Question

According to Maryland Motor Vehicle Administration [MVA] data, Gary Turgeon, a clerk at the Beltsville, Maryland, MVA location, assists six customers per hour, on average.

a. Determine the probability the amount of time Gary takes to assist the next customer is less than 12 minutes.
b. Determine the probability the amount of time Gary takes to assist the next customer exceeds 20 minutes.
c. Determine the probability the amount of time Gary takes to assist the next customer is between 8 and 15 minutes (in the interval 8 to 15 minutes)
d. Determine the probability the amount of time Gary takes to assist the next customer is either less than 14 minutes or greater than 22 minutes.

Answer 1

Explanation:

6 customers per hour
(given mean)

Pdf is given by:
`f(x)=(1)/(\beta) e^(-x / \beta), 0<x`

Pdf:
`f(x)=(1)/(10) e^(-x / 10), \quad 0<x,[tex] and zero otherwise. </p><p>Cdf: [tex]P(X \leq x)=1-e^(-x / \beta)`

`P(X \leq x)=1-e^((-x / 10)) \quad` for
`x>0 \Longrightarrow P(X>x)=e^(-x / 10)`

a)
`\begin{array}{c} </p><p>P(X<12)=1-\exp (-12 / \beta)=1-e^(-1.2)=1-0.301194=0.698806 \approx 0.699 \\ </p><p>P(X<12)=0.699 \end{array}`

&\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}