Question

The number of typing errors made by a typist has a Poisson distribution with an average of seven errors per page. If more than seven 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

Answer:

0.599  is the probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

We are given the following in the question:

The number of typing errors made by a typist has a Poisson distribution.

$\lambda =7$

The probability is given by:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

We have to find the probability that a  randomly selected page does not need to be retyped

P(less than or equal to 7 mistakes in a page)

$P( x \leq 7) = P(x =1) + P(x=1) +...+ P(x=6) + P(x = 7)\\\\= \displaystyle\frac{7^0 e^{-7}}{0!} + \displaystyle\frac{7^1 e^{-7}}{1!} +...+ \displaystyle\frac{7^6 e^{-7}}{6!} + \displaystyle\frac{7^7 e^{-7}}{7!} \\\\= 0.59871\approx 0.599$

Thus, 0.599  is the probability that a randomly selected page does not need to be retyped.