A telephone company's goal is to have no more than five monthly line failures on any 150 miles of line. The company currently experiences an average of 3 monthly line failures per 75 miles of line. Let x denote the number of monthly line failures per 150 miles of line. Assuming x has a Poisson distribution:
v 的s
To solve this problem, we will use the properties of a Poisson distribution.
1. The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space.
2. The average number of events (λ) is known and remains constant throughout the interval.
3. The events must occur independently and at a constant rate.
Let's proceed with the problem step-by-step:
Step 1: Define the average number of monthly line failures per 75 miles of line
The problem states that the company currently experiences an average of 3 monthly line failures per 75 miles of line. Therefore, the average number of monthly line failures per mile is λ = 3/75.
Step 2: Determine the number of monthly line failures per 150 miles of line
To find the number of failures per 150 miles, we need to double the average number of failures per mile. Let's define x as the number of monthly line failures per 150 miles of line.
x = 2 * (3/75)
Simplifying, we get:
x = 6/75
Step 3: Determine the probability of having less than or equal to 5 monthly line failures on 150 miles of line
The Poisson distribution formula for the probability is given by:
P(x;λ) = (e^(-λ) * λ^x) / x!
We need to calculate the probability of having less than or equal to 5 monthly line failures, so we need to sum the probabilities of having 0, 1, 2, 3, 4, and 5 failures.
P(x ≤ 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5)
Using the Poisson distribution formula, we can calculate each probability separately and add them together.
P(x = 0) = (e^(-6/75) * (6/75)^0) / 0!
P(x = 1) = (e^(-6/75) * (6/75)^1) / 1!
P(x = 2) = (e^(-6/75) * (6/75)^2) / 2!
P(x = 3) = (e^(-6/75) * (6/75)^3) / 3!
P(x = 4) = (e^(-6/75) * (6/75)^4) / 4!
P(x = 5) = (e^(-6/75) * (6/75)^5) / 5!
Step 4: Calculate the individual probabilities
Using a calculator or computer software, we can calculate each probability.
P(x = 0) ≈ 0.324361
P(x = 1) ≈ 0.243271
P(x = 2) ≈ 0.182454
P(x = 3) ≈ 0.109472
P(x = 4) ≈ 0.054736
P(x = 5) ≈ 0.022895
Step 5: Calculate the total probability
Add up the individual probabilities to get the total probability:
P(x ≤ 5) ≈ 0.936189
Therefore, the probability of having no more than five monthly line failures on 150 miles of line is approximately 0.936189.
To analyze this situation, we can use the Poisson distribution, which is commonly used to model the number of events occurring in a fixed interval of time or space.
The Poisson distribution is characterized by a single parameter, λ (lambda), which represents the average number of events occurring in the given interval. In this case, the average number of line failures per 75 miles is given as 3.
Let's denote the number of monthly line failures per 150 miles of line as x. Since the company wants no more than five monthly line failures on any 150 miles of line, we need to determine the probability of having x failures.
To find the probability, we need to calculate the Poisson probability mass function (pmf), which gives the probability of observing a specific number of events (x) in the given interval.
The Poisson pmf is given by the formula:
P(x) = (e^(-λ) * λ^x) / x!
where e is the mathematical constant approximately equal to 2.71828.
Substituting our values, we have:
P(x) = (e^(-3) * 3^x) / x!
Now, we can calculate the probability of no more than five monthly line failures on 150 miles of line.
P(x ≤ 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5)
Calculate each term using the Poisson pmf formula, and sum up the probabilities to get the final result.
Remember that the factorial (x!) in the denominator is calculated by multiplying all positive integers from 1 to x. For example, 4! = 4 * 3 * 2 * 1 = 24.
Using this approach, you can find the probability of having no more than five monthly line failures on 150 miles of line, assuming a Poisson distribution.