dan815
  • dan815
what is the expected standard deviation on a fair coin, for heads with respect to the number of trials? Like suppose there are 120 heads out of 200 flips, is that within the range of expected standard deviation or not? How can we determine how far off it is from the deviations and stuff?
Mathematics
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schrodinger
  • schrodinger
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taramgrant0543664
  • taramgrant0543664
I am really bad at trying to explain standard deviation so I tried to find you a link specific to coin flipping: http://www.stat.ucla.edu/~cochran/stat10/winter/lectures/lect8.html I hope this helps you!
dan815
  • dan815
okay, thank you! I will give it a read
dan815
  • dan815
I am interested in trying to find a simple relation between the number of samples and the expected standard deviation or standard error. I thought of trying to find a relationship with a brute force method at first. If you were to consider 2 flips of a coin, heads or tails. You can expect a big fluctuation in the results and it's intuitive, as well as the data shows. For example 2 out of the 4 cases would give you 1H and 1T, so there is a 50% chance of getting heads or tails. However 3 out of the 4 choices contain at least 1 Head, TH,HT, HH. So one can expect 25% of the time it will be no heads. Which is a huge standard deviation when compared proportionally to the expected value of heads. It seems intuitive to assume this effect is lessened as the number of trials increase, but I am trying to find a simple relationship to the standard deviation in proportion to the sample size in an efficient manner. Here are 2 main ways, I am considering, if anyone can help link these to the actual Standard Error formulas or find a more easy way to arrive at the closed form solutions to these ways, it would be great! Method 1 Binomial Distribution taking 1 STD to contain 68% of the possibilities let P(x) be the probability of x success for T trials \[ \sum_{k=0}^{N} P(x\pm k)=0.68\] solving for N gives the range on the standard deviation. for instance, in the example question of 200 trials, 100 +/- N would tell us the expected range for 1 STD. The probability of x success out of T trials where P is the rate of success is given by \[P(x) = P^x*(1-P)^{T-x} * \binom{T}{x}\\\] Suppose now x is taken to be the expected value for the number of success, Then we can see this relationship for 1 standard deviation. \[ \sum_{k=0}^{N} P(x\pm k)=0.68 \\ \sum_{k=0}^{N} P(x+k)=0.34\\ \sum_{k=0}^{N} P^{x+k}*(1-P)^{T-(x+k)} * \binom{T}{x+k} =0.34\] you can solve for N to see the standard deviation, instead of trying to find a closed form for this, I tried method 2 which is an approximate for it, I think.. Method 2 Normal Distribution Formula is given by \[P(x)=\frac {e^{1/2*(x-u/\sigma)^2}}{\sigma\sqrt{2\pi}}\] \[\sigma =STD\\x=successes \\u= expected. value \] This part im not sure about, because I am not sure I can exactly do this as there is technically no end value on a normal distribution graph but, suppose we can do: For T trials in total \[ \int_{0}^{T} P(x) dx= 1 \\ \int_{0}^{T}\frac {e^{1/2*(x-u/\sigma)^2}}{\sigma\sqrt{2\pi}} dx =1 \] and solving for the sigma with numerical methods

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dan815
  • dan815
I think a more legal way to use the normal distribution integral would be to integrate from x-N to x+N and set that integral equal to 0.68
dan815
  • dan815
@Michele_Laino
dan815
  • dan815
or the bounds on the integral X to X+infinity and the integral = 0.5
dan815
  • dan815
@ganeshie8
anonymous
  • anonymous
Hi dan, I think Alan Turing was interested in something similar when he came up with the Central Limit theorem.
dan815
  • dan815
Oh wow so thats what CLT was about, thank you for mentioning that, I'll read into it more
dan815
  • dan815
I knew I saw a neat derivation or formula for this question somewhere, I couldnt remember what it was related to
anonymous
  • anonymous
Note: You might also be interested in this relationship, Gaussian error function and normal distribution (see "Related functions" section of this article: https://en.wikipedia.org/wiki/Error_function/
ganeshie8
  • ganeshie8
@perl
ganeshie8
  • ganeshie8
@Miracrown
ganeshie8
  • ganeshie8
@ikram002p
ikram002p
  • ikram002p
standard deviation is what you got theoretically, in your case mean of p(head)=number of trials * P(h)=200*0.5=100 also standard of h will be 100 -.-
ikram002p
  • ikram002p
now your trials is 20 far from S.D so find standard deviation ERROR and find t test to check out it its allowed or not

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