What is the mean of X having the following density function? f(x) = 1/4sqrt(2pi) e-(x-10)2/32 for -∞ < x < ∞

Option a: 10. Solution: Mean of Normal Density Function We are given the PDF: f(x) = 1/4sqrt(2pi) e-(x-10)2/32 First Principles: The PDF of a normal distribution with mean mu and standard deviation sigma is: f(x) = 1/sigmasqrt(2pi) e-(x-mu)2/2sigma2 Comparing the terms: 1. Inside the exponent, the numerator is (x-10)2 . Comparing to (x-mu)2 , we get: mu = 10 2. The exponent denominator is 2sigma2 = 32 sigma2 = 16 sigma = 4 . 3. The leading coefficient is 1/4sqrt(2pi) , which matches 1/sigmasqrt(2pi) when sigma = 4 . Thus, the mean of the distribution is mu = 10 . This matches Option A. Hence, Option A is the correct answer.

What is the mean of X having the following density function? $f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}}$ for $-\infty < x < \infty$

The correct answer is Option a: 10. **Mean of Normal Density Function** We are given the PDF: $$f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}}$$ **First Principles:** The PDF of a normal distribution with mean $\mu$ and standard deviation $\sigma$ is: $$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ Comparing the terms: 1. Inside the exponent, the numerator is $(x-10)^2$. Comparing to $(x-\mu)^2$, we get: $$\mu = 10$$ 2. The exponent denominator is $2\sigma^2 = 32 \implies \sigma^2 = 16 \implies \sigma = 4$. 3. The leading coefficient is $\frac{1}{4\sqrt{2\pi}}$, which matches $\frac{1}{\sigma\sqrt{2\pi}}$ when $\sigma = 4$. Thus, the mean of the distribution is $\mu = 10$. This matches Option A. Hence, **Option A** is the correct answer.

Theoretical DistributionsMCQMTP March 22Question 3589 of 230

All Questions

What is the mean of X having the following density function? $\displaystyle f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}}$ for $\displaystyle -\infty < x < \infty$

Options

A10

C40

DNone of these

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 10

All Options:

A10
B4
C40
DNone of these

Detailed Solution & Explanation

**Mean of Normal Density Function** We are given the PDF:

f(x) = \frac{1}{4\sqrt{2\pi}} e^{-\frac{(x-10)^2}{32}}

**First Principles:** The PDF of a normal distribution with mean

\displaystyle \mu

and standard deviation

\displaystyle \sigma

is:

f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}

Comparing the terms: 1. Inside the exponent, the numerator is

\displaystyle (x-10)^2

. Comparing to

\displaystyle (x-\mu)^2

, we get:

\mu = 10

2. The exponent denominator is

\displaystyle 2\sigma^2 = 32 \implies \sigma^2 = 16 \implies \sigma = 4

. 3. The leading coefficient is

\displaystyle \frac{1}{4\sqrt{2\pi}}

, which matches

\displaystyle \frac{1}{\sigma\sqrt{2\pi}}

when

\displaystyle \sigma = 4

. Thus, the mean of the distribution is

\displaystyle \mu = 10

. This matches 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

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.

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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}}f(x)=42π​1​e−32(x−10)2​ for −∞<x<∞\displaystyle -\infty < x < \infty−∞<x<∞