Answer:
A Poisson distribution, with λ=5.3
[tex]P(x) = \frac{\lambda^{x} *e^{-\lambda}}{x!}[/tex]
Step-by-step explanation:
For this type of random variables, a Poisson distribution is the most adecuate.
If the number of occurrences in ten minutes is 5.3, we can define the parameter λ=5.3 and the probability function as:
[tex]P(x) = \frac{\lambda^{x} *e^{-\lambda}}{x!} =\frac{5.3^{x} *e^{-5.3}}{x!} \approx\frac{0.005*5.3^{x}}{x!}[/tex]
