Calculation of probability for random variable The number of the daily car accidents in Luna-land is a random variable with mean 50 and standard deviation 5. Find the probability for the total accidents in the following 25 days to be $\leq 1300$.
I am only familiar with combinatorics, not distributions.
I did a bit of reading around and found that probably this must be approached by the normal distribution.
So we want $P (X\leq 1300)$?
(this is not homework!)
Thank you!
EDIT: I found the formula for the central limit theorem.
So, it seems I must calculate $\frac {(1300-25*50)}{\frac{5}{\sqrt25}}$?
 A: Assuming that you can use CLT (25 is a little borderline) or Assuming normality in the daily car accidents, the correct way to calculate this probability is
$$\mathbb{P}[X\leq 1300]=\mathbb{P}\left[Z\leq \frac{1300-1250}{\sqrt{625}}\right]=\Phi(2)=97.725\%$$
A: Suppose $X_i$ denotes the number of accidents on day $i$ and suppose you have $\mathbb{E}[X_i]=\mu$ and $\text{Var}(X_i)=\sigma^2<\infty$. What you are interested in is $S_{25}=\sum_{i=1}^{25}X_i$
By the CLT you have that $\frac{S_{25}-25\mu}{\sqrt{25}\sigma}$ has approximately $\mathcal{N}(0,1)$ distribution (of course, if you are willing to assume that $25$ is a reasonably large enough number for your purposes) so you have that
$$\mathbb{P}(S_{25}\leq1300)=\mathbb{P}\Big(\frac{S_{25}-25\mu}{\sqrt{25}\sigma} \leq \frac{1300-25\mu}{\sqrt{25}\sigma}\Big)=\mathbb{P}\Big(Z \leq \frac{1300-25\mu}{\sqrt{25}\sigma}\Big)$$
for $Z\sim\mathcal{N}(0,1)$. What's left is to plug in the values given and use software to get the exact probability.
If you do that, you end up with the expression $\mathbb{P}(Z \leq 2)$ which should be approximately $0.975$.
