Question

Customers arrive at a drive-through teller window of a bank. They stay in line when the teller is busy. The service time is exponentially distributed with a mean of four minutes.
a. What is the probability that the next customer in line will take longer than seven minutes to be served? (2 Marks) b. What is the probability that the next customer in line will take less than eight minutes to be served? (2 Marks) c. What is the probability that the next customer in line will take between three and six minutes to be served? (4 Marks)

Answer 1

Defining lambda.

Density function of the exponential distribution is $f_X(x)=\begin{cases} \lambda e^{-\lambda x} \quad ,x \geq0 \\ 0 \qquad , x < 0\end{cases}$ with mean $\dfrac{1}{\lambda}.$

Task says that mean is 4 minutes, so $\lambda = \frac{1}{4}$.

a.

Solution

$\int\limits_7^\infty f_X(x)\;dx = \int\limits_7^\infty \frac{1}{4}\cdot e^{-\frac{1}{4}x}\;dx = -e^{-\frac{1}{4}x} |_7^\infty = e^{-\frac{7}{4}} \approx 0.1738$

Answer: 0.1738

b.

Solution

$\int\limits_{-\infty}^8 f_X(x)\;dx = \int\limits_0^8 \frac{1}{4}\cdot e^{-\frac{1}{4}x}\;dx = -e^{-\frac{1}{4}x} |_0^8= 1 - e^{-2} \approx 0.8647$

Answer: 0.8647

c.

Solution

$\int\limits_3^6 f_X(x)\;dx = \int\limits_3^6 \frac{1}{4}\cdot e^{-\frac{1}{4}x}\;dx = -e^{-\frac{1}{4}x} |_3^6= e^{-\frac{3}{4}} - e^{-\frac{6}{4}} \approx 0.2492$

Answer: 0.2492