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.
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