Answer
(a) The probability that a customer waits less than a second is 0.465 which is 46.5%
(b) The probability that a customer waits more than 3 seconds is 0.153 which is 15.3%
(c) The minimum approval time for the slowest 5% of transactions is 4.79 seconds.
Work Step by Step
$f(t) = \frac{1}{\mu}e^{-t/\mu},$ where $t \geq 0$
$f(t) = \frac{1}{1.6}e^{-t/1.6}$
(a) (i) We can find $P(T \lt 1)$:
$P(T \lt 1) = \int_{0}^{1}f(t)~dt$
$P(T \lt 1) = \int_{0}^{1}\frac{1}{1.6}e^{-t/1.6}~dt$
$P(T \lt 1) = -e^{-t/1.6}~\vert_{0}^{1}$
$P(T \lt 1) = 1-e^{-1/1.6}$
$P(T \lt 1) = 0.465$
The probability that a customer waits less than a second is 0.465 which is 46.5%
(b) We can find $P(T \gt 3)$:
$P(T \gt 3) = \int_{3}^{\infty}f(t)~dt$
$P(T \gt 3) = \int_{3}^{\infty}\frac{1}{1.6}e^{-t/1.6}~dt$
$P(T \gt 3) = -e^{-t/1.6}~\vert_{3}^{\infty}$
$P(T \gt 3) = 0+e^{-3/1.6}$
$P(T \gt 3) = 0.153$
The probability that a customer waits more than 3 seconds is 0.153 which is 15.3%
(c) To find the minimum time $t$, we can find $t$ such that $P(T \gt t) = 0.05$:
$P(T \gt t) = 0.05$
$\int_{t}^{\infty}f(t)~dt = 0.05$
$\int_{t}^{\infty}\frac{1}{1.6}e^{-t/1.6}~dt = 0.05$
$-e^{-t/1.6}~\vert_{t}^{\infty} = 0.05$
$0+e^{-t/1.6} = 0.05$
$-t/1.6 = ln(0.05)$
$t = -1.6~ln(0.05)$
$t = 4.79$
The minimum approval time for the slowest 5% of transactions is 4.79 seconds.
