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 May 13th, 2012, 02:12 PM #1 Newbie   Joined: Mar 2012 Posts: 12 Thanks: 0 Poisson Distribution - find mean Hi, have this problem i need to solve. The number of clients that go to a store per hour uses a Poisson distribution. It's known that the probability of arriving 15 customers in an hour is 0.031 and, also the probability of arriving 16 customers in an hour is 0.062. What is the mean of arriving customers per hour in the store. I'm having trouble interpreting the problem. I don't know what is given. If it's P(X=15)=0,031, or P(X <=15)=0,031 or P(X >= 15)=0,031 Any help appreciated. Thanks.
May 13th, 2012, 04:29 PM   #2
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Re: Poisson Distribution - find mean

Quote:
 Originally Posted by mdroid If it's P(X=15)=0,031
It's this one. There aren't any words like "more than" or "at least" or "no more than" or anything like that which would tell you that it's anything but an equal. They're saying the probably that "15 people" arrive.

So we're saying $.031\=\ \frac{\mu e^{-\mu}}{15!}$

(Now if you need help with the equation, you'll need someone else cause I have no idea how to solve for the mu. We always just used the charts. I looked on the chart, but never found 15 & 16 together matching. So unless you have a much bigger chart, I'm assuming you're using the equation.)

May 13th, 2012, 06:10 PM   #3
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Re: Poisson Distribution - find mean

Quote:
 Originally Posted by Erimess So we're saying $.031\=\ \frac{\mu e^{-\mu}}{15!}$
I believe it's $.031\=\ \frac{\mu^{15} e^{-\mu}}{15!}$

 May 13th, 2012, 06:12 PM #4 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Poisson Distribution - find mean First, note that the probability of We're told that $0.031= \frac{\mu^{15} e^{-\mu}}{15!}$ (1) and $0.062=\frac{\mu^{16} e^{-\mu}}{16!}$ (2) and we need to solve for $\mu$. From 2, we can rewrite the equation to get $0.062=\frac{\mu^{15} . \mu e^{-\mu}}{16.15!}$ $0.062= \frac{\mu^{15} e^{-\mu}}{15!}.\frac{\mu}{16}$ (3) Substitute (1) into (3) yields $0.062=0.031.\frac{\mu}{16}$ $\mu=\frac{0.062\times16}{0.031}=32$ Edit: Ah! You didn't ask for this (how to solve for the mean)!
 May 14th, 2012, 12:35 PM #5 Newbie   Joined: Mar 2012 Posts: 12 Thanks: 0 Re: Poisson Distribution - find mean Yes, i just wanted help interpreting the problem... but thanks.
May 15th, 2012, 12:10 PM   #6
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Re: Poisson Distribution - find mean

Quote:
Originally Posted by Isbell
Quote:
 Originally Posted by Erimess So we're saying $.031\=\ \frac{\mu e^{-\mu}}{15!}$
I believe it's $.031\=\ \frac{\mu^{15} e^{-\mu}}{15!}$
Sorry, you're right. I didn't transfer from paper to Latex very well.

May 15th, 2012, 08:29 PM   #7
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Re: Poisson Distribution - find mean

Quote:
 Originally Posted by Erimess Sorry, you're right. I didn't transfer from paper to Latex very well.
Got it and no worries, OK?

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