Question

Identify the type of continuous random variable—uniform,normal, or exponential—described by each of the following probability density functions:

a.$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{{\displaystyle \mathbf{-}\frac{\mathbf{x}}{\mathbf{7}}}}}{\mathbf{7}}\mathbf{;}\mathit{x}\mathbf{>}\mathit{o}$

b.$\mathit{f}\left(x\right)\mathit{=}\frac{\mathit{1}}{\mathit{20}}\mathit{;}\mathit{5}\mathit{<}\mathit{x}\mathit{<}\mathit{25}$

c.$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{=}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{.}\mathbf{5}{\left[\left(x-10\right)/5\right]}^{\mathbf{2}}}}{\mathbf{5}\sqrt{\mathbf{2}\mathbf{\pi }}}$

Short answer

1. X is an exponential random variable.
2. X is a uniform random variable.
3. X is a normal random variable.

Step-by-step solution

Given information

X is arandom variable.

Identifying the random variable when f(x)=e-x77;x>o

a.

$$\begin{gathered} f\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{{\displaystyle -\frac{\mathrm{x}}{7}}}}{7};x>o \\ =\frac{1}{\theta }{e}^{-\frac{x}{\theta }} \end{gathered}$$

Where,$\theta =7$ and$x>0$

The p.d.f of an exponential distribution is$f\left(\mathrm{x}\right)==\frac{1}{\theta }{e}^{-\frac{x}{\theta }};x>o$

Hence, X is an exponential random variable.

Identifying the random variable when f(x)=120;5<x<25

b.

$$\begin{gathered} f\mathit{\left(}x\mathit{\right)}=\frac{\mathit{1}}{\mathit{20}}\mathit{;}\mathit{5}\mathit{<}x\mathit{<}\mathit{25} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=\frac{1}{25-5} \\ =\frac{1}{\mathrm{d}-\mathrm{c}} \end{gathered}$$

Where, d= 25 and c = 5

The p.d.f of uniform distribution is$f\mathit{\left(}x\mathit{\right)}=\frac{1}{\mathrm{d}-\mathrm{c}}\mathit{;}c\mathit{\le }x\mathit{\le }d$

Hence, X is a uniform random variable.

Identifying the random variable when f(x)=e-.5[(x-10)/5]252π

c.

$$\begin{gathered} f\left(\mathrm{x}\right)\mathit{=}\frac{{\mathrm{e}}^{-.5{[(\mathrm{x}-10)/5]}^2}}{5\sqrt{2\mathrm{\pi }}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{1}}{\mathit{\sigma }\sqrt{\mathit{2}\mathit{\pi }}}{e}^{\mathit{-}\frac{\mathit{1}}{\mathit{2}}{\left[\left(x-\mu \right)/\sigma \right]}^{\mathit{2}}} \end{gathered}$$

Where ,$\mu =10$ and$\sigma =5$

The p.d.f of uniform distribution is$f\left(\mathrm{x}\right)\mathit{=}\frac{\mathit{1}}{\mathit{\sigma }\sqrt{\mathit{2}\mathit{\pi }}}{e}^{\mathit{-}\frac{\mathit{1}}{\mathit{2}}{\mathit{[}\mathit{(}\mathit{x}\mathit{-}\mathit{\mu }\mathit{)}\mathit{/}\mathit{\sigma }\mathit{]}}^{\mathit{2}}}$

Hence, X is a normal random variable.