Question

The time between arrivals of customers at an automatic teller machine is an exponential random variable with a mean of 4 minutes. Round yours answers to 4 decimal places. (a) What is the probability that more than three customers arrive in 10 minutes

Answer 1

Answer:

0.7788 = 77.88% probability that more than three customers arrive in 10 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question, we have that:

The time between arrivals of customers at an automatic teller machine is an exponential random variable with a mean of 4 minutes, and we have three customers, which means that $m = 3*4 = 12, \mu = \frac{1}{12} = 0.0833$

(a) What is the probability that more than three customers arrive in 10 minutes

This is P(X > 3). So

$P(X > 3) = e^{-0.0833*3} = 0.7788$

0.7788 = 77.88% probability that more than three customers arrive in 10 minutes