Question

Autos arrive at a toll plaza located at the entrance to a bridge at a rate of 50 per minute during the​ 5:00-to-6:00 P.M. hour. Determine the following probabilities assuming that an auto has just arrived. a. What is the probability that the next auto will arrive within 6 seconds ​(0.1 ​minute)? b. What is the probability that the next auto will arrive within 3 seconds ​(0.05 ​minute)? c. What are the answers to​ (a) and​ (b) if the rate of arrival of autos is 60 per​ minute? d. What are the answers to​ (a) and​ (b) if the rate of arrival of autos is 30 per​ minute?

Answer 1

Answer:

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%.