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Dmitry_Shevchenko [17]
3 years ago
13

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 th

e 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?
Business
1 answer:
Dimas [21]3 years ago
3 0

Answer:

The probability will be "0.26938".

Explanation:

The given value is:

Mean

\mu = 1.8

By using Poisson probability formula, we get

⇒  P(X = x) = \frac{ (e-\mu\times \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

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