Theoretical DistributionsMCQPYQ June 24Question 3430 of 230
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For a binomial distribution the mean and standard deviation are 10\displaystyle 10 and 3\displaystyle 3 respectively. Find the value of n\displaystyle n.

Options

A30\displaystyle 30
B9\displaystyle 9
C90\displaystyle 90
D100\displaystyle 100
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Correct Answer

✅ Option d — 100\displaystyle 100

All Options:

  • A30\displaystyle 30
  • B9\displaystyle 9
  • C90\displaystyle 90
  • D100\displaystyle 100

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Detailed Solution & Explanation

Given:Mean=np=10andSD=sqrtnpq=3.\n\nStep1:FindtheVariance.\nVariance=(SD)2=32=9\nSonpq=9.\n\nStep2:Findq.\nfractextVariancetextMean=fracnpqnp=q\n\nq=frac910=0.9\n\nStep3:Findp.\np=1−q=1−0.9=0.1\n\nStep4:Findn.\nnp=10\nn=frac10p=frac100.1=100\n\nStep5:Verification.\nnpq=100times0.1times0.9=9\nsqrtnpq=sqrt9=3✓\n\nTherefore,n=100.\n\nHence,∗∗OptionD∗∗isthecorrectanswer.\displaystyle Given: Mean = np = 10 and SD = \\sqrt{npq} = 3.\n\nStep 1: Find the Variance.\nVariance = (SD)^2 = 3^2 = 9\nSo npq = 9.\n\nStep 2: Find q.\n\\frac{\\text{Variance}}{\\text{Mean}} = \\frac{npq}{np} = q\n\nq = \\frac{9}{10} = 0.9\n\nStep 3: Find p.\np = 1 - q = 1 - 0.9 = 0.1\n\nStep 4: Find n.\nnp = 10\nn = \\frac{10}{p} = \\frac{10}{0.1} = 100\n\nStep 5: Verification.\nnpq = 100 \\times 0.1 \\times 0.9 = 9\n\\sqrt{npq} = \\sqrt{9} = 3 ✓\n\nTherefore, n = 100.\n\nHence, **Option D** 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

Standard Deviation

A measure of the dispersion or spread of values in a data set around the mean. It is the square root of Variance. A lower standard deviation indicates data points are clustered close to the mean; a higher value indicates greater spread.

More Questions from Theoretical Distributions

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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,

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