Question

). An expert statistical typist averages about one error for every four pages typed. If the number of errors follow a Poisson distribution, then (i). What is the average errors per page? ( 2 pts). (ii). What is the probability that the typist will make no errors at all typing the next page? ( 3 pts). (iii). What is the probability that the typist will make two errors typing the next page? ( 3 pts). (iv). What is the probability that the typist will make at most four errors typing the next page? ( 5 pts). (v). What is the probability that the typist will make between five and eight errors inclusive typing the next page? ( 5 pts). (vi). What is the expected number of errors to be made by the typist if he/she averages about 5 errors for every 10 pages typed.

Answer 1

The number of errors here could be modelled as:

that is an average of 0.25 error per page.

a) Therefore 0.25 is the average number of error per page.

b) Probability that the typist makes no error on the next page is computed here as:

Therefore 0.7788 is the required probability here.

c) The required probability here is computed as:

Therefore 0.0243 is the required probability here.

d) The required probability here is computed as:

Therefore approximate probability here is 1.

e) The probability here is computed as:

f) Expected number of errors per page here is computed as: = 5/10 = 0.5 error.