
#1
Dec2712, 03:05 AM

P: 24

Alright my brother posed this question to me tonight. You have 7 coins (for this sake we will go with the "normal" coins, penny, nickel, dime and quarter) and we are trying to find the probability that a person has over a dollar in coins.
So basically I did it and just wanted to confirm my answer before telling him tomorrow morning. You just do: E(x)=((1*(1/4))+(5*(1/4))+(10*(1/4))+(25*(1/4))) E(x)= 41/5 = 10.25 7*E(x)=71.75 Var(x)=E(x^2)E(x)^2 Var(x)=187.75105.0625 Var(x)=82.6875 7*Var(x)=578.8125 Now I believe that I can do normal Z score stuff to find probability with the continuity correction. So we are looking for P(Z≥100) so we'll flip it into 1P(Z≤100). Then we will add the continuity correction to make it 1P(Z>100.5) and calculate it from there. Am I right so far? So I go: (71.75100.5)/(24.058) Z≈1.195 And thus, using R normal probability calculator we are looking at a final answer of .1160379. Did I do that all right? Thanks! 



#2
Dec2712, 03:22 AM

P: 4,570

Hey bloynoys.
I used R on the assumption that E[X] and Var[X} where calculated correctly (the method you used is good) and assuming the normality condition holds (which it hopefully should for 7 observations) then R gave me: > 1  pnorm(100,71.75,24.058) [1] 0.1201483 However I would suggest you use continuity correction since you have discrete units instead of continuous ones (in terms of 1, 5, 25, etc units). Are you familiar with continuity correction? 



#3
Dec2712, 03:25 AM

P: 24

Yeah I tried to apply the continuity correction to the z calculation, so it would basically move the mean to 100.5 which is what gave me the answer that I had in the first post. Does that look right to you?




#4
Dec2712, 03:39 AM

P: 4,570

7 coins  probability over a dollar
It does, but the only thing I am thinking about is whether the distribution isn't skewed enough to make it normal.
One way you could check this is to simulate the distribution of the sum and see if its normal. It might sound anal, but it's just a nice thing to do for peace of mind. 



#5
Dec2712, 03:48 AM

P: 24

Is there an easy way to simulate this in R or similar? It would be interesting to see what kind of distribution it comes out as. I would hope the pennies are able to "normal" it out but hard to know.




#6
Dec2712, 04:05 AM

P: 4,570

Are you aware of sampling techniques like EM or Metropolis Hastings?




#7
Dec2712, 04:28 AM

P: 24

I am not. I am about to start my final semester of undergraduate stats education (have stats 2, stochastic processes and multivariate analysis this semester) but so far we haven't gotten to things like that. Any reading or hints you would suggest as I try to learn about these things? Not a big deal in terms of this question but seems like a good skill to understand going forward. They seem very cool from a quick perusing of the wikis.




#8
Dec2712, 04:32 AM

P: 4,570

The wiki pages outline the algorithms:
http://en.wikipedia.org/wiki/Rejection_sampling http://en.wikipedia.org/wiki/Metropo...ings_algorithm 



#9
Dec2712, 12:36 PM

P: 24

Alright, I wanted to check this fast so what I did is just write a simple program in Java to generate random numbers between 1 and 4 and then if statements to create the cent amount after 7 coins. I ran that 50000 times and got this histogram. Do you think this is approximately normal? It doesn't look entirely great and maybe a slight skew right, what do you think?




#10
Dec2712, 12:47 PM

P: 24

Yeah pretty clear it isn't close enough to normal. From just checking how many coin amounts are over 100 a few times it is always between 8 and 9 percent pretty far off from the 1112 that it would be with the normal distribution.




#11
Dec2712, 05:40 PM

P: 4,570

You might want to consider using a distribution with a heavier tail or simulate say 50,000 times and use the generated empirical distribution (i.e. the one you just simulated).




#12
Dec2712, 07:36 PM

Mentor
P: 14,476

In this case, a brute force approach works quite nicely. There are 4^{7} or 16384 equiprobable outcomes, so just check each one. Here's a quick and dirty perl script:




#13
Dec2812, 10:26 PM

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#14
Dec2812, 10:27 PM

P: 4,570





#15
Dec3112, 01:37 PM

P: 8

Someone (haruspex) on this thread mentioned the right thing. In the real world, those coins are not equally probable. Guess it all depends on whether the coins are selected at random or are they real "change" from transactions.




#16
Jan113, 06:16 AM

Mentor
P: 14,476

You are missing a key point by focusing on realism. That point is that a normal distribution is not a good model for this process. 



#17
Jan213, 03:02 AM

P: 523





#18
Jan413, 12:32 AM

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