Question

On weekend nights, a large urban hospital has an average of 4.8 emergency arrivals per hour. Let X be the number of arrivals per hour on a weekend night at this hospital. Assume that successive arrivals are random and independent. What is the probability P(X < 3)?

Answer 1

P(X < 3) = 0.14254

Explanation:

We have only the mean, which means that the Poisson distribution is used to solve this question.

Poisson distribution:

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) = (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 interval.

On weekend nights, a large urban hospital has an average of 4.8 emergency arrivals per hour.

This means that
`\mu = 4.8`

What is the probability P(X < 3)?

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

So

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-4.8)*4.8^(0))/((0)!) = 0.00823`

`P(X = 1) = (e^(-4.8)*4.8^(1))/((1)!) = 0.03950`

`P(X = 2) = (e^(-4.8)*4.8^(2))/((2)!) = 0.09481`

So

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00823 + 0.03950 + 0.09481 = 0.14254`

P(X < 3) = 0.14254