Question

The time between arrivals of customers at the drive-up window of a bank follows an exponential probability distribution with a mean of 10 minutes.
A. What is the probability that the arrival time between customers will be 7 minutes or less?
B. What is the probability that the arrival time between customers will be between 3 and 7 minutes?

Answer 1

Answer:

a) 50.34% probability that the arrival time between customers will be 7 minutes or less.

b) 24.42% probability that the arrival time between customers will be between 3 and 7 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:

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

Mean of 10 minutes:

This means that $m = 10, \mu = \frac{1}{10} = 0.1$

A. What is the probability that the arrival time between customers will be 7 minutes or less?

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

$P(X \leq 7) = 1 - e^{-0.1*7} = 0.5034$

50.34% probability that the arrival time between customers will be 7 minutes or less.

B. What is the probability that the arrival time between customers will be between 3 and 7 minutes?

$P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3)$

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

$P(X \leq 7) = 1 - e^{-0.1*7} = 0.5034$

$P(X \leq 3) = 1 - e^{-0.1*3} = 0.2592$

$P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 3) = 0.5034 - 0.2592 = 0.2442$

24.42% probability that the arrival time between customers will be between 3 and 7 minutes