Answer:
The answer is "0.6227 and 0.5971".
Step-by-step explanation:
[tex]\to p(x\leq 2)=p(0)+p(1)+p(2)[/tex]
[tex]=e^{-2.2}(1+2.2^{1}+\frac{2.2^2}{2!})\\\\=0.6227[/tex]
[tex]\to x \sim emp(\frac{1}{2.2})\\\\\to f(x)=\frac{1}{2.2} emp1-\frac{x}{2.2}[/tex]
[tex]\to p(x \leq 2)=\int^{2}_{0} \frac{1}{2.2} emp1(-\frac{x}{2.2}\ dx[/tex]
[tex]=1-emp1-2|2.2\\\\=0.5971[/tex]
