Respuesta :
Answer:
a) [tex]P(X=5) = 15.63 \%[/tex]
b) [tex]P(X=9) = 6.88 \%[/tex]
c) [tex]P(X<3) = 67.65 \%[/tex]
Explanation:
a) What is the probability that 5 messages are received in a given minute?
The Poisson distribution is given by
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}[/tex]
Where x is the number of messages received in some time interval t.
mean rate = λ = 4 per minute
number of messages = x = 5
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=5) = e^{-4}\frac{4^{5}}{5!}\\\\P(X=5) = 0.1563\\\\P(X=5) = 15.63 \%[/tex]
b) What is the probability that 9 messages are received in 1.5 minutes?
mean rate = λ = 4*1.5 = 6
number of messages = x = 9
[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=9) = e^{-6}\frac{6^{9}}{9!}\\\\P(X=9) = 0.0688\\\\P(X=9) = 6.88 \%[/tex]
c) What is the probability that fewer than 3 messages are received in a period of 30 seconds?
mean rate = λ = 0.5*4 = 2
number of messages = x < 3
[tex]P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)\\\\P(X=0) = e^{-2}\frac{2^{0}}{0!} = 0.1353\\\\ P(X=1) = e^{-2}\frac{2^{1}}{1!} = 0.2706\\\\ P(X=2) = e^{-2}\frac{2^{2}}{2!} = 0.2706\\\\ P(X<3) =0.1353+0.2706+ 0.2706\\\\P(X<3) = 0.6765\\\\P(X<3) = 67.65 \%[/tex]
A) The probability that 5 messages are received in a given minute is; 15.629%
B) The probability that 9 messages are received in 1.5 minutes is; 6.884%
C) The probability that fewer than 3 messages are received in a period of 30 seconds is; 67.668%
We are told that the number of hits follows a poisson distribution with mean of 4 per minutes.
Formula for poisson distribution is;
P(X = x) = [tex]e^{-\lambda}(\frac{\lambda^{x}}{x!})[/tex]
where;
x is number of items in counting
λ is the mean rate
A) We want to find the probability that 5 messages are received in a given minute. Thus;
λ = 4 × 1 = 4
x = 5
Thus;
P(X = 5) = [tex]e^{-4}(\frac{4^{5}}{5!})[/tex]
P(X = 5) = 0.15629
P(X = 5) = 15.629%
B) We want to find the probability that 9 messages are received in 1.5 minutes. Thus;
λ = 4 × 1.5 = 6
x = 9
Thus;
P(X = 9) = [tex]e^{-6}(\frac{6^{9}}{9!})[/tex]
P(X = 9) = 0.06884
P(X = 9) = 6.884%
C) We want to find the probability that 3 messages are received in 30 seconds. 30 seconds is 0.5 minutes. Thus;
λ = 4 × 0.5 = 2
x = 3
Thus;
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
From online Poisson distribution calculator, we have;
P(X < 3) = 0.13534 + 0.27067 + 0.27067
P(X < 3) = 0.67668
P(X < 3) = 67.668%
Read more about Poisson distribution at; https://brainly.com/question/7879375
