Respuesta :
Answer:
1.
a. 2
b. 0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.
c. 0.1428 = 14.28% probability that delays will occur.
2.
a. 0.4512 = 45.12% probability that the service time is one minute or less.
b. 0.6988 = 69.88% probability that the service time is two minutes or less.
c. 0.3012 = 30.12% probability that the service time is more than two minutes.
Step-by-step explanation:
Poisson distribution:
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.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
Question 1:
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 5 minutes:
[tex]\mu = 0.4*5 = 2[/tex]
So 2 is the answer.
Question b:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707[/tex]
[tex]P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707[/tex]
[tex]P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1805[/tex]
0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.
Question c:
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)[/tex]
The values we have in item b, so:
[tex]P(X \leq 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.
Question 2:
[tex]\mu = 0.6[/tex]
a. What is the probability that the service time is one minute or less?
[tex]P(X \leq 1) = 1 - e^{-0.6} = 0.4512[/tex]
0.4512 = 45.12% probability that the service time is one minute or less.
b. What is the probability that the service time is two minutes or less?
[tex]P(X \leq 2) = 1 - e^{-0.6(2)} = 1 - e^{-1.2} = 0.6988[/tex]
0.6988 = 69.88% probability that the service time is two minutes or less.
c. What is the probability that the service time is more than two minutes?
[tex]P(X > 2) = e^{-1.2} = 0.3012[/tex]
0.3012 = 30.12% probability that the service time is more than two minutes.
