Find the probabilty of 25 random people X is the weight of one person, $X \sim N(\mu =78,\sigma =13.15 )$.
If I choose randomly 25 people, what is the probability that the average of their weights will be $86$ ?
I define $\displaystyle Z = \frac{X-\mu}{\sigma} = \frac{X-78}{13.15}$ and now $Z\sim N(0,1)$.
$\displaystyle \mathbb{P}(X>86) = 1- \mathbb{P}(X<86) = 1 - \mathbb{P}(Z< \frac{86-78}{13.15}) = 1 - \mathbb{P}(Z<0.6) = 1- \Theta(0.6) = 0.28$.
But, how can I calculate the average weights of 25 random people ?
Thanks.
 A: Let the weights of our people be $X_1,X_2,\dots, X_n$, where $n=25$. Then the average weight $\bar{X}$ is given by
$$\bar{X}=\frac{1}{n}(X_1+\cdots +X_n).$$
Under our assumptions, $\bar{X}$ has normal distribution, mean $\mu$, and standard deviation $\frac{\sigma}{\sqrt{n}}$. In our case the mean of $\bar{X}$ is $78$, and the standard deviation is $\frac{13.15}{5}$.
Now you can compute any probability you wish, in the usual way. Note that the probability that $\bar{X}$ is exactly $86$ is in principle $0$.  But if what you are interested in is $\Pr(\bar{X}\gt 86)$, that is done in the style you used in your calculation, the only thing that changes is that instead of standard deviation $13.15$ we use standard deviation $\frac{13.15}{5}$.
A: The probability will be zero, since $\mathbb{P}(Y=y)=0$ for any $y \in \mathbb{R}$ when $Y$ is a continuous random variable, and the normalized sum of independent normal random variables (i.e. the average of independent Gaussians), is a continuous random variable.
If we define $Y= \displaystyle\sum_{i=1}^{25} \frac{X_i}{25}$ to be the average of the weights of the 25 people, we can however find a non-zero value for $$\mathbb{P}(Y \le 86)$$ which technically is different than what the question asks for $\mathbb{P}(Y=86)=0$.
Anyway, when we say that we choose 25 people "randomly", I assume that to mean that the weight of all of the people are assumed to be independent. This has the benefit that it allows us to have a simple form for the sum of normal random variables (i.e. because they are independent).
Specifically, the sum of the $X_i$ should be distributed as $$\displaystyle\sum_{i=1}^{25} X_i \sim \mathcal{N}(25 \cdot 78, 5 \cdot 13.15)$$
(see for example: here)
Now, normalizing by $25$, (i.e. dividing by 25, since $Y=\frac{1}{25} \cdot \sum_{i=1}^{25} X_i$) we get that $$Y \sim \mathcal{N}(78, \frac{13.15}{5})$$ Since the mean of $nY=\sum_{i=1}^{25} X_i$ is $n \mu$, but the mean of $Y=\frac{n\mu}{n}=\mu$, and the standard deviation of $nY$ is $\sqrt{n}\sigma$ since the variance is $n\sigma^2$ (because it is the sum of 25 copies of $X$, which has variance $\sigma^2$), and then the standard deviation of $Y=\frac{1}{n^2}\cdot nY$ is $\sqrt{\frac{1}{n^2}\cdot n \sigma^2}= \sqrt{\sigma^2}{n}=\frac{\sigma}{\sqrt{n}}$ (See for example here).
Therefore $\mathbb{P}(Y \le 86)=\mathbb{P}(Z < \frac{5(86-78)}{13.5})$.
