A service station has a pump that distributes diesel fuel to automobiles. The station owner estimates that only about 3.2 cars use the diesel pump every 2 hours. Assume the arrivals of diesel pump users are Poisson distributed.
a. What is the probability that three cars will arrive to use the diesel pump during a 1-hour period?
b. Suppose the owner needs to shut down the diesel pump for half an hour to make repairs. However, the owner hates to lose any business. What is the probability that no cars will arrive to use the diesel pump during a half-hour period?
c. Suppose five cars arrive during a 1-hour period to use the diesel pump. What is the probability of five or more cars arriving during a 1-hour period to use the diesel pump?
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Determining m:
3.2 cars every 2 hours
1/2(3.2) cars every hour = 1.6
1/4(3.2) cars every 1/2 hour = 0.8
For a):
x = 3
m = 1.6
Substitute into the formula and find the probability.
For b):
x = 0
m = 0.8
Substitute into the formula and find the probability.
For c):
x = 0,1,2,3,4
m = 1.6
Find the probabilities and add for a total. Take that value and subtract from 1.
I hope this will help get you started.
To solve these problems, we can use the Poisson probability formula:
P(X=k) = (e^(-λ) * λ^k) / k!
Where:
P(X=k) is the probability of k events occurring,
λ is the average rate of events occurring in the given time period,
e is Euler's number (approximately 2.71828),
k is the number of events occurring.
a. To find the probability that three cars will arrive to use the diesel pump during a 1-hour period, we need to find P(X=3) when λ=3.2.
P(X=3) = (e^(-3.2) * 3.2^3) / 3!
Evaluating this expression:
P(X=3) = (2.71828^(-3.2) * 3.2^3) / (3*2*1)
P(X=3) ≈ 0.201
Therefore, the probability that three cars will arrive to use the diesel pump during a 1-hour period is approximately 0.201.
b. To find the probability that no cars will arrive to use the diesel pump during a half-hour period, we need to find P(X=0) when λ=1.6 (half of 3.2).
P(X=0) = (e^(-1.6) * 1.6^0) / 0!
Since 0! is equal to 1 (zero factorial is defined as 1), this simplifies to:
P(X=0) = e^(-1.6)
Evaluating this expression:
P(X=0) ≈ 0.201
Therefore, the probability that no cars will arrive to use the diesel pump during a half-hour period is approximately 0.201.
c. To find the probability of five or more cars arriving during a 1-hour period given that five cars have already arrived, we need to find P(X>=5) when λ=5.
P(X>=5) = 1 - P(X<5)
To find P(X<5), we can calculate the sum of P(X=k) for k=0, 1, 2, 3, 4.
P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
Using the Poisson probability formula, we can calculate each term and sum them:
P(X<5) = (e^(-5) * 5^0) / 0! + (e^(-5) * 5^1) / 1! + (e^(-5) * 5^2) / 2! + (e^(-5) * 5^3) / 3! + (e^(-5) * 5^4) / 4!
Evaluating this expression:
P(X<5) ≈ 0.493
Therefore, the probability of five or more cars arriving during a 1-hour period given that five cars have already arrived is approximately 0.493.
To answer these questions, we need to use the Poisson distribution formula. The formula for the Poisson distribution is:
P(x;λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x;λ) represents the probability of observing x events in a given time period, when the average number of events in that time period is λ.
- e is Euler's number, approximately 2.71828.
- x is the number of events we want to calculate the probability for.
- λ is the average number of events in the given time period.
Now let's solve the questions step by step:
a. What is the probability that three cars will arrive to use the diesel pump during a 1-hour period?
Here, we need to calculate P(x=3;λ) where λ is the average number of cars arriving in a 2-hour period. From the given information, we know that 3.2 cars arrive every 2 hours. To determine the average number of cars arriving in one hour, we divide λ by 2:
λ = 3.2 / 2 = 1.6
Now we can use the Poisson distribution formula:
P(x=3;λ=1.6) = (e^(-1.6) * 1.6^3) / 3!
Using a calculator or Python programming, evaluate the expression to find the solution.
b. What is the probability that no cars will arrive to use the diesel pump during a half-hour period?
To find the probability of no cars arriving in a half-hour period, we need to calculate P(x=0;λ), where λ is the average number of cars arriving in a 2-hour period. Again, divide the average by 2 to get the average number of cars arriving in half an hour:
λ = 3.2 / 2 = 1.6
Now, use the Poisson distribution formula:
P(x=0;λ=1.6) = (e^(-1.6) * 1.6^0) / 0!
Evaluate the expression to find the probability.
c. What is the probability of five or more cars arriving during a 1-hour period to use the diesel pump?
For this question, we need to calculate the sum of probabilities for x ≥ 5. We can calculate P(X=5;λ), P(X=6;λ), P(X=7;λ), and so on until we reach a small enough probability that can be considered negligible. Then sum up all these probabilities.
P(X=5;λ) = (e^(-λ) * λ^5) / 5!
P(X=6;λ) = (e^(-λ) * λ^6) / 6!
P(X=7;λ) = (e^(-λ) * λ^7) / 7!
...
Continue calculating these probabilities until you reach a negligible probability (usually less than 0.001). Then add up all these probabilities to get the final result.
Keep in mind that for all these calculations, you need to use a calculator or programming language that can handle exponentials and factorials, as they can be quite cumbersome to calculate manually.