Respuesta :
Answer:
a) 3.47% probability that there will be exactly 15 arrivals.
b) 58.31% probability that there are no more than 10 arrivals.
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.
If the mean number of arrivals is 10
This means that [tex]\mu = 10[/tex]
(a) that there will be exactly 15 arrivals?
This is P(X = 15). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347[/tex]
3.47% probability that there will be exactly 15 arrivals.
(b) no more than 10 arrivals?
This is [tex]P(X \leq 10)[/tex]
[tex]P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045[/tex]
[tex]P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045[/tex]
[tex]P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023[/tex]
[tex]P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076[/tex]
[tex]P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189[/tex]
[tex]P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378[/tex]
[tex]P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631[/tex]
[tex]P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901[/tex]
[tex]P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126[/tex]
[tex]P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251[/tex]
[tex]P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251[/tex]
[tex]P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831[/tex]
58.31% probability that there are no more than 10 arrivals.
