Random Sampling w/ Poisson Distribution?
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>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!
Favorite Answer
For lamda(mean)=3.8
P(x=0)=0.022
P(x>1)=1-p(x<=1) (same as 1-p(x=0)-p(x=1) =1.0-0.107=0.893 P(x<=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.
