Answer:
The answer to this question can be defined as follows:
Explanation:
[tex]\ Given \ values:\\\\\ let \ x \ is \ a \ poisson \ (x). Then,\\\\\ The \ probability \ mass \ function \ of \ x \ is,\\\\p(x= x) = c^{-\lambda} \frac{\lambda^n}{x!}, \\\\x = 0, 1, 2\\ \ consider \ the \ ratios \\\\\frac{p(x=i)}{p(x=i-1)} = \frac{e^{-\lambda}\frac{\lambda!}{i!}}{e^{-\lambda} \frac{ \lambda^{i-1}}{(i-1)!}} = \frac{\lambda}{i} \\\\[/tex]
[tex]\If \frac{\lambda}{i} > 1 = \frac{p(x-i)}{p(x=i-1)} >1 \\\\p(x=i)>p(x=i-1)\\\\\ Therefore,\ the \ probability \ mass \ method \ x\ increased \ unit \ by \ \ \frac{\lambda}{i} > 1\\\\\lambda > i\\\\\ If \ \ \frac{\lambda}{i}<1 \ = \frac{p(x-i)}{p(x=i-1)} <1 \\\\p(x=i)<p(x=i-1)[/tex]
[tex]\ The \ probability \ mass \ method \ x \ decreased \ by \ \ \frac{\lambda}{i}<1\\\\\lambda < i \\\\ \ The \ highest \ value \ counters \ when \\\\ i=[\lambda]\\\\\ If \ \lambda \ is \ number \ then \ the \ highest \ value \ occured \ by \\\\ \ i= \lambda -1 \ and\ i=\lambda[/tex]