Average annual value lost: $7,390.65First-year depreciation: $3,000.00Total depreciation: $11,125.89Total depreciation percentage: 55.63%Value of vehicle at end of ownership period: $8,874.11
see attachment for graph

7 0

3 years ago

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

Kamila [148]

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

5 0

3 years ago