At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.60 per day on a typical Wednesday. Let X be the number of cancellations on a particular Wednesday.
a) What is the probability that no cancellations will occur on a particular Wednesday? (Round your answer to 4 decimal places.)
b) What is the probability that exactly three cacellation will occur on a particular wednesday? (Round your answer to 4 decimal places.)
c) What is the probability that more than five cacellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)
d) What is the probability that five or more cacellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)
To solve this problem, we can use the Poisson distribution formula because it is appropriate for modeling the number of events occurring in a fixed interval of time or space when events occur randomly and independently.
The formula for the Poisson distribution is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- X is the random variable representing the number of events.
- k is the specific number of events we are interested in.
- λ (lambda) is the average rate of events (mean).
Given that the mean rate of appointment cancellations is 1.60 per day on a typical Wednesday, we can use this value as λ.
a) To find the probability that no cancellations will occur on a particular Wednesday (X = 0):
P(X = 0) = (e^(-1.60) * 1.60^0) / 0!
Since 0! is equal to 1, the formula simplifies to:
P(X = 0) = e^(-1.60)
To calculate this, we can use a calculator or the value of e (approximately 2.71828) to obtain:
P(X = 0) ≈ 0.2019
So, the probability that no cancellations will occur on a particular Wednesday is approximately 0.2019.
b) To find the probability that exactly three cancellations will occur on a particular Wednesday (X = 3):
P(X = 3) = (e^(-1.60) * 1.60^3) / 3!
Calculating this using a calculator or the value of e results in:
P(X = 3) ≈ 0.2137
So, the probability that exactly three cancellations will occur on a particular Wednesday is approximately 0.2137.
c) To find the probability that more than five cancellations will occur on a particular Wednesday:
P(X > 5) = 1 - P(X ≤ 5)
Given that the Poisson distribution is infinite, we need to calculate the probabilities for X = 0, 1, 2, 3, 4, and 5, and then subtract their sum from 1 to find the probability for X > 5.
Using the Poisson distribution formula for each value of X and summing the probabilities, we get:
P(X > 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))
Substituting the respective values and calculating gives:
P(X > 5) ≈ 0.0141
So, the probability that more than five cancellations will occur on a particular Wednesday is approximately 0.0141.
d) To find the probability that five or more cancellations will occur on a particular Wednesday:
P(X ≥ 5) = P(X = 5) + P(X > 5)
Using the probabilities calculated earlier, we can find:
P(X ≥ 5) = P(X = 5) + P(X > 5)
P(X ≥ 5) = P(X = 5) + (1 - P(X ≤ 5))
P(X ≥ 5) = P(X = 5) + 0.0141
From part b, we know P(X = 5) is approximately 0.1336.
Thus:
P(X ≥ 5) = 0.1336 + 0.0141
P(X ≥ 5) ≈ 0.1477
So, the probability that five or more cancellations will occur on a particular Wednesday is approximately 0.1477.
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