1 answer

The mean number of errors per page made by a member of the word processing pool...

Question:

The mean number of errors per page made by a member of the word processing pool for a large company is thought to be 1.8 with the number of errors distributed according to a Poisson distribution. If a page is​ examined, what is the probability that more than two errors will be​ observed?

The probability that more than two errors will be observed is ___________?


Answers

Solution :

Given that ,

mean = \mu = 1.8

Using poisson probability formula,

P(X = x) = (e-\mu * \mu x ) / x!

P(X > 2) = 1 - P(X \leq 2)

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

= 1 - (e-1.8 * 1.80) / 0! + e-1.8 * 1.81) / 1! + e-1.8 * 1.82) / 2! )

= 1 - 0.73062

= 0.26938

Probability = 0.26938

The probability that more than two errors will be observed is 0.26938 .

.

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