Respuesta :
Answer:
0.8987
Explanation:
Exponential random variable
P(X=x) = λ e^(-λ.x)
A mean of a request every 0.5 seconds translates to 0.5 seconds between each request.
Rate parameter = (1/0.5) = 2
λ = 2
5 requests in a 2-second time interval translates to (2/5) seconds between requests, that is, x = 0.40
P(X=x) = λ e^(-λ.x)
P(X=0.4) = 2 e^(-2×0.4)
P(X=0.4) = 2 e⁻⁰•⁸ = 2 × 0.4493 = 0.8987
Hope this Helps!!!
Answer:
Probability that there will be exactly 5 requests is 0.156
Explanation:
Poisson probability will be used to solve this problem
[tex]P(X=x) =\frac{\lambda^{x}e^{-\lambda} }{x!}[/tex]...............(1)
[tex]\lambda = \frac{2}{0.5} \\\lambda = 4[/tex]
Probability that there will be exactly 5 requests in a 2 second time interval
x = 5
Substituting the values of x and λ into equation (1)
P(X = 5) = [tex]\frac{4^{5} e^{-4} }{5!}[/tex]
P(X=5) = 0.156
