Respuesta :
Answer:
0.2103 = 21.03% 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 interval.
Mean of 0.6 times a day
7 day week, so [tex]\mu = 7*0.6 = 4.2[/tex]
What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.
[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^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150[/tex]
[tex]P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630[/tex]
[tex]P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323[/tex]
So
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103[/tex]
0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.
