Anyone good with Statistics and Probabilities?
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I'm a clinical trial statistician... Mostly do FDA stuff for pharma these days, but it's good to dust off the mental gear and do some hand calculations every now and then.Gallo Ref 3.1 : Bryston 4b SST : Musical fidelity CD Pre : VPI HW-19
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UNC, would you like to do another one? It will be the last one, I promise. I think it involves using Poisson. I am doing fine in the other areas of my Stats course so far, its just these probability parts.
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Sure, throw it out there.Gallo Ref 3.1 : Bryston 4b SST : Musical fidelity CD Pre : VPI HW-19
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Okay, here it is.
Records show that there is an average of three accidents each day in a certain city between 2 and 4 pm. Assume the accidents are independant.
a) Find the probability that there will be exactly one accident between 2 and 3 pm on a particular day.
b) Find the probability that there will be exactly one accident between 2 and 4 pm on a particular day.
I know that this uses Poisson. Because in class we have not covered any other way of solving this. -
Poisson has just one parameter: lambda. This is the average number of events per time period. Technically part A can't be solved since you don't know anything about 2-3pm vs 3-4pm... but we'll run with this anyway.
Sooooooooo. p(x)=[(lambda^x) *(e^-lambda)]/x!
For part B, lambda=3, so you get [(3^1)*(e^-3)]/1!
For part A, you should say that you can't assume anything about the distribution of accidents within the time period, but then calculate it using lamba=1.5Gallo Ref 3.1 : Bryston 4b SST : Musical fidelity CD Pre : VPI HW-19
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So for
part a, if I have 3 accidents in a 2 hour period, then the average is 1.5 per hour, if solved, it should look something like
t=time=x=1 for the time interval, u=avg=lambda=3/2=1.5
P(x = 1) = [(3/2)^1 * (e^-3)] / 1! = .0747 or 7.47 percent
part b, this time x = 2 for time interval, so lambda = 3
I get .2240 or 22.40 percent. Is this correct? I got a zero grade for this. -
No, X is the number of events that you want to know the probability of. The two formulas will be:
A)
[(1.5^1)*(e^-1.5)]/1! = 33.4%
[(3^1)*(e^-3)]/1! =14.9%Gallo Ref 3.1 : Bryston 4b SST : Musical fidelity CD Pre : VPI HW-19
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oh my.. I see where I am messing up.
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You all lost me at Hello.
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Might want to reel in the ego there, son. Three things here:
1)It's an extremely basic problem
2) The _exactly_ part makes it easier, not harder
3) It's generally bad form to just give out the answer- make him solve something so he'll remember it when test time comes around.
stebesplace- you were doing it the hard way, but if you worked through the expected value and did 0-100 you would have gotten the same answer- just with a LOT more work.
How is this extremely basic if you have to use combinations? An easy probability problem would be something like what are the odds you roll a 6 three times in a row. I gave all of the formulas and defined terms, isn't that the same as giving the answer. Of course he has to practice other problems and actual calculations to prepare for a test. I was just trying to help and joking around with a liar liar quote, no need to get all serious. who put bees in your bonnet?HT Setup: Onkyo 706; Rotel RB-1075; Rti A3; Csi A6; Fxi3's
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My view about statistics (I slept through most of it in college):
Everything is a 50% chance...it either happens or it doesn't!
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