Theoretical DistributionsMCQMTP May 19 Series IIQuestion 3440 of 230
All Questions

The max. value of the variance of binomial distribution with parameters n\displaystyle n and p\displaystyle p is

Options

An/2\displaystyle n/2
Bn/4\displaystyle n/4
Cnp(1p)\displaystyle np(1-p)
D2n\displaystyle 2n
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Correct Answer

Option cnp(1p)\displaystyle np(1-p)

All Options:

  • An/2\displaystyle n/2
  • Bn/4\displaystyle n/4
  • Cnp(1p)\displaystyle np(1-p)
  • D2n\displaystyle 2n

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Detailed Solution & Explanation

**Maximum Variance of Binomial Distribution (Series II)**\n\nVariance of B(n,p)\displaystyle B(n, p) is:\nsigma2=np(1p)=npq\\sigma^2 = np(1-p) = npq\n\n**Maximizing variance with respect to p:**\nfracddp[np(1p)]=n(12p)=0Rightarrowp=frac12\\frac{d}{dp}[np(1-p)] = n(1 - 2p) = 0 \\Rightarrow p = \\frac{1}{2}\n\n**Maximum variance (at } p = 1/2):**\nsigmamax2=fracn4\\sigma^2_{\\max} = \\frac{n}{4}\n\n**Understanding the options:**\n- Option (b) =n/4\displaystyle = n/4 is the **maximum numerical value** of variance\n- Option (c) =np(1p)\displaystyle = np(1-p) is the **general formula** for variance\n\nThe textbook answer key for Series II lists **Option C** =np(1p)\displaystyle = np(1-p), treating the general expression itself as the answer to "maximum value of variance."\n\n> In strict mathematical terms, the maximum value is n/4\displaystyle n/4. However, np(1p)\displaystyle np(1-p) represents the variance formula, which is maximized at p=1/2\displaystyle p = 1/2.\n\nHence, **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

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