Anyone good with Statistics and Probabilities?

unc2701

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.

[Deleted User]

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.

unc2701

Sure, throw it out there.

[Deleted User]

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.

unc2701

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=1.5

[Deleted User]

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.

unc2701

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%

[Deleted User]

oh my.. I see where I am messing up.

avelanchefan

You all lost me at Hello.

cdn5003

unc2701 wrote: »

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?

croutons

My view about statistics (I slept through most of it in college):

Everything is a 50% chance...it either happens or it doesn't!