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2 years ago

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At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.5 per day on a typical Wednesday. Let

X be the number of cancellations on a particular Wednesday. (a) Justify the use of the Poisson model. Cancellations are not independent. Cancellations are independent and similar to arrivals. Most likely cancellations arrive independently. (b) What is the probability that no cancellations will occur on a particular Wednesday

1 answer:

Kamila [148]2 years ago

5 0

Answer:

a) Cancellations are independent and similar to arrivals.

b) 22.31% probability that no cancellations will occur on a particular Wednesday

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Mean rate of 1.5 per day on a typical Wednesday.

This means that $\mu = 1.5$

(a) Justify the use of the Poisson model.

Each wednesday is independent of each other, and each wednesday has the same mean number of cancellations.

So the answer is:

Cancellations are independent and similar to arrivals.

(b) What is the probability that no cancellations will occur on a particular Wednesday

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.5}*(1.5)^{0}}{(0)!} = 0.2231$

22.31% probability that no cancellations will occur on a particular Wednesday

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