Let X₁, X₂ be two independent normal random variables with means µ₁, µ₂ and standard deviations σ₁, σ₂, respectively. Consider Y = X₁ – X₂; µ₁=µ₂ =1, σ₁=1,σ₂= 2. Then,

The correct option is: Y is normally distributed with mean 0 and variance 5.

To understand why this is the correct choice, let us break down the problem step-by-step using the properties of normal random variables.

Step 1: Distribution of Y
We are given that $X_1$ and $X_2$ are independent normal random variables. A fundamental property of normal distributions is that any linear combination of independent normally distributed random variables is also normally distributed. Since $Y=X_1-X_2$ is a linear combination of $X_1$ and $X_2$, we can conclude that $Y$ must be normally distributed.

Step 2: Finding the Mean of Y
Using the linearity of expectation, the expected value (mean) of $Y$ is given by:

$E\left[Y\right]=E[X_1-X_2]=E\left[X_1\right]-E\left[X_2\right]$

Substituting the given values ${\mu }_1=1$ and ${\mu }_2=1$:

$E\left[Y\right]=1-1=0$

Thus, the mean of $Y$ is 0.

Step 3: Finding the Variance of Y
Since $X_1$ and $X_2$ are independent random variables, the variance of their difference is the sum of their individual variances:

$\mathrm{Var}\left(Y\right)=\mathrm{Var}(X_1-X_2)=\mathrm{Var}\left(X_1\right)+\mathrm{Var}\left(X_2\right)$

Using the relation between variance and standard deviation, $\mathrm{Var}\left(X\right)={\sigma }^2$:

$\mathrm{Var}\left(X_1\right)={\sigma }_1^2=1^2=1$

$\mathrm{Var}\left(X_2\right)={\sigma }_2^2=2^2=4$

Now we calculate the variance of $Y$:

$\mathrm{Var}\left(Y\right)=1+4=5$

Thus, the variance of $Y$ is 5.

Combining these findings, we establish that $Y$ is normally distributed with a mean of 0 and a variance of 5.