The mean of a Poisson distribution coincides with its rate parameter, so [tex]\lambda=4[/tex]. We have a PMF of
[tex]f_X(x)=\dfrac{e^{-4}4^x}{x!}[/tex]
So,
a. [tex]\mathbb P(2\le X\le5)=\dfrac{e^{-4}4^2}{2!}+\cdots+\dfrac{e^{-4}4^5}{5!}\approx0.6936[/tex]
b. [tex]\mathbb P(X\ge3)=\dfrac{e^{-4}4^3}{3!}+\dfrac{e^{-4}4^4}{4!}+\cdots\approx0.7619[/tex]
c. [tex]\mathbb P(X\le3)=\dfrac{e^{-4}4^0}{0!}+\cdots+\dfrac{e^{-4}4^3}{3!}\approx0.4335[/tex]
