Question

Assume that the Poisson distribution applies to the number of births at a particular hospital during a randomly selected day. Assume that the mean number of births per day at this hospital is 13.4224. Find the probability that in a day, there will be at least 1 birth.

Answer 1

Answer:

0.9999985  = 99.99985% probability that in a day, there will be at least 1 birth.

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 interval.

Assume that the mean number of births per day at this hospital is 13.4224.

This means that $\mu = 13.4224$

Find the probability that in a day, there will be at least 1 birth.

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = \frac{e^{-13.4224}*13.4224^{0}}{(0)!} = 0.0000015$

Then

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000015 = 0.9999985$

0.9999985  = 99.99985% probability that in a day, there will be at least 1 birth.