Anewspaper finds that the mean number of typographical errors per page is fourfour. find the probability that (a) exactly fivefive typographical errors are found on a page, (b) at most fivefive typographical errors are found on a page, and (c) more than fivefive typographical errors are found on a page.

Using the Poisson distribution, it is found that there is a:

a) 0.1563 = 15.63% probability that exactly five typographical errors are found on a page.b) 0.7852 = 78.52% probability that at most five typographical errors are found on a page.c) 0.2148 = 21.48% probability that more than five typographical errors are found on a page.

We have only the mean, hence the Poisson distribution is used to solve this question.

Poisson distributionIn a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

The parameters are:

x is the number of successese = 2.71828 is the Euler number is the mean in the given interval.

In this problem, the mean is of 4, hence .

Item a:

This is P(X = 5), hence:

0.1563 = 15.63% probability that exactly five typographical errors are found on a page.

Item b:

This is:

Hence:

Then:

0.7852 = 78.52% probability that at most five typographical errors are found on a page.

Item c:

0.2148 = 21.48% probability that more than five typographical errors are found on a page.

You can learn more about the Poisson distribution at link

For this question, we are given a mean/average rate, that is 4 errors per page;
This suggests we need to use the Poisson distribution, so:
Let X be the random variable, the number of errors per page;
X~Po(4) ← X has a Poisson distribution with mean (λ) of 4
a)
What we want is the probability that X = 5, so we use the formula for Poisson:

= 0.156 (to 3 s.f.)
b)
What we want to find is the probability that X ≤ 5, so we have to use the cumulative tables:
P(X ≤ 5) = 0.7851
c)
What we want to find is the probability that X > 5, so we have to do a bit of manipulation and then use the cumulative tables;
The tables give values for ≤, so we cannot directly look up a value for >, thus the manipulation:
P(X > 5) = 1 - P(X ≤ 5)
= 1 - (0.7581)
= 0.2419