Theoretical DistributionsMCQMTP Nov 21Question 3584 of 230
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What is the mean of X having the following density function f(x)=142πe(x10)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 dNone of these

All Options:

  • A10
  • B4
  • C40
  • DNone of these

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Detailed Solution & Explanation

**Mean of Normal Density Function** The given probability density function is: f(x)=142πe(x10)232f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}} **First Principles:** The general formula for the PDF of a normal distribution with mean μ\displaystyle \mu and standard deviation σ\displaystyle \sigma is: f(x)=1σ2πe(xμ)22σ2f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} Comparing the term inside the exponent: (x10)232=(xμ)22σ2-\frac{(x-10)^2}{32} = -\frac{(x-\mu)^2}{2\sigma^2} From this, we directly identify: - Mean μ=10\displaystyle \mu = 10 - Exponent denominator: 2σ2=32    σ2=16    Standard Deviation σ=4\displaystyle 2\sigma^2 = 32 \implies \sigma^2 = 16 \implies \text{Standard Deviation } \sigma = 4 **Conclusion:** The mean of the random variable X\displaystyle X is μ=10\displaystyle \mu = 10. This matches Option A. *Note:* The textbook key lists Option D, which is incorrect since the mean is clearly 10 (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.

Key Concepts to Understand

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