Respuesta :
Answer:
0.40006
Step-by-step explanation:
Let as assume that X is a Poisson distributed random variable with parameter μ = 5.
Poisson distribution Formula:
[tex]P(X=x)=\dfrac{\mu^xe^{-\mu}}{x!}[/tex]
where, μ is mean.
We need to find the value of P(2 ≤ X ≤ 4).
[tex]P(2\leq X\leq 4)=P(X=2)+P(X=3)+P(X=4)[/tex]
[tex]P(2\leq X\leq 4)=\dfrac{5^2e^{-5}}{2!}+\dfrac{5^3e^{-5}}{3!}+\dfrac{5^4e^{-5}}{4!}[/tex]
[tex]P(2\leq X\leq 4)=0.08422+0.14037+0.17547[/tex]
[tex]P(2\leq X\leq 4)=0.40006[/tex]
Therefore the value of P(2 ≤ X ≤ 4) is 0.40006.
Using the Poisson distribution, it is found that P(2 ≤ X ≤ 4) = 0.4001.
What is the Poisson distribution?
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- 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 problem, the mean is [tex]\mu = 5[/tex], and:
[tex]P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-5}5^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}5^{3}}{(3)!} = 0.1404[/tex]
[tex]P(X = 4) = \frac{e^{-5}5^{4}}{(4)!} = 0.1755[/tex]
Then:
[tex]P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0842 + 0.1404 + 0.1755 = 0.4001[/tex]
Hence P(2 ≤ X ≤ 4) = 0.4001.
More can be learned about the Poisson distribution at https://brainly.com/question/13971530
