After correcting 50 pages of a book, the proof reader finds that there are on the average, 2 errors per 5 pages. How many pages would one expect to find With 0,1,2,3 and 4 errors, in 1000 pages of the first print of the book?

Question

(8)

0.015

(Given 40.6703) (Fit a Poisson distribution

Answer 1

To find the number of pages expected to have 0 errors, we first calculate the average number of errors per page:

2 errors / 5 pages = 0.4 errors per page

Then, we use the Poisson distribution formula to find the probability of having 0 errors on a page:

P(X = 0) = (e^(-0.4) * 0.4^0) / 0!

P(X = 0) = e^(-0.4)

P(X = 0) = 0.6703

Now, for 1000 pages, we calculate the expected number of pages with 0 errors:

Expected pages with 0 errors = 1000 * 0.6703

Expected pages with 0 errors = 670.3

Similarly, for 1 error:
P(X = 1) = (e^(-0.4) * 0.4^1) / 1!

P(X = 1) = e^(-0.4) * 0.4

P(X = 1) = 0.2681

Expected pages with 1 error = 1000 * 0.2681 = 268.1

For 2 errors, P(X = 2) = (e^(-0.4) * 0.4^2) / 2!

P(X = 2) = e^(-0.4) * 0.4^2 / 2

P(X = 2) = 0.0670

Expected pages with 2 errors = 1000 * 0.0670 = 67

For 3 errors:
P(X = 3) = (e^(-0.4) * 0.4^3) / 3!

P(X = 3) = e^(-0.4) * 0.4^3 / 6

P(X = 3) = 0.0134

Expected pages with 3 errors = 1000 * 0.0134 = 13.4

For 4 errors:
P(X = 4) = (e^(-0.4) * 0.4^4) / 4!

P(X = 4) = e^(-0.4) * 0.4^4 / 24

P(X = 4) = 0.0020

Expected pages with 4 errors = 1000 * 0.0020 = 2

Therefore, the expected number of pages with 0, 1, 2, 3, and 4 errors in 1000 pages of the first print of the book are 670.3, 268.1, 67, 13.4, and 2 respectively.