As i understand it the exercise asks "what is the distribution of a r.v.$Z$ subtracted by a value for a mean". To me it makes sense that $Z$ would have a mean $µ_{n+1}-\bar{X_n}$, and as it happens that mean would be $0$. The variance of said r.v. should be the same as for all the others. However, my teacher's solution is:
Thus, he seems to think that $\bar{X_n}$ is a r.v. with a variance of $9/10*15^2$ because he subtracts the variances. But in that case:
Would the variance not be $\frac{11}{10}*15^2$ as you cannot subtract variances but instead add them?
Let $X_n$ be the sum of the first $10$ candidates: $$X_n=\sum_{i=1}^{10}X_i$$ This is the sum of $10$ independent Normal Random variables. So by standard principles of the sum of independent Normal Random variables, we have the distribution of $X_n$ is Normal with mean $\sum \mu=10 \cdot \mu$ and variance $\sum15^2=10 \cdot15^2$.
By the scaling property of the Normal distribution, we have: $$\bar X_n=\frac 1nX_n$$
Then $\bar X_n$ has a Normal distribution with mean $\frac 1{10}\cdot 10\cdot \mu=\mu$ and variance $\left( \frac 1{10} \right)^2\cdot10\cdot 15^2=\frac {15^2}{10}$
And by definition the distribution of $X_{n+1}$ is Normal with mean $\mu$ and variance $15^2$ as in the question.
So then again from the properties of the sum of $2$ Normal Random variables, we have the distribution of $X_{n+1}- \bar X_n$ is Normal with mean $\mu-\mu=0$ and variance $15^2+\frac {15^2}{10}=15^2(1+\frac 1{10})$. The variance comes fromt he fact that $X_{n+1}$ and $\bar X_n$ are independent since the sum of the first $10$ $X_i$ are independent from the $X_{n+1}$$^{th}$ observation.
So $X_{n+1}- \bar X_n$~$\mathcal N(0,\frac {11}{10}15^2)$
I have used the standard properties of Normal Random variables here. So I believe you are correct.
Please let me know if I can clarify!