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Sagot :
Answer:
a) 2
b)
0.1353 = 13.53% probability that exactly 0 customers will arrive during a five-minute period.
0.2707 = 27.07% probability that exactly 1 customer will arrive during a five-minute period.
0.2707 = 27.07% probability that exactly 2 customers will arrive during a five-minute period.
0.1805 = 18.05% probability that exactly 3 customers will arrive during a five-minute period.
c) 0.1428 = 14.28% probability that delays will occur
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 interval.
a. What is the mean or expected number of customers that will arrive in a five-minute period?
0.4 customers per minute, so for five minutes, this is [tex]\mu = 5*0.4 = 2[/tex]
b. Assume that the Poisson probability distribution can be used to describe the arrival process. Use the arrival rate in part (a) and compute the probabilities that exactly 0, 1, 2, and 3 customers will arrive during a five-minute period.
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]
0.1353 = 13.53% probability that exactly 0 customers will arrive during a five-minute period.
[tex]P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707[/tex]
0.2707 = 27.07% probability that exactly 1 customer will arrive during a five-minute period.
[tex]P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707[/tex]
0.2707 = 27.07% probability that exactly 2 customers will arrive during a five-minute period.
[tex]P(X =3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1805[/tex]
0.1805 = 18.05% probability that exactly 3 customers will arrive during a five-minute period.
c. Delays are expected if more than three customers arrive during any five-minute period. What is the probability that delays will occur?
This is:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which:
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1353 + 0.2707 + 0.2707 + 0.1805 = 0.8572[/tex]
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8572 = 0.1428[/tex]
0.1428 = 14.28% probability that delays will occur
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