Respuesta :
Answer:
10.16% probability that at most 6 trains will pass his house in a 12-hour time period
Step-by-step explanation:
We are dealing with the mean during a period of time, so we use the Poisson distribution to solve this question.
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.
Colby has noticed that 21 trains pass by his house daily (24 hours)
We are working with 12 hour periods(half of 24), so [tex]\mu = \frac{21}{2} = 10.5[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
What is the probability that at most 6 trains will pass his house in a 12-hour time period
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-10.5}*(10.5)^{0}}{(0)!} = 0[/tex]
[tex]P(X = 1) = \frac{e^{-10.5}*(10.5)^{1}}{(1)!} = 0.0003[/tex]
[tex]P(X = 2) = \frac{e^{-10.5}*(10.5)^{2}}{(2)!} = 0.0015[/tex]
[tex]P(X = 3) = \frac{e^{-10.5}*(10.5)^{3}}{(3)!} = 0.0053[/tex]
[tex]P(X = 4) = \frac{e^{-10.5}*(10.5)^{4}}{(4)!} = 0.0139[/tex]
[tex]P(X = 5) = \frac{e^{-10.5}*(10.5)^{5}}{(5)!} = 0.0293[/tex]
[tex]P(X = 6) = \frac{e^{-10.5}*(10.5)^{6}}{(6)!} = 0.0513[/tex]
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0 + 0.0003 + 0.0015 + 0.0053 + 0.0139 + 0.0293 + 0.0513 = 0.1016[/tex]
10.16% probability that at most 6 trains will pass his house in a 12-hour time period
