Question

The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability that there are 3 or less errors in 100 pages?

Answer 1

Answer: the probability that there are 3 or less errors in 100 pages is

0.65

Explanation:

The formula for poisson distribution is expressed as

P(x = r) = (e^- µ × µ^r)/r!

Where

µ represents the mean of the theoretical distribution.

r represents the number of successes of the event.

From the information given,

µ = 0.03 errors per page

Therefore, in 100 pages,

µ = 0.03 × 100 = 3 errors per page

For the probability that there are 3 or less errors in 100 pages, it is expressed as

P(x ≤ 3) = P(x = 0) + P(x = 1) + P( x = 2) + P(x = 3)

Therefore,

P(x = 0) = (e^- 3 × 3^0)/0! = 0.0497

P(x = 1) = (e^- 3 × 3^1)/1! = 0.149

P(x = 2) = (e^- 3 × 3^2)/2! = 0.224

P(x = 3) = (e^- 3 × 3^3)/3! = 0.224

P(x ≤ 3) = 0.0497 + 0.149 + 0.224 + 0.224 = 0.65

:

Answer 2

The probability that there are 3 or less errors in 100 pages is 0.648.

Explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is,
`\lambda` = E(x)

= n × p

= 100 × 0.03

= 3

Therefore the we find the probability that there are 3 or less errors on the page as

P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

Using the formula for Poisson distribution for P(x = X ) =
`(e^(-\lambda)\lambda^X)/(X!)`

Therefore P( X ≤ 3) =
`(e^(-3) 3^0)/(0!) + (e^(-3) 3^1)/(1!) + (e^(-3) 3^2)/(2!) + (e^(-3) 3^3)/(3!)`

= 0.05 + 0.15 + 0.224 + 0.224

= 0.648

The probability that there are 3 or less errors in 100 pages is 0.648.