Standard Deviation Question

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I can't locate it just now but I remember blackjack hero Arnold Snyder giving a very simple method of calculating Standard Deviation
from googling I found out that trying to calculate it can be quite complicated

I would like to know if this is correct - what I remember


if you flip a coin 100 times and it ends up 60 heads and 40 tails then:

it varied from 50/ 50 by 20

to get the standard deviation of this series of 100 flips you get the square root of 20 - which is 4.47

you then divide that by 2 to get 2.235.............................and that is the standard deviation of the results of your 100 coin flip series if the results are 60/40 - if not exactly - a very good approximation


is this correct?



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Quote: lilredrooster

................................


I can't locate it just now but I remember blackjack hero Arnold Snyder giving a very simple method of calculating Standard Deviation
from googling I found out that trying to calculate it can be quite complicated

I would like to know if this is correct - what I remember


if you flip a coin 100 times and it ends up 60 heads and 40 tails then:

it varied from 50/ 50 by 20

to get the standard deviation of this series of 100 flips you get the square root of 20 - which is 4.47

you then divide that by 2 to get 2.235.............................and that is the standard deviation of the results of your 100 coin flip series if the results are 60/40 - if not exactly - a very good approximation


is this correct?



*

The standard deviation for the proportion of heads in one hundred fair-coin spins is: 0.5 / sqrt(100), which is 0.05.

The mean, or average, number of heads for one spin is 0.5. And 0.5 is the standard deviation for one spin because 0 heads is 0.5 head below the mean, and 1 head is 0.5 head above the mean.

In your example of 60 heads and 40 tails, 60/100 = 0.6, which is 0.1 above the mean. This corresponds to 0.1 / 0.05 = 2 standard deviations above the mean.

If we approximate 100 spins with the “normal distribution,” we would figure the probability of having at least 60% heads, or 2 SDs above the mean, is about 2.5%.

So, be sure to use 0.5 (not 2) for the standard deviation for one spin, and to use the square root of the number of observations (not the excess heads.)

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thanks ChesterDog

The average is np and the standard deviation = SQRT(npq). For a coin flip p=q=1/2.

So 60 heads from 100 tosses is 2 SDs from the norm. Av = 100*(1/2) = 50. SD = SQRT ( 100 * 1/2 * 1/2 ) = 5.

https://ned.ipac.caltech.edu/level5/Berg/Berg3.html