Theoretical DistributionsMCQPYQ Dec 23Question 3555 of 230
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If 'x' and 'y' are independent normal variate with mean and SD μ1,μ2\displaystyle \mu_1, \mu_2 and σ1,σ2\displaystyle \sigma_1, \sigma_2 respectively, then for z=x+y\displaystyle z = x + y which also follows normal distribution mean and SD are:

Options

AMean =μ1+μ2\displaystyle = \mu_1 + \mu_2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
BMean =\displaystyle = (μ1+μ2)/2\displaystyle (\mu_1 + \mu_2) / 2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
CMean =μ1−μ2\displaystyle = \mu_1 - \mu_2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
DMean =(μ1−μ2)/2\displaystyle = (\mu_1 - \mu_2) / 2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
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Correct Answer

✅ Option a — Mean =μ1+μ2\displaystyle = \mu_1 + \mu_2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}

All Options:

  • AMean =μ1+μ2\displaystyle = \mu_1 + \mu_2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
  • BMean =\displaystyle = (μ1+μ2)/2\displaystyle (\mu_1 + \mu_2) / 2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
  • CMean =μ1−μ2\displaystyle = \mu_1 - \mu_2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}
  • DMean =(μ1−μ2)/2\displaystyle = (\mu_1 - \mu_2) / 2, SD =σ12+σ22\displaystyle = \sqrt{\sigma_1^2 + \sigma_2^2}

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Detailed Solution & Explanation

**Sum of Independent Normal Random Variables** Let X\displaystyle X and Y\displaystyle Y be independent normal random variables: - X∼N(μ1,σ12)\displaystyle X \sim N(\mu_1, \sigma_1^2) - Y∼N(μ2,σ22)\displaystyle Y \sim N(\mu_2, \sigma_2^2) Let Z=X+Y\displaystyle Z = X + Y. By the reproductive property of normal distribution, the sum of independent normal random variables is also normally distributed: Z∼N(μZ,σZ2)Z \sim N(\mu_Z, \sigma_Z^2) **Step 1: Find the Mean of Z\displaystyle Z** Using the linearity of expectation: μZ=E[Z]=E[X+Y]=E[X]+E[Y]=μ1+μ2\mu_Z = E[Z] = E[X + Y] = E[X] + E[Y] = \mu_1 + \mu_2 **Step 2: Find the Variance of Z\displaystyle Z** Since X\displaystyle X and Y\displaystyle Y are independent, the covariance between them is zero. Therefore: σZ2=Var(Z)=Var(X+Y)=Var(X)+Var(Y)=σ12+σ22\sigma_Z^2 = \text{Var}(Z) = \text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) = \sigma_1^2 + \sigma_2^2 **Step 3: Find the Standard Deviation of Z\displaystyle Z** The standard deviation (SD) is the square root of the variance: SD=σZ=σ12+σ22\text{SD} = \sigma_Z = \sqrt{\sigma_1^2 + \sigma_2^2} Thus, the mean and standard deviation of Z=X+Y\displaystyle Z = X + Y are μ1+μ2\displaystyle \mu_1 + \mu_2 and σ12+σ22\displaystyle \sqrt{\sigma_1^2 + \sigma_2^2} respectively. 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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