Question

An exponential distribution is formed by the waiting times for a patient in the emergency room. The average waiting time for this distribution is 38 minutes. Given that it has already taken 37 minutes, what is the probability that the wait time will be more than an additional 41 minutes? Round your answer to three decimal places.

Answer 1

Answer:

0.207 = 20.7% probability that the wait time will be more than an additional 41 minutes

Step-by-step explanation:

To solve this question, we should understand the exponential distribution and conditional probability.

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:

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

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

The average waiting time for this distribution is 38 minutes.

This means that $m = 38, \mu = \frac{1}{38} = 0.0263$

Given that it has already taken 37 minutes, what is the probability that the wait time will be more than an additional 41 minutes?

Event A: Taking more than 37 minutes.

Event B: More than 37 + 41 = 78 minutes.

Probability of taking more than 37 minutes:

$P(A) = P(X \leq 37) = 1 - e^{-0.0263*37} = 0.6221$

More than 37 minutes and more than 78 minutes:

The intersection is more than 78 minutes, so:

$P(A \cap B) = P(X > 78) = e^{-0.0263*78} = 0.1286$

The probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1286}{0.6221} = 0.207$

0.207 = 20.7% probability that the wait time will be more than an additional 41 minutes