Question

Assume that the number of network errors experienced in a day on a local area network (LAN) is distributed as a Poisson random variable. The mean number of network errors experienced in a day is 2.5 Complete parts (a) through (d) below.

A. What is the probability that in any given day zero network errors will occur?

B. What is the probability that in any given day exactly one network error will occur?

C. What is the probability that in any given day two or more network errors will occur?

D. What is the probability that in any given day fewer than three network errors will occur?

Answer #1

µ = 2.5

This is a poisson distribution.

P(X = x) = e^{-µ} * µ^{x} / x!

a) P(X = 0) = e^{-2.5} * 2.5^{0} / 0! =
0.0821

b) P(X = 1) = e^{-2.5} * 2.5^{1} / 1! =
0.2052

c) P(X > 2) = 1 - (P(X = 0) + P(X = 1))

= 1 - (e^{-2.5} * 2.5^{0} / 0! + e^{-2.5} *
2.5^{1} / 1!)

= 1 - 0.2873

= 0.7127

d) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= e^{-2.5} * 2.5^{0} / 0! + e^{-2.5} *
2.5^{1} / 1! + e^{-2.5} * 2.5^{2} / 2!

= 0.5438

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