uncertainty with distribution I have a question which probably solves with Poisson distribution. Could you help me to solve that?
This is the question:

Police say that in a city, a home is burgled every day. If burglaries are randomly distributed in time, what would be the rate of burglaries have to be yield 80 percent that this statement is true for a given year?
Can we say that $$p=\lambda e^\lambda x$$ and consider the $p=0.8$ and $x=1$ and find the $\lambda$?

 A: If I understand the problem correctly, you have $X\sim\mathsf{Pois}(\lambda)$ and you seek $\lambda$ such that
$$P(X \ge 365) = 1 - P(X \le 364) = 0.8.$$
Normal approximation to Poisson. For an approximate answer (I guess what is expected) you can use
the fact that, for large $\lambda,$ one has $X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu=\lambda, \sigma=\sqrt{\lambda}).$
So, using a continuity correction, you have
$$P(X < 364.5) \approx 
P\left(Z = \frac{X-\lambda}{\sqrt{\lambda}} < \frac{364.5-\lambda}{\sqrt{\lambda}}=c\right) = 0.2,$$
where $Z$ is standard normal and $c$ is chosen so that $P(Z < c) = 0.2.$
You can get $c$ from printed tables of the standard normal CDF, and solve the last equality inside parentheses in the displayed equation for
$\lambda.$
Exact computation in R. Also, you can get an exact value of $\lambda=381$ (to the nearest integer)
by doing a 'grid search' in R. It stands to reason that $\lambda$ has to
be somewhat larger than $365.$ [In R, ppois is a Poisson CDF.]
lam = 360:1000           # integers 350 to 1000
pr = 1 - ppois(364, lam)
min(lam[pr >=.8])        # read [ ] as "such that"
[1] 381

1 - ppois(364,381)       # verification
[1] 0.8003664

Figure:
x = 300:475;  PDF = dpois(x, 381)
hdr = "POIS(381) with Normal Approx."
plot(x, PDF, type="h", col="blue", main=hdr)
 curve(dnorm(x, 381, sqrt(381)), add=T, lwd=2, col="orange")
 abline(h=0, col="green2")
 abline(v=364.5, lty = "dashed") 


