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 . The interval between patient arrivals in an emergency room  is a random variable with exponential density function p(t)=0.125e0.125t(t in minutes). What is the average time between pa-  tient arrivals? What is the probability of two patients arriving within 3 min of each other? \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.

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

p(t)=0.125e0.125tp(t) = 0.125e^{-0.125t}

where tt is in minutes.

Now the average time between patient arrivals is

0tp(t) dt=00.125te0.125t dt\int_{0}^{\infty} tp(t)\ dt = \int_{0}^{\infty} 0.125te^{-0.125t}\ dt

Let u=tu = t and dv=0.125e0.125t dtdv = 0.125e^{-0.125t}\ dt. Then du=1du = 1 and v=e0.125tv = -e^{-0.125t}. Now using integration by parts, we get

00.125te0.125t dt=te0.125t0+0e0.125t dt=limRRe0.125R0+(10.125e0.125t)0=limR8e0.125R8(01)=0+8=8\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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