Respuesta :
Answer:
4.47% probability that 2 or more people arrive in the next 10 minutes
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 time interval.
Mean of 2 customers per hour.
What's the probability that 2 or more people arrive in the next 10 minutes?
10 minutes, so [tex]\mu = \frac{2*10}{60} = 0.3333[/tex]
Either less than two people arrive, or more than two do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.3333}*(0.3333)^{0}}{(0)!} = 0.7165[/tex]
[tex]P(X = 1) = \frac{e^{-0.3333}*(0.3333)^{1}}{(1)!} = 0.2388[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.7165 + 0.2388 = 0.9553[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9553 = 0.0447[/tex]
4.47% probability that 2 or more people arrive in the next 10 minutes
The probability that 2 or more people arrive in the next 10 minutes is;
P(X ≥ 2) = 0.0447
This is a poisson distribution problem with the formula;
P(X = x) = [e^(-μ) × μˣ]/x!
Where;
x is number of successes
μ is the mean in the time interval given
Since we have a mean of 2 customers per hour and want to find the probability that 2 or more people arrive in the next 10 minutes. Thus;
μ = (2 × 10)/60
μ = 0.333
Thus, probability that 2 or more people arrive in the next 10 minutes is;
P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))
P(X = 0) = [e^(-0.333) × 0.333⁰]/0!
P(X = 0) = 0.7165
P(X = 2) = [e^(-0.333) × 0.333²]/2!
P(X = 2) = 0.2388
Thus;
P(X ≥ 2) = 1 - (0.7165 + 0.2388)
P(X ≥ 2) = 0.0447
Read more at; https://brainly.com/question/16928843
