If X and Y are independent Normal Variables with mean 100 and 80 respectively and standard deviation as 4 and 3 respectively. What is the distribution of $\displaystyle (X+Y)$?

**Distribution of Sum of Independent Normal Variables** Let $\displaystyle X$ and $\displaystyle Y$ be independent normal random variables: $$X \sim N(\mu_X, \sigma_X^2) \quad$$$\displaystyle \text{and}$ \quad Y \sim N(\mu_Y, \sigma_Y^2) Here, we are given: - Mean of $\displaystyle X$, $\displaystyle \mu_X = 100$ - Standard deviation of $\displaystyle X$, $\displaystyle \sigma_X = 4 \implies$$\displaystyle \text{Variance }$ \sigma_X^2 = 4^2 = 16 - Mean of $\displaystyle Y$, $\displaystyle \mu_Y = 80$ - Standard deviation of $\displaystyle Y$, $\displaystyle \sigma_Y = 3 \implies$$\displaystyle \text{Variance }$ \sigma_Y^2 = 3^2 = 9 Since $\displaystyle X$ and $\displaystyle Y$ are independent normal variables, their sum $\displaystyle X+Y$ is also normally distributed: $$X+Y \sim N(\mu_{X+Y}, \sigma_{X+Y}^2)$$ **Step 1: Calculate the Mean of $\displaystyle (X+Y)$** $$\mu_{X+Y} = E[X+Y] = E[X] + E[Y]$$ $$\mu_{X+Y} = 100 + 80 = 180$$ **Step 2: Calculate the Variance of $\displaystyle (X+Y)$** Since $\displaystyle X$ and $\displaystyle Y$ are independent, their variances add directly: $$\sigma_{X+Y}^2 =$$$\displaystyle \text{Var}$(X+Y) = $\displaystyle \text{Var}$(X) + $\displaystyle \text{Var}$(Y) $$\sigma_{X+Y}^2 = 16 + 9 = 25$$ **Step 3: Calculate the Standard Deviation of $\displaystyle (X+Y)$** $$\sigma_{X+Y} =$$$\displaystyle \sqrt{\text{Var}(X+Y)} = \sqrt{25} = 5$ **Step 4: Identify the correct parameters** A normal distribution is characterized by its mean and variance $\displaystyle ($$\displaystyle \mu$, $\displaystyle \sigma^2$): $$\text{Parameters}=(180,25)$$$\displaystyle \text{Parameters}$ = (180, 25) This matches Option B. *Note:* Option A $\displaystyle (180, 7)$ represents the common error of adding standard deviations directly ($\displaystyle 4 + 3 = 7$), which is incorrect as standard deviations do not add for independent variables. Hence, **Option B** is the correct answer.