## Related questions with answers

$\left. \begin{array} { l } { \text { . The interval between patient arrivals in an emergency room } } \\ { \text { is a random variable with exponential density function } p ( t ) = } \\ { 0.125 e ^ { - 0.125 t } ( t \text { in minutes). What is the average time between pa- } } \\ { \text { tient arrivals? What is the probability of two patients arriving within } 3 } \\ { \text { min of each other? } } \end{array} \right.$

Solution

VerifiedGiven that the interval between patient arrivals in an emergency room is a random variable with exponential density function

$p(t) = 0.125e^{-0.125t}$

where $t$ is in minutes.

Now the average time between patient arrivals is

$\int_{0}^{\infty} tp(t)\ dt = \int_{0}^{\infty} 0.125te^{-0.125t}\ dt$

Let $u = t$ and $dv = 0.125e^{-0.125t}\ dt$. Then $du = 1$ and $v = -e^{-0.125t}$. Now using integration by parts, we get

$\begin{align*} \int_{0}^{\infty} 0.125te^{-0.125t}\ dt & = te^{-0.125t}\bigg|_{0}^{\infty} + \int_{0}^{\infty} e^{-0.125t}\ dt\\ \\ & = \lim\limits_{R \to \infty} Re^{-0.125R} - 0 + \bigg(-\dfrac{1}{0.125}e^{-0.125t}\bigg)_{0}^{\infty}\\ \\ & = \lim\limits_{R \to \infty} -8e^{-0.125R}- 8(0 - 1)\\ \\ & = 0 +8\\ \\ & = 8 \end{align*}$

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