Question

The number of typing errors made by a typist has a Poisson distribution with an average of six errors per page. If more than six 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? (Round your answer to three decimal places.)

Answer 1

Answer:

The probability that a randomly selected page does not need to be retyped is P(x≤6)=0.607.

Step-by-step explanation:

We have a Poisson distribution with a parameter that is equal to the average of six errors per page:

$\lambda=6$

Then, the probability of having k errors per page can be calculated as:

$P(x=k)=6^{k} \cdot e^{-6}/k!$

Then, the probability of a randomly selected page does not need to be retyped (6 error or less) is:

$P(x\leq6)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\P(3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\P(4)=6^{4} \cdot e^{-6}/4!=1296*0.0025/24=0.134\\\\P(5)=6^{5} \cdot e^{-6}/5!=7776*0.0025/120=0.161\\\\P(6)=6^{6} \cdot e^{-6}/6!=46656*0.0025/720=0.161\\\\\\P(x\leq6)=0.002+0.015+0.045+0.089+0.134+0.161+0.161\\\\P(x\leq6)=0.607$