Theoretical DistributionsMCQMTP Nov 19Question 3572 of 230
All Questions

For the normal distribution density function f(x)=Ke(x6)28\displaystyle f(x) = K e^{-\frac{(x-6)^2}{8}}, the mean and variance are.

Options

A2:2
B3:4
C4:5
D6:4
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Correct Answer

Option d6:4

All Options:

  • A2:2
  • B3:4
  • C4:5
  • D6:4

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Detailed Solution & Explanation

**Mean and Variance of normal distribution** We are given the probability density function: f(x)=Ke(x6)28f(x) = K e^{-\frac{(x-6)^2}{8}} **Step 1: Compare with standard normal PDF** The standard form of a normal distribution's PDF is: f(x)=1σ2πe(xμ)22σ2f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} Comparing the terms in the exponent: (x6)28=(xμ)22σ2-\frac{(x-6)^2}{8} = -\frac{(x-\mu)^2}{2\sigma^2} From this, we identify: - Mean μ=6\displaystyle \mu = 6 - Exponent denominator: 2σ2=8    Variance σ2=4\displaystyle 2\sigma^2 = 8 \implies \text{Variance } \sigma^2 = 4 **Step 2: Conclusion** The mean of the distribution is 6\displaystyle 6 and the variance is 4\displaystyle 4. Written in format `mean:variance`, it is `6:4`, which corresponds to Option D. *Note:* The textbook key lists Option A, which is a printing/key error. Hence, **Option D** 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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