Respuesta :
Answer:
a) [tex]P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}[/tex]
b) f(2) = 0.04462
c) f(1) = 0.01487
d) [tex]P(X \geq 2) = 0.93803[/tex]
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 successes
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
In this question:
[tex]\mu = 6[/tex]
a. Write the appropriate Poisson probability function.
Considering [tex]\mu = 6[/tex]
[tex]P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}[/tex]
b. Compute f (2).
This is P(X = 2). So
[tex]P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462[/tex]
So f(2) = 0.04462
c. Compute f (1).
This is P(X = 1). So
[tex]P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}[/tex]
[tex]P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487[/tex]
So f(1) = 0.01487.
d. Compute P(x≥2)
This is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248[/tex]
[tex]P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487[/tex]
[tex]P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462[/tex]
Then
[tex]P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803[/tex]
So
[tex]P(X \geq 2) = 0.93803[/tex]
