Question

The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than three errors appear on a given page, the typist must retype the whole page. What is the probability that a randomly selected page does not need to be retyped

Answer 1

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

Explanation:

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

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given interval.

Poisson distribution with an average of three errors per page

This means that
`\mu = 3`

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)`

In which

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

`P(X = 0) = (e^(-3)*3^(0))/((0)!) = 0.0498`

`P(X = 1) = (e^(-3)*3^(1))/((1)!) = 0.1494`

`P(X = 2) = (e^(-3)*3^(2))/((2)!) = 0.2240`

`P(X = 3) = (e^(-3)*3^(3))/((3)!) = 0.2240`

Then

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472`

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.