Respuesta :
Answer:
0.3216 = 32.16% probability that, in any seven-day week, the computer will crash less than 3 times
Step-by-step explanation:
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 time interval
Mean of 0.5
7-day week, so [tex]\mu = 7*0.5 = 3.5[/tex]
What is the probability that, in any seven-day week, the computer will crash less than 3 times?
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.5}*(3.5)^{0}}{(0)!} = 0.0302[/tex]
[tex]P(X = 1) = \frac{e^{-3.5}*(3.5)^{1}}{(1)!} = 0.1057[/tex]
[tex]P(X = 2) = \frac{e^{-3.5}*(3.5)^{2}}{(2)!} = 0.1850[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0302 + 0.1057 + 0.1857 = 0.3216[/tex]
0.3216 = 32.16% probability that, in any seven-day week, the computer will crash less than 3 times
