Poisson Normal Approximation Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities: P(X<64) so, P(X<64) = P( Z < 64-64/(64)^(1/2)) = P(Z < 0) = 0.5 The text book says the answer is 0.4761 what am I doing wrong??
It seems like every answer I get is slightly off from what the textbook answer is.
Sir/Ma'am please show your Working for the question you have posted.
I did
P(X<64) = P( Z < 64-64/(64)^(1/2)) = P(Z < 0) = 0.5
there it is again
are we trying to approximate it with a binomial?
no this is a Poisson distribution approximation
oops sorry for poorly worded question
http://www.stattler.com/article/poisson-approximation-binomial-probabilities im reading this for a refresher
that is the table im using to get my approximations of the normal distribution, not that it is needed for the question because P(Z<0) = 0.5
The textbook is saying, P(X<64) = P(Z<-0.06)
All my answer are like a few decimals off every time.
it may be that your using a normal distribution instead of a binomial distribution ... but thats just a thought at the moment
are you wanting to approx it with a binomial?
No this question is a poisson distribution approximation using the normal distribution
nothing to do with binomials, I edited my question to avoid confusion sorry again about that
use a continuity correction
oh yeah but why? Isn't it greater than?
there is no equality so why do I have to correct?
you are going from a discrete distribution to a continuous one
So you always have to correct?
you should
Ok thank you so much, sorry for the bother, I should have tried correcting before asking this question ugh
@Zarkon it says right in the textbook, To correct P(X<x) = P(Z<x+0.5) To Correct P(X>x) = P(Z>x - 0.5) For this problem, P(X<64), therefore, P(X<64) = P(X<64+0.5) = P(Z<0.063) = 0.532922 SO the text book is in fact wrong.
I'm pretty frustrated atm ugh
\[P(X<64) =P(X\le 63)=P(X\le 63+.5)=P(X\le63.5)=P(Z\le -.0625)\]
Ok I think I finally get this stuff
good
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