Theoretical DistributionsMTP March 22Question 3987 of 230
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What is the mean of X having the following density function? f(x)=142πe−(x−10)232\displaystyle f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}} for −∞<x<∞\displaystyle -\infty < x < \infty

Options

A10
B4
C40
DNone of these
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Correct Answer

✅ Option a — 10

All Options:

  • A10
  • B4
  • C40
  • DNone of these

Detailed Solution & Explanation

**Mean from Normal Probability Density Function** The standard probability density function (pdf) of a normal random variable X\displaystyle X is: f(x)=frac1sigmasqrt2pie−frac(x−mu)22sigma2f(x) = \\frac{1}{\\sigma\\sqrt{2\\pi}} e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}} Comparing the given function: f(x)=frac14sqrt2pie−frac(x−10)232f(x) = \\frac{1}{4\\sqrt{2\\pi}} e^{-\\frac{(x-10)^2}{32}} with the standard form: 1. **From the exponent term:** −frac(x−10)232=−frac(x−mu)22sigma2-\\frac{(x-10)^2}{32} = -\\frac{(x-\\mu)^2}{2\\sigma^2} This directly gives the mean: mu=10\\mu = 10 And the variance term: 2sigma2=32impliessigma2=16impliessigma=42\\sigma^2 = 32 \\implies \\sigma^2 = 16 \\implies \\sigma = 4 2. **From the constant coefficient:** frac1sigmasqrt2pi=frac14sqrt2piimpliessigma=4\\frac{1}{\\sigma\\sqrt{2\\pi}} = \\frac{1}{4\\sqrt{2\\pi}} \\implies \\sigma = 4 This is consistent with the standard deviation derived from the exponent. Thus, the mean (mu\displaystyle \\mu) of the distribution is 10\displaystyle 10, which corresponds to Option A. Hence, **Option A** 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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