Calculation of probability for random variable

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 $0.975$.

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

$$\mathbb{P}[X\leq 1300]=\mathbb{P}\left[Z\leq \frac{1300-1250}{\sqrt{625}}\right]=\Phi(2)=97.725\%$$