Theoretical DistributionsMCQMTP Dec 2023 Series IQuestion 3598 of 230
All Questions

The mean deviation about median of standard normal variate is

Options

A0.675σ\displaystyle 0.675\sigma
B0.675\displaystyle 0.675
C0.80σ\displaystyle 0.80\sigma
D0.80\displaystyle 0.80
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Correct Answer

✅ Option c — 0.80σ\displaystyle 0.80\sigma

All Options:

  • A0.675σ\displaystyle 0.675\sigma
  • B0.675\displaystyle 0.675
  • C0.80σ\displaystyle 0.80\sigma
  • D0.80\displaystyle 0.80

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Detailed Solution & Explanation

**Mean Deviation of Standard Normal Variate** For any normal distribution, the Mean Deviation (MD) about the mean (or median) is given by the relation: textMeanDeviation=sqrtfrac2pisigmaapprox0.79788sigmaapprox0.80sigma\\text{Mean Deviation} = \\sqrt{\\frac{2}{\\pi}} \\sigma \\approx 0.79788 \\sigma \\approx 0.80\\sigma For a standard normal variate, the standard deviation is sigma=1\displaystyle \\sigma = 1. Substituting this value: textMeanDeviationapprox0.80times1=0.80\\text{Mean Deviation} \\approx 0.80 \\times 1 = 0.80 *Note on discrepancy:* Option C represents the general formula (0.80sigma\displaystyle 0.80\\sigma) while Option D represents the specific numerical value for standard normal (0.80\displaystyle 0.80). The textbook key lists Option C (0.80sigma\displaystyle 0.80\\sigma) as correct, using the general algebraic formula representation. 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.

Key Concepts to Understand

Median

The middle value of a data set when arranged in ascending or descending order. It divides the distribution into two equal halves. For even n: Median = average of (n/2)th and (n/2 + 1)th values. It is NOT affected by extreme values.

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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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