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

The probability will be "0.26938".

Step-by-step explanation:

The given value is:

Mean

`\mu = 1.8`

By using Poisson probability formula, we get

⇒
`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`