Respuesta :
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] 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 [tex]\mu = 13.4224[/tex]
Find the probability that in a day, there will be at least 1 birth.
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-13.4224}*13.4224^{0}}{(0)!} = 0.0000015[/tex]
Then
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000015 = 0.9999985 [/tex]
0.9999985 = 99.99985% probability that in a day, there will be at least 1 birth.
