a. The probability that the next auto will arrive within 6 seconds (0.1 minute) is 99.33%.
b. The probability that the next auto will arrive within 3 seconds (0.05 minute) is 91.79%.
c. What are the answers to (a) and (b) if the rate of arrival of autos is 60 per minute?
For c(a.), the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 99.75%.
For c(b.), the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 99.75%.
d. What are the answers to (a) and (b) if the rate of arrival of autos is 30 per minute?
For d(a.), the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 95.02%.
For d(b.), the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 77.67%.
Step-by-step explanation:
a. What is the probability that the next auto will arrive within 6 seconds (0.1 minute)?
Assume that x represents the exponential distribution with parameter v = 50,
Given this, we can therefore estimate the probability that the next auto will arrive within 6 seconds (0.1 minute) as follows:
P(x < x) = 1 – e^-(vx)
Where;
v = parameter = rate of autos that arrive per minute = 50
x = Number of minutes of arrival = 0.1 minutes
Therefore, we specifically define the probability and solve as follows:
P(x ≤ 0.1) = 1 – e^-(50 * 0.10)
P(x ≤ 0.1) = 1 – e^-5
P(x ≤ 0.1) = 1 – 0.00673794699908547
P(x ≤ 0.1) = 0.9933, or 99.33%
Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is 99.33%.
b. What is the probability that the next auto will arrive within 3 seconds (0.05 minute)?
Following the same process in part a, x is now equal to 0.05 and the specific probability to solve is as follows:
P(x ≤ 0.05) = 1 – e^-(50 * 0.05)
P(x ≤ 0.05) = 1 – e^-2.50
P(x ≤ 0.05) = 1 – 0.0820849986238988
P(x ≤ 0.05) = 0.9179, or 91.79%
Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is 91.79%.
c. What are the answers to (a) and (b) if the rate of arrival of autos is 60 per minute?
For c(a.) Now we have:
v = parameter = rate of autos that arrive per minute = 60
x = Number of minutes of arrival = 0.1 minutes
Therefore, we specifically define the probability and solve as follows:
P(x ≤ 0.1) = 1 – e^-(60 * 0.10)
P(x ≤ 0.1) = 1 – e^-6
P(x ≤ 0.1) = 1 – 0.00247875217666636
P(x ≤ 0.1) = 0.9975, or 99.75%
Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 99.75%.
For c(b.) Now we have:
v = parameter = rate of autos that arrive per minute = 60
x = Number of minutes of arrival = 0.05 minutes
Therefore, we specifically define the probability and solve as follows:
P(x ≤ 0.05) = 1 – e^-(60 * 0.05)
P(x ≤ 0.05) = 1 – e^-3
P(x ≤ 0.05) = 1 – 0.0497870683678639
P(x ≤ 0.05) = 0.950212931632136, or 95.02%
Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 99.75%.
d. What are the answers to (a) and (b) if the rate of arrival of autos is 30 per minute?
For d(a.) Now we have:
v = parameter = rate of autos that arrive per minute = 30
x = Number of minutes of arrival = 0.1 minutes
Therefore, we specifically define the probability and solve as follows:
P(x ≤ 0.1) = 1 – e^-(30 * 0.10)
P(x ≤ 0.1) = 1 – e^-3
P(x ≤ 0.1) = 1 – 0.0497870683678639
P(x ≤ 0.1) = 0.950212931632136, or 95.02%
Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 95.02%.
For d(b.) Now we have:
v = parameter = rate of autos that arrive per minute = 30
x = Number of minutes of arrival = 0.05 minutes
Therefore, we specifically define the probability and solve as follows:
P(x ≤ 0.05) = 1 – e^-(30 * 0.05)
P(x ≤ 0.05) = 1 – e^-1.50
P(x ≤ 0.05) = 1 – 0.22313016014843
P(x ≤ 0.05) = 0.7767, or 77.67%
Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 77.67%.
