Respuesta :
Answer:
0.0204 = 2.04% probability that fewer than three tickets are written on a randomly selected day.
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.
Poisson distribution with a mean of 7.5.
This means that [tex]\mu = 7.5[/tex].
Find the probability that fewer than three tickets are written on a randomly selected day.
This is:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-7.5}*(7.5)^{0}}{(0)!} = 0.0006[/tex]
[tex]P(X = 1) = \frac{e^{-7.5}*(7.5)^{1}}{(1)!} = 0.0042[/tex]
[tex]P(X = 2) = \frac{e^{-7.5}*(7.5)^{2}}{(2)!} = 0.0156[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0006 + 0.0042 + 0.0156 = 0.0204[/tex]
0.0204 = 2.04% probability that fewer than three tickets are written on a randomly selected day.
