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?

Answer 1

Answer:

The probability will be "0.26938".

Explanation:

The given value is:

Mean

$\mu = 1.8$

By using Poisson probability formula, we get

⇒  $P(X = x) = \frac{ (e-\mu\times \mu x )}{x!}$

⇒  $P(X > 2) = 1 - P(X \leq 2)$

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

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

⇒  $1 - 0.73062$

⇒  $0.26938$