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Mathematics 21 Online
OpenStudy (australopithecus):

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??

OpenStudy (australopithecus):

It seems like every answer I get is slightly off from what the textbook answer is.

OpenStudy (goformit100):

Sir/Ma'am please show your Working for the question you have posted.

OpenStudy (australopithecus):

I did

OpenStudy (australopithecus):

P(X<64) = P( Z < 64-64/(64)^(1/2)) = P(Z < 0) = 0.5

OpenStudy (australopithecus):

there it is again

OpenStudy (amistre64):

are we trying to approximate it with a binomial?

OpenStudy (australopithecus):

no this is a Poisson distribution approximation

OpenStudy (australopithecus):

oops sorry for poorly worded question

OpenStudy (amistre64):

http://www.stattler.com/article/poisson-approximation-binomial-probabilities im reading this for a refresher

OpenStudy (australopithecus):

OpenStudy (australopithecus):

OpenStudy (australopithecus):

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

OpenStudy (australopithecus):

The textbook is saying, P(X<64) = P(Z<-0.06)

OpenStudy (australopithecus):

All my answer are like a few decimals off every time.

OpenStudy (amistre64):

it may be that your using a normal distribution instead of a binomial distribution ... but thats just a thought at the moment

OpenStudy (amistre64):

are you wanting to approx it with a binomial?

OpenStudy (australopithecus):

No this question is a poisson distribution approximation using the normal distribution

OpenStudy (australopithecus):

nothing to do with binomials, I edited my question to avoid confusion sorry again about that

OpenStudy (zarkon):

use a continuity correction

OpenStudy (australopithecus):

oh yeah but why? Isn't it greater than?

OpenStudy (australopithecus):

there is no equality so why do I have to correct?

OpenStudy (zarkon):

you are going from a discrete distribution to a continuous one

OpenStudy (australopithecus):

So you always have to correct?

OpenStudy (zarkon):

you should

OpenStudy (australopithecus):

Ok thank you so much, sorry for the bother, I should have tried correcting before asking this question ugh

OpenStudy (australopithecus):

@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.

OpenStudy (australopithecus):

OpenStudy (australopithecus):

I'm pretty frustrated atm ugh

OpenStudy (zarkon):

\[P(X<64) =P(X\le 63)=P(X\le 63+.5)=P(X\le63.5)=P(Z\le -.0625)\]

OpenStudy (australopithecus):

Ok I think I finally get this stuff

OpenStudy (zarkon):

good

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