Question

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?

Answer 1

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!

Answer 2

okay, thank you! I will give it a read

Answer 3

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

Answer 4

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

Answer 5

@Michele_Laino

Answer 6

or the bounds on the integral X to X+infinity and the integral = 0.5

Answer 7

@ganeshie8

Answer 8

Hi dan, I think Alan Turing was interested in something similar when he came up with the Central Limit theorem.

Answer 9

Oh wow so thats what CLT was about, thank you for mentioning that, I'll read into it more

Answer 10

I knew I saw a neat derivation or formula for this question somewhere, I couldnt remember what it was related to

Answer 11

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/

Answer 12

@perl

Answer 13

@Miracrown

Answer 14

@ikram002p

Answer 15

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 -.-

Answer 16

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