It can be shown that if $x_1, x_2, \dots, x_n$ are $N(\mu, \sigma^2)$, then $x̃$ is $N(\mu, \frac{\sigma^2}{n})$
Here we define $x̃=\frac{x_1+x_2+\dots+x_n}{n}$ (the average).
The question is:
If $x_1, x_2, \dots, x_{25}$ are $N(65, \sigma^2 = 5^2)$, then find $P(x̃ > 72)$.
It is implied that all $x_i$ are iid (independently identically distributed).
How do I calculate this probability?
Would it be as follows:
$x̃$ is $N(65, 5^2/25)=N(65,1)$
Thus, $P(x̃ > 72) = 1 - P(x̃ \leq 72) = 1 - \phi(\frac{72-65}{1})=1-\phi(7)\approx 0$
where $\phi$ is how I calculate the probability using the CDF.
I don't think this is right though. If the mean is about $65$, then I do atleast except some $70+$ grades.
The random variable $X_i$ is distributed as $X_i\sim \mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) \gg 0$
What is the variance of $\frac1n\sum_{i=1}^n X_i$, where $X_i$ are iid?
$Var\left( \frac1n\sum_{i=1}^n X_i\right)=\frac1{n^2}\cdot Var\left(\sum_{i=1}^n X_i\right)=\frac{1}{n^2}\cdot \left( Var(X_1)+Var(X_2)+\ldots + Var(X_n) \right)$
$=\frac{1}{n^2}\cdot \left(n\cdot Var(X_1) \right)=\frac{\sigma^2}{n}=\frac{25}{25}=1$. Therefore $\frac1{25}\sum_{i=1}^{25} X_i=\overline X_{25}\sim \mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(\overline X_{25}> 72)\approx 0$.
The pdf of $\overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.