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?

Respuesta :

Answer:

The probability will be "0.26938".

Explanation:

The given value is:

Mean

[tex]\mu = 1.8[/tex]

By using Poisson probability formula, we get

⇒  [tex]P(X = x) = \frac{ (e-\mu\times \mu x )}{x!}[/tex]

⇒  [tex]P(X > 2) = 1 - P(X \leq 2)[/tex]

⇒  [tex]1 - (P(X = 0) + P(X = 1) + P(X = 2))[/tex]

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

⇒  [tex]1 - 0.73062[/tex]

⇒  [tex]0.26938[/tex]