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Problem 106: A computing system manager states that the rate of interruptions to the internet service is 0.2 per week. Use the Poisson distribution to find the probability of a. one interruption in 3 weeks
b. at least two interruptions in 5 weeks
c. at most one interruption in 15 weeks

Problem 106: A computing system manager states that the rate of interruptions to the internet service is 0.2 per week. Use the Poisson distribution to find the probability of a. one interruption in 3 weeks
b. at least two interruptions in 5 weeks
c. at most one interruption in 15 weeks


Solution:


Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$

a. one interruption in 3 weeks
Given that $x=1, \lambda = \alphaT = 0.2\dot 3 = 0.6$
$f(1;0.6 )=\frac{{0.6}^{1}{e}^{-0.6}}{1!} = 0.329$


b. at least two interruptions in 5 weeks
Given that $x\ge 2, \lambda = \alphaT = 0.2dot 5 = 1$
$P(x\ge 2) = 1-P(x<2) = 1-[P(x=0)+P(x=1)]$
$P(x=0) = f(0;1)=\frac{{1}^{0}{e}^{-1 }}{0!} = 0.367879$
$P(x=1) = f(1;1)=\frac{{1}^{1}{e}^{-1 }}{1!} = 0.367879$
$P(x\ge 2) = 1-[P(x=0)+P(x=1)] = 1-(0.367879+0.367879) = 1-0.7357 = 0.264241.$

c. at most one interruption in 15 weeks
Given that $x\le 1, \lambda = \alphaT = 0.2dot 15 = 3$
$P(x\le 1) = [P(x=0)+P(x=1)]$
$P(x=0) = f(0;3)=\frac{{3}^{0}{e}^{-3 }}{0!} = 0.049787$
$P(x=1) = f(1;3)=\frac{{3}^{1}{e}^{-3 }}{1!} = 0.149361$
$P(x\le 1) = P(x=0)+P(x=1) = 0.049787+0.149361 =0.199.$

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