Let $X_{(i)} = ( i = 1,2, \ldots, n+1)$ be a random sample of size $n+1$ that is produced from a normal population. Let $M$ be the sample mean of the first $n$ random variables in this random sample.
Find the probability that $$P(X_{(n+1)} - M < 0.5)$$
Variance $= 1$ and Sample Size $n = 40$.
I have been stuck in this question for hours since I cannot find a formula to compute $X_{(n+1)}$. I tried computing the sample mean including $(n+1)$ and then figure out something between two sample means but I went nowhere.
Am I interpreting this question wrong, isn't $X_{(n+1)}$ just a data value or does that notation mean its the sample mean of $n+1$ values? I cannot see how this problem could be solved if $X_{(n+1)}$ is a value ....
The random variables $X_{n+1}$ and $M$ have the same mean. Let $Y=X_{+1}-M$. Then $Y$ has mean $0$.
The random variable $Y$ is a linear combination of independent normals, so it is normal.
Note that $X_{n+1}$ has variance $1$ and $M$ has variance $\frac{1}{40}$. Thus $Y$ has variance $1+(-1)^2\frac{1}{40}=\frac{41}{40}$.
Now that we know the mean and the variance of the normally distributed random variable $Y$, we can find $\Pr(Y\lt 0.5)$ in the usual way.