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? Round your answer to four decimal places (e.g. 98.7654).

Answer 1

Answer:

Therefore the required probability is 0.6472.

Step-by-step explanation:

Poisson distribution: A Poisson distribution is discrete distribution.

Let X be a discrete variable the number of event. Let λ be the mean of X.

Then the probability of k event is

$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

Here mean of each page is =0.03

Mean of 100 pages = (0.03×100)= 3

λ = 3

The required probability

P(X≤3)  

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

$=\frac{3^0e^{-3}}{0!}$ $+\frac{3^1e^{-3}}{1!}$$+\frac{3^2e^{-3}}{2!}$$+\frac{3^3e^{-3}}{3!}$

$=e^{-3}(1+3+\frac{9}{2}+\frac{27}{6})$

$=e^{-3}(13)$

≈ 0.6472

Therefore the required probability is 0.6472