Respuesta :
Answer: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318
Step-by-step explanation:
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:
Since we have given that
[tex]\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5[/tex]
Using the poisson process, we get that
(a) less than or equal to 12
probability= [tex]P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980[/tex]
(b) equal to 13
probability= [tex]P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013[/tex]
(c) greater than 12
probability= [tex]P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020[/tex]
(d) equal to 20
probability= [tex]P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026[/tex]
(e) between 10 and 15, inclusively
probability=[tex]P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318[/tex]
Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318
Using the Poisson distribution, the probability of x inquires in a particular 50-millisecond stretch will be:
[tex]P(X = x) = \frac{e^{-5}5^{x}}{(x)!}[/tex]
In this problem, we are given the mean during an interval, which means that the Poisson distribution is used.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
We are given an average rate of 0.1 inquires per millisecond.
50-millisecond interval, thus [tex]\mu = 50(0.1) = 5[/tex]
Then, the probability of x inquires in a particular 50-millisecond stretch will be:
[tex]P(X = x) = \frac{e^{-5}5^{x}}{(x)!}[/tex]
A similar problem is given at https://brainly.com/question/24098004
