Respuesta :
Answer:
F(t<0.1 ) = 0.91791
Step-by-step explanation:
Solution:
- Let X be an exponential RV denoting time t in hours from start of interval to until first log-on that arises from Poisson process with the rate λ = 25 log-ons/hr. Its cumulative density function is given by:
F(t) = 1 - e ^ ( - 25*t ) t > 0
A) In this case we are interested in the probability that it takes t = 6/60 = 0.1 hrs until the first log-on. F ( t < 0.1 hr ), we have:
F(t<0.1 ) = 1 - e ^ ( - 25*0.1 )
F(t<0.1 ) = 0.91791
Using the Poisson distribution, it is found that there is a:
a) 0.0821 = 8.21% probability that there are no logons in an interval of 6 minutes.
b) [tex]1.3887 \times 10^{-11}[/tex] probability that the distance between two log-ons be more than one hour.
We are given only the mean, hence, the Poisson distribution is used.
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.
Item a:
Mean of 25 log-ons per hour, hence, in 6 minutes, the mean is of [tex]\mu = \frac{6}{60} \times 25 = 2.5[/tex]
The probability is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2.5}(2.5)^{0}}{(0)!} = 0.0821[/tex]
0.0821 = 8.21% probability that there are no logons in an interval of 6 minutes.
Item b:
This is the probability of no log-ons in one hour, which is P(X = 0) with [tex]\mu = 25[/tex], then:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-25}(25)^{0}}{(0)!} = 1.3887 \times 10^{-11}[/tex]
[tex]1.3887 \times 10^{-11}[/tex] probability that the distance between two log-ons be more than one hour.
To learn more about the Poisson distribution, you can take a look at https://brainly.com/question/17037644
