7 accidents per 30 days on average, but variance is 12. Why Poisson distribution can't be used here? Let's say a city has 7 accidents per 30 days on average.
We can use Poisson distribution formula where $\lambda = 7$
$$\frac 1 {n!} (\lambda t)^n e^{-\lambda t}$$
But in Poisson distribution the lambda equals variance. In this case variance is $12.$ So we can't use Poisson.
But why we can't use Poisson distribution, what does it mean variance is not equal to lambda?
 A: *

*You observe a mean of 7 per day with a variance of 12. 

*You imagine that the process is (should be) a Poisson process.


These two are contradicting since for a Poisson distribution of the data you would expect equal mean and variance.
So one of the two premisses must be false. 


*

*It could be that the measurements are not correct. Or possibly the data set is only small such that it is incorrect to interpret the values $\mu=7$ and $\sigma^2=12$ as good estimates for the parameters of the distributiuon.
I see you have a related question on the math.stackexchange where you mention that the data are accidents per month. I can imagine that with this low frequency of sampling you do not have a large data set. Say you only have data of a single year (12 points), and if the data is hypotetically distributed according to a Poisson distribution with $\lambda=7$, then the distribution for the means and variance of your sample would look like the image below:

So while the mean and variance of the Poisson distribution are be equal, the same is not true for the particular samples that are sampled from a Poisson distribution.

*It could be (and I believe it is likely) that you do not have a (single) Poisson process. This would namely be the situation that the probability of an accident is constant in time.
An example how you could have a different situation is when you have the situation that the weekends have a lower rate of accidents (and probably there will be more different types of variations, day/night, winter/summer, peak hours, holidays, etc). In this simple example you get a mixture of two Poisson distributions: $$f(k) = \frac{2}{7} \frac{(\lambda-5a)^k e^{-(\lambda-5a)}}{k!}   + \frac{5}{7} \frac{(\lambda+2a)^k e^{-(\lambda+2a)}}{k!}$$ When $\lambda = 7$ and $a=\sqrt{0.5}$, then you have a mean 7 and variance 12. 
Thus a situation with $\lambda_\text{weekdays} \approx 8.41$ and $\lambda_\text{weekend}  \approx 3.46$ could explain your observation. But, of course, this is only one of many examples how the probability for accidents is not constant in time and you should investigate this further. If you look for overdispersion you will be able to find hints how to deal with your problem.
