Question

Assume that xis a random variable best described by a uniform distribution with $\mathit{c}\mathbf{=}\mathbf{10}$and$\mathit{d}\mathbf{=}\mathbf{90}$.

a. Find$\mathit{f}\left(x\right)$.

b. Find the mean and standard deviation of x.

c. Graph the probability distribution for xand locate its mean and theintervalon the graph.

d. Find$\mathit{P}\left(x\le 60\right)$.

e. Find$\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\ge }\mathbf{90}\mathbf{)}$.

f. Find$\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{80}\mathbf{)}$.

g. Find$\mathit{P}\mathbf{(}\mathit{\mu }\mathbf{-}\mathit{\sigma }\mathbf{\le }\mathit{x}\mathbf{\le }\mathit{\mu }\mathbf{+}\mathit{\sigma }\mathbf{)}$.

h. Find$\mathit{P}\mathbf{(}\mathit{x}\mathbf{>}\mathbf{75}\mathbf{)}$.

Short answer

a.$f\left(x\right)=\frac{1}{80};10\le x\le 90$

b. The mean is$\mathrm{\mu }=50$ and the standard deviation is $\sigma =23.094$.

c. The graph is a symmetrical curve.

d.$P(x\le 60)=0.625$

e.$P(x\ge 90)=0$

f.$P(x\le 80)=0.875$

g. $P(\mu -\sigma <x<\mu +\sigma )=0.57735$

h.$P(x>75)=0.57735$

Step-by-step solution

Given information

X is auniformrandom variable.

$$\begin{gathered} c=10 \\ d=90 \end{gathered}$$

Calculate f(x)

a.

The p.d.f of X is

$$\begin{gathered} f\left(x\right)=\frac{1}{d-c};c\le x\le d \\ =\frac{1}{90-10} \\ =\frac{1}{80};10\le x\le 90 \end{gathered}$$

Hence, $\mathit{f}\left(x\right)\mathit{=}\frac{\mathit{1}}{\mathit{80}}\mathit{;}\mathit{10}\mathit{\le }\mathit{x}\mathit{\le }\mathit{90}$.

Calculate mean and standard deviation

b.

Mean=$\mu$

$$\begin{gathered} \mu =\frac{c+d}{2} \\ =\frac{\left(10+90\right)}{2} \\ =\frac{100}{2} \\ =50 \end{gathered}$$

Standard deviation=$\sigma$

$$\begin{gathered} \sigma =\frac{d-c}{\sqrt{12}} \\ =\frac{90-10}{\sqrt{12}} \\ =\frac{80}{\sqrt{12}} \\ =23.094 \end{gathered}$$

Hence, the mean is$\mathit{\mu }\mathbf{=}\mathbf{50}$and standard deviation is $\mathit{\sigma }\mathit{=}\mathit{23}\mathit{.}\mathit{094}$.

Plot the graph

c.

$$\begin{gathered} \mathrm{\mu }\pm 2\mathrm{\sigma }=50\pm \left(2\times 23.094\right) \\ 50\pm 46.188 \\ =\left(3.812,96.188\right) \end{gathered}$$

Calculate P(x≤60)

d.

$$\begin{gathered} P\left(x\le 60\right)={\int }_{10}^{60}f\left(x\right)dx \\ ={\int }_{10}^{60}\frac{1}{80}dx \\ =\frac{1}{80}{\int }_{10}^{60}dx \\ =\frac{1}{80}\times 50 \\ =\frac{5}{8} \\ =0.625 \end{gathered}$$

Hence, $\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{60}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{625}$.

Calculate P(x≥90)

e.

$$\begin{gathered} P\left(x\ge 90\right)=1-P\left(x<90\right) \\ =1-{\int }_{10}^{90}f\left(x\right)dx \\ =1-{\int }_{10}^{90}\frac{1}{80}dx \\ =1-\left(\frac{1}{80}\times 80\right) \\ =1-1 \\ =0 \end{gathered}$$

Hence, $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\ge }\mathbf{90}\mathbf{)}\mathbf{=}\mathbf{0}$.

Calculate P(x≤80)

f.

$$\begin{gathered} P\left(x\le 80\right)=1-{\int }_{80}^{90}f\left(x\right)dx \\ =1-{\int }_{80}^{90}\frac{1}{80}dx \\ =1-\left(\frac{1}{80}\times 10\right) \\ =1-\frac{1}{8} \\ =0.875 \end{gathered}$$

Hence, $\mathit{P}\left(x\le 80\right)\mathit{=}\mathit{0}\mathit{.}\mathit{875}$.

Calculate P(μ-σ≤x≤μ+σ)

g.

$$\begin{gathered} P\left(\mu -\sigma <x<\mu +\sigma \right)={\int }_{\mu -\sigma }^{\mu +\sigma }f\left(x\right)dx \\ ={\int }_{26.906}^{73.094}\frac{1}{80}dx \\ =\frac{1}{80}\times 46.188 \\ =0.57735 \end{gathered}$$

Hence,role="math" localid="1660281305747" $\mathit{P}\mathbf{(}\mathbf{\mu }\mathbf{-}\mathbf{\sigma }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\mu }\mathbf{+}\mathbf{\sigma }\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{57735}$.

Calculate

h.

$$\begin{gathered} P\left(x>75\right)={\int }_{75}^{90}f\left(x\right)dx \\ ={\int }_{75}^{90}\frac{1}{80}dx \\ =\frac{1}{80}\times 15 \\ =\frac{3}{16} \\ =0.57735 \end{gathered}$$

Hence, $\mathit{P}\left(x>75\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{57735}$.