Random Sampling w/ Poisson Distribution?

Hi, I have this problem for my statistics class and I can't figure it out. I have the answer, but I don't understand. If you can help, that would be awesome!!





Scientists at a university designed an experiment to measure x, the number of times a reader’s eye fixated on a word before moving past that word. X was found to have a mean of 3.8. Suppose one of the readers in the experiment is randomly selected and assume that x has a Poisson distribution.





a. P ( x=0) =


b. P (x%26gt;1) =


c. P (x less than or equal to 2) =





I used the Poisson formula, plugging in x, but it's not working.





Answers I got:


a. .0224


b. 0850


c. 7.3212





The right answers:


a. .050


b. .801


c. .423





What am I doing wrong?





Thanks!!! :) 10 points for Best Answer!

Random Sampling w/ Poisson Distribution?
Poisson distribution


For lamda(mean)=3.8


P(x=0)=0.022


P(x%26gt;1)=1-p(x%26lt;=1) (same as 1-p(x=0)-p(x=1)


=1.0-0.107=0.893


P(x%26lt;=2)=p(x=0)+p(x=1)+p(x=2)=0.731


The following tables were used. You'll need to download Adobe Reader to calculate the cumulative probabilities. If you're using a different table, you'll get the same results.


The results are different from the right answers.


Your answer 7.3212 is incorrect as the probability cannot exceed 1.

garden centre