Question

Identify the type of random variable—binomial, Poisson or hypergeometric—described by each of the following probability distributions:

a.$\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{5}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{5}}}{\mathbf{x}\mathbf{!}}\mathbf{;}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$

b.$\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left(\begin{array}{c}6\\ x\end{array}\right){\left(.2\right)}^{\mathbf{x}}{\left(.8\right)}^{\mathbf{6}\mathbf{-}\mathbf{x}}\mathbf{;}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathbf{6}$

c.$\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{10}\mathbf{!}}{\mathbf{x}\mathbf{!}\left(10-x\right)\mathbf{!}}{\left(.9\right)}^{\mathbf{x}}{\left(.1\right)}^{\mathbf{10}\mathbf{-}\mathbf{x}}\mathbf{:}\mathit{x}\mathit{=}\mathit{0}\mathit{,}\mathit{1}\mathit{,}\mathit{2}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}\mathit{10}$

Short answer

a. .The given variable is a poisson random variable.

b. .The given variable is a binomial random variable.

c. .The given variable is also a binomial random variable.

Step-by-step solution

Given information

X is a random variable.

Identifythe type of random variable when p(x)=.5xe-5x!;x=0,1,2,...

a.

$$\begin{gathered} p\left(x\right)=\frac{.5^x{e}^{-5}}{x!};x=0,1,2,... \\ =\frac{{\lambda }^x{e}^{-\lambda }}{x!} \end{gathered}$$

where$\lambda =0.5,x=0,1,2,...$

Since the probability distribution of a poisson random variable is

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

Hence, X is a poisson random variable.

Identify the type of random variable when p(x)=(6x)(.2)x(.8)6-x;x=0,1,2,...,6

b.

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\left(\begin{array}{c}6\\ \mathrm{x}\end{array}\right){(.2)}^{\mathrm{x}}{(.8)}^{6-\mathrm{x}};x=0,1,2,...,6 \\ =\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x};x=0,1,2,...n \end{gathered}$$

where$\mathrm{n}=6,\mathrm{p}=0.2,\mathrm{q}=0.8,\mathrm{x}=0,1,...,6$

Since the probability distribution of a binomial random variable is

$p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x};x=0,1,2,...n$

Hence, X is a binomial random variable.

Identify the type of random variable when p(x)=10!x!(10-x)!(.9)x(.1)10-x:x=0,1,2,...,10 

c.

$$\begin{gathered} p\left(\mathrm{x}\right)=\frac{10!}{\mathrm{x}!(10-\mathrm{x})!}{(.9)}^{\mathrm{x}}{(.1)}^{10-\mathrm{x}}:x\mathit{=}\mathit{0}\mathit{,}\mathit{1}\mathit{,}\mathit{2}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}\mathit{10} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{n}\mathit{!}}{\mathit{x}\mathit{!}\left(n-x\right)\mathit{!}}{p}^{\mathit{x}}{q}^{\mathit{n}\mathit{-}\mathit{x}}\mathit{;}x\mathit{=}\mathit{0}\mathit{,}\mathit{1}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}n \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\left(\begin{array}{c}n\\ x\end{array}\right)p^{\mathit{x}}{q}^{\mathit{n}\mathit{-}\mathit{x}} \end{gathered}$$

where$\mathrm{n}=10,\mathrm{p}=0.9,\mathrm{q}=0.1,\mathrm{x}=0,1,...,10$

Since the probability distribution of a binomial random variable is

$p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x};0,1,2,...,n$

Hence, X is a binomial random variable.