Question

The commute time for people in a city has an exponential distribution with an average of 0.5 hours. What is the probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours? Answer: (round to 3 decimal places)

Answer 1

Answer:

0.314 = 31.4% probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours

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:

$m = 0.5, \mu = \frac{1}{0.5} = 2$

What is the probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours?

$P(0.4 \leq X \leq 1) = P(X \leq 1) - P(X \leq 0.4)$

In which

$P(X \leq 1) = 1 - e^{-2} = 0.8647$

$P(X \leq 0.4) = 1 - e^{-2*0.4} = 0.5507$

So

$P(0.4 \leq X \leq 1) = P(X \leq 1) - P(X \leq 0.4) = 0.8647 - 0.5507 = 0.314$

0.314 = 31.4% probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours