Question

The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. a. Find the value of λ. b. What is the probability that a person waits for less than 3 minutes?

Answer 1

Answer:

a) 0.25

b) 52.76% probability that a person waits for less than 3 minutes

Step-by-step explanation:

Exponential distribution:

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

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

In which $\lambda = \frac{1}{m}$ 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:

$m = 4$

a. Find the value of λ.

$\lambda = \frac{1}{m} = \frac{1}{4} = 0.25$

b. What is the probability that a person waits for less than 3 minutes?

$P(X \leq 3) = 1 - e^{-0.25*3} = 0.5276$

52.76% probability that a person waits for less than 3 minutes