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Suppose the number of, lets say flowers, are distributed randomly in a field of $1120$ sq. meters. If there are $96$ flowers in this field, then the mean is $96/1120$ for a Poisson distribution. Assuming this if we chose a random area of $20$ sq. meters and call the amount of flowers $X$, what are the possible values of $X$ and what is the value of the parameter of the distribution of $X$ in $20$ sq. meters?

idriskameni
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1 Answers1

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The Poisson distribution is denoted by $$P(X=x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$$ Where $X$ is the random variable, in your case number of flowers in a given area (by selecting your area you implicitly define the $\lambda$) and $x\in\mathbb{N}_+\cup\{0\}$ is a given realization of your random variable.

Notice that the range of this random variable is $\{0,1,\cdots\}$.

For instance it could be useful to normalise everything to areas of $1$ sq. meter. Hence if your parameter $\lambda$ (that as you said represents the mean)is $\frac{96}{1120}$ it means that in average there are $0.085$ flowers per sq.meter.

If you consider an area of $20$ sq.meters you have that your new $\lambda$ is nothing more than $20\times\frac{96}{1120}$. (Because if in average the number of flowers per sq.meter is $0.085$ then in 20 sq. meters this average is $20\times 0.085$).

Notice furthermore that the range of this new random variable is again $\{0,1,\cdots\}$ meaning that you could have any number (non-negative integers) of flowers in this field.

Chaos
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  • Thanks. So if we were to calculate the probability of there being only one flower, for instance, the mean in the formula would be 20 x 0.085 and the value of x would be 1 – Olivetti Apr 03 '19 at 03:54