Theoretical DistributionsMCQMTP Dec 2022 Series IQuestion 3537 of 230
All Questions

What is the first quartile of x\displaystyle x having the following probability density function? f(x)=172πe−(x−10)272\displaystyle f(x) = \frac{1}{\sqrt{72\pi}} e^{-\frac{(x-10)^2}{72}} for −∞<x<∞\displaystyle -\infty < x < \infty

Options

A4\displaystyle 4
B5\displaystyle 5
C5.95\displaystyle 5.95
D6.75\displaystyle 6.75
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Correct Answer

✅ Option c — 5.95\displaystyle 5.95

All Options:

  • A4\displaystyle 4
  • B5\displaystyle 5
  • C5.95\displaystyle 5.95
  • D6.75\displaystyle 6.75

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Detailed Solution & Explanation

**First Quartile of a Normal Distribution** **Given Probability Density Function (PDF):** f(x)=frac1sqrt72pie−frac(x−10)272quadtextfor−infty<x<inftyf(x) = \\frac{1}{\\sqrt{72\\pi}} e^{-\\frac{(x-10)^2}{72}} \\quad \\text{for } -\\infty < x < \\infty **Step 1: Identify Parameters of the Normal Distribution** The standard probability density function of a normal distribution N(mu,sigma2)\displaystyle N(\\mu, \\sigma^2) is: f(x)=frac1sigmasqrt2pie−frac(x−mu)22sigma2f(x) = \\frac{1}{\\sigma \\sqrt{2\\pi}} e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}} By comparing the given PDF to the standard form: 1. The mean is mu=10\displaystyle \\mu = 10. 2. The denominator in the exponent is 2sigma2=72impliessigma2=36impliestextS.D.sigma=6\displaystyle 2\\sigma^2 = 72 \\implies \\sigma^2 = 36 \\implies \\text{S.D. } \\sigma = 6. 3. The normalization factor is: sigmasqrt2pi=6sqrt2pi=sqrt72pi\\sigma \\sqrt{2\\pi} = 6 \\sqrt{2\\pi} = \\sqrt{72\\pi} This matches the coefficient perfectly, confirming that mu=10\displaystyle \\mu = 10 and sigma=6\displaystyle \\sigma = 6. **Step 2: Calculate the First Quartile (Q1\displaystyle Q_1)** For a normal distribution, the first quartile Q1\displaystyle Q_1 is located at: Q1=mu−0.6745sigmaQ_1 = \\mu - 0.6745 \\sigma Substitute the values mu=10\displaystyle \\mu = 10 and sigma=6\displaystyle \\sigma = 6: Q1=10−0.6745times6Q_1 = 10 - 0.6745 \\times 6 Q1=10−4.047=5.953approx5.95Q_1 = 10 - 4.047 = 5.953 \\approx 5.95 **Discrepancy Note:** The mathematical derivation yields Q1approx5.95\displaystyle Q_1 \\approx 5.95, which corresponds to **Option C**. The textbook answer key has a typographical error, listing **Option A** (4\displaystyle 4) as the correct answer. We have presented the correct mathematical derivation leading to Option C. 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

Function

A relation from a set A (Domain) to a set B (Codomain) such that every element in A is associated with exactly one element in B. Written as f: A→B. Types include Linear, Quadratic, Polynomial, and Exponential functions.

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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For a Poisson distribution,

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