# How is the standard deviation in this coin tossing experiment calculated?

I'm reading Hypothesis Testing: The Basics, there is such an experiment:

So, we have a coin. Our null hypothesis is that this coin is fair. We flip it 100 times and it comes up heads 51 times. Do we know whether the coin is biased or not?

The author mentioned Central Limit Theorem and said the random variable is the proportion of heads in our sample of 100 coin flips. In our case, it is equal to 0.51.

But by the central limit theorem we also know that p approximates a normal distribution. This means we can estimate the standard deviation of p as $$\sigma=\sqrt{\frac{p(1-p)}{N}}$$

Why the standard deviation is calculated using this equation? According to Wikipedia, the std of a binomial distribution is $$\sqrt{np(1-p)}$$. What am I missing?

• The binomial is a count of successes; you're not looking at the count. Jan 18, 2019 at 9:47

If numerous random samples or repetitions of the same size $$n$$ are taken, the distribution of possible values of $$\hat{p}$$ is approximately a normal curve distribution with mean $$p$$ and $$SD=\sqrt{\frac{p(1-p)}{N}}$$. Have a look here. And, for the distinction from binomial distribution please have a look at this thread.
EDIT: To the explanation asked in the comment. $$\hat{p} = X/n$$, where $$X$$ is a binomial distribution with parameters $$n$$ and $$p$$. $$n$$ is the number of trials and $$p$$ is the probability of success. We are interested in the fraction of successes.
$$E(\hat{p}) = E(X/n) = np/n = p$$ and $$Var(\hat{p}) = Var(X/n) = Var(X)/n^2 = np(1-p)/n^2 = p(1-p)/n$$. So, $$SD=\sqrt{\frac{p(1-p)}{N}}$$
• Thank you for the answer. Could you please explain a bit more on why there is $\frac{1}{N}$ in the SD calculation? Jan 21, 2019 at 5:38
The variance for the Binomial distribution is in fact equal to $$np(1-p)$$, however you are estimating the variance for the proportion $$p$$, which is a Bernoulli trial, which has variance $$p(1-p)$$.