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 2.4 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   .

​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 2.4

Using poisson probability formula,

P(X = x) = (e^- * ^x ) / x!

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

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

= 1 - (e^-2.4 * 2.4⁰) / 0! -  (e^-2.4 * 2.4¹⁾ / 1! - e^-2.4 * 2.4²⁾ / 2!

= 1 - 0.56971

= 0.43029

Probability = 0.4303

The probability that more than two errors will be observed is 0.4303