A survey of downtown Chicago found that the typical city block had an average of five rats per block. This was true irrespective of whether or not there was construction occurring on that block. If this average is applicable across all city blocks, what is the probability that, in a sample of two city blocks taken together, there will be a total of five or fewer rats for those two blocks?

In my view, it is a Poisson distribution. Here the average is given, i.e., $\lambda=5$. (I have looked at the table even.) But I'm not sure how to solve for two city blocks when taken together.

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    $\begingroup$ Iconoclast, welcome to Cross Validated. Your problem seems like a homework problem, so you should tag it as "self-study" and then explain as much of your solution as you can. You are on the right track. You need to find distribution of the sum of two Poisson random variables. Anything you tell us will help us answer, since the answer could involve moment generating functions, or a double sum, depending on what you've been taught so far. To get you started, if you expect 5 rats on one block, how many would you expect on two, given counts per block are independent and identically distributed? $\endgroup$ – Peter Leopold Sep 15 at 5:31
  • $\begingroup$ @Iconoclast Per your title -- you're not adding the distributions, you're adding the random variables. When they're independent, the operation applied to the pmfs is convolution, rather than addition. $\endgroup$ – Glen_b -Reinstate Monica Sep 15 at 8:13

Let's suppose your model is correct and in each city block, if you take a survey by un unspecified method, your probability of finding $n$ rats is $P(X=n)$ where $X \sim Poisson(\lambda)$ and $\lambda = 5$.

Then, according to the hypotheses in you question the probability of observing rats in each block is i.i.d. (independent and identically distributed); for each block we have $X_i \sim Poisson(5)$.

If we visit two blocks, the number of rats we are going to see is the random variable $Y \sim X_1 + X_2$.

For a property of the Poisson distribution, check your book, $Y$ will be again a Poisson distribution $Y \sim Poisson(\lambda_1 + \lambda_2)$ which in our case is $Y \sim Poisson(10)$.

Finally, you want to find the probability of observing 5 or fewer rats, that is $P(Y \le 5)$. You can compute that in the way you prefer. Using R it is

> ppois(5, 10)
[1] 0.06708596
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    $\begingroup$ Nicely explained (+1). But be mindful that it's not always helpful over the long run to give complete and explicit answers to hwk problems, especially when OP has not shown any engagement toward an answer. (Also, I'd say not always unhelpful either.) $\endgroup$ – BruceET Sep 15 at 6:23
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    $\begingroup$ Hi, thank you, it is the first time I answer in this group, I don't know your habits. I understand your concern, it is true. There is the possiblity the student is asking just to evade thinking. Let's hope this student will check the formula on the book and who know, maybe he will try to do the proof;) $\endgroup$ – Nicola Mingotti Sep 15 at 7:17

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