Question

2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected passenger will wait the following times for a CTfastrak bus. b. Between 5 and 10 minutes. c. Exactly 7.5922 minutes. d. Exactly 5 minutes. e. Between 15 and 25 minutes.

Answer 1

Answer:

b. 0.25

c. 0.05

d. 0.05

e. 0.25

Step-by-step explanation:

if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

$P(x)=\frac{1}{b-a}=\frac{1}{20-0} =0.05$

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

$P(X<x)=\frac{x-a}{b-a}=\frac{x-0}{20-0}=\frac{x}{20}$

Then, the probability that a randomly selected passenger will wait:

b. Between 5 and 10 minutes.

$P(5<x<10) = P(x<10) - P(x<5)\\P(5<x<10) = \frac{10}{20} -\frac{5}{20}=0.25$

c. Exactly 7.5922 minutes

$P(7.5922)=0.05$

d. Exactly 5 minutes

$P(5)=0.05$

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

$P(15<x<25)=P(15<x<20) \\P(15<x<20)=P(x<20) - P(x<15)\\P(15<x<20) = \frac{20}{20} -\frac{15}{20}=0.25$