Question

Given that xis a poisson random variable, compute$\mathit{p}\left(x\right)$for each of the following cases:

a.$\mathbf{}\mathit{\lambda }\mathbf{=}\mathbf{2}\mathbf{,}\mathit{x}\mathbf{=}\mathbf{3}$

b.$\mathbf{\lambda }\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{4}$

c.$\mathbf{}\mathit{\lambda }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathit{x}\mathbf{=}\mathbf{2}$

Short answer

a. The probability distribution$p\left(x\right)$ is $\frac{4}{3}{e}^{-2}$.

b. The probability distribution$p\left(x\right)$ is $\frac{1}{24}{e}^{-1}$.

c. The probability distribution$p\left(x\right)$ is $0.125\times e^{-0.5}$.

Step-by-step solution

Given information

X is a Poisson random variable.

Compute the probability distribution p(x) when λ=2,x=3

a.

For a Poisson random variable, the probability distribution

$p\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!}:x=0,1,2,...$

Here,$\lambda =2,x=3$

$$\begin{gathered} p\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!} \\ =\frac{2^3{e}^{-2}}{3!} \\ =\frac{8\times e^{-2}}{6} \\ =\frac{4}{3}{e}^{-2} \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$ is $\frac{4}{3}{e}^{-2}$.

Compute the probability distribution p(x) when λ=1,x=4

b.

Here,$\lambda \mathit{=}\mathit{1}\mathit{,}x\mathit{=}\mathit{4}$

$$\begin{gathered} p\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!} \\ =\frac{1^4{e}^{-1}}{4!} \\ =\frac{1}{24}{e}^{-1} \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$is $\frac{1}{24}{e}^{-1}$.

Compute the probability distribution p(x) when λ=0.5,x=2

c.

Here,$\lambda =0.5,x=2$

$$\begin{gathered} p\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!} \\ =\frac{{\left(0.5\right)}^2{e}^{-0.5}}{2!} \\ =0.125\times e^{-0.5} \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$ is $0.125\times e^{-0.5}$.