle \frac{4}{25}
D35\displaystyle \frac{3}{5}
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Correct Answer

✅ Option b — 125\displaystyle \frac{1}{25}

All Options:

  • A425\displaystyle \frac{4}{25}
  • B125\displaystyle \frac{1}{25}
  • C425\displaystyle \frac{4}{25}
  • D35\displaystyle \frac{3}{5}

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Detailed Solution & Explanation

For a Poisson distribution with parameter m\displaystyle m:\n- **Mean** =m\displaystyle = m\n- **Variance** =m\displaystyle = m, so **S.D.** =m\displaystyle = \sqrt{m}\n\n**Given condition:**\nm+m=625m + \sqrt{m} = \frac{6}{25}\n\n**Substitution:** Let m=t\displaystyle \sqrt{m} = t, so m=t2\displaystyle m = t^2\nt2+t=625t^2 + t = \frac{6}{25}\n25t2+25t=625t^2 + 25t = 6\n25t2+25t−6=025t^2 + 25t - 6 = 0\n\n**Applying the quadratic formula:**\nt=−25±625+60050=−25±122550=−25±3550t = \frac{-25 \pm \sqrt{625 + 600}}{50} = \frac{-25 \pm \sqrt{1225}}{50} = \frac{-25 \pm 35}{50}\n\nTaking the positive root (since t=m≥0\displaystyle t = \sqrt{m} \geq 0):\nt=−25+3550=1050=15t = \frac{-25 + 35}{50} = \frac{10}{50} = \frac{1}{5}\n\n**Finding m\displaystyle m:**\nm=t2=(15)2=125m = t^2 = \left(\frac{1}{5}\right)^2 = \boxed{\frac{1}{25}}\n\nHence, **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.

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