Step-by-step explanation:
6 customers per hour [tex]=1[tex] customer per 10 minutes
[tex]X \sim \operatorname{Exp}(\beta=10) \quad[/tex] (given mean)
Pdf is given by: [tex]f(x)=\frac{1}{\beta} e^{-x / \beta}, 0<x[/tex]
Pdf: [tex]f(x)=\frac{1}{10} e^{-x / 10}, \quad 0<x,[tex] and zero otherwise.
Cdf: [tex]P(X \leq x)=1-e^{-x / \beta}[/tex]
[tex]P(X \leq x)=1-e^{(-x / 10)} \quad[/tex] for [tex]x>0 \Longrightarrow P(X>x)=e^{-x / 10}[/tex]
a)[tex] \begin{array}{c}
P(X<12)=1-\exp (-12 / \beta)=1-e^{-1.2}=1-0.301194=0.698806 \approx 0.699 \\
P(X<12)=0.699 \end{array}[/tex]
[tex]\begin{aligned}
&\begin{array}{c}
P(X>20)=\exp (-20 / \beta)=e^{-2} \approx 0.135335 \approx 0.1353 \\
P(X>20)=0.1353
\end{array}\\[/tex]
&\text { c) }\\
&P(8<X<15)\\
&\begin{array}{l}
=P(X<15)-P(X<8) \\
=(1-\exp (-15 / \beta))-(1-\exp (-8 / \beta)) \\
=\exp (-8 / \beta)-\exp (-15 / \beta) \\
=e^{-0.8}-e^{-1.5} \\
=0.449329-0.22313 \approx 0.226199 \approx 0.2262 \\
\quad P(8<X<15)=0.2262
\end{array}
\end{aligned}
\begin{aligned}
&\text { d) }\\
&\begin{array}{l}
P(14<X<22) \\
=P(X<22)-P(X<14) \\
=(1-\exp (-22 / \beta))-(1-\exp (-14 / \beta)) \\
=\exp (-14 / \beta)-\exp (-22 / \beta) \\
=e^{-1.4}-e^{-2.2} \\
=0.246597-0.110803 \approx 0.135794 \approx 0.1358 \\
P(X<14 \cup X>22)=1-P(14<X<22)=1-0.1358=0.8642 \\
P(X<14 \cup X>22)=0.8642
\end{array}
\end{aligned}
