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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>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!

Top 1 Answers

Favorite Answer

Poisson distribution

For lamda(mean)=3.8


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.


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