# Two different standard deviations?

I'm reading this book and on page 419 the authors state the following theorem: I'm really confused. What is the standard deviation? $$\sigma$$ or $$\sigma/\sqrt{n}$$?

• Interesting. My understanding is that the equation, $\frac{\sigma}{\sqrt{n}}$, is the standard error. Sep 21 '20 at 3:30

The first distribution is the population, which has a mean $$\mu$$ and a standard deviation $$\sigma$$. Let's assume the height of all humans is normal distributed, with $$(\mu, \sigma) = (1.5, 0.5)$$ in metres.
From that population, you draw one sample of size $$n$$, and compute its mean -- that's called the sample mean. Say you measure the height of $$n=100$$ people, and get a mean height for that one sample of 1.57.
Repeat many times: you'll get a distribution of sample means, which is the second distribution -- and that has a mean of $$\mu$$ and a standard deviation of $$\sigma / \sqrt n$$.
Standard deviation of what? It really depends on the random variable you consider. With that being said, let $$X_i\stackrel{\text{i.i.d.}}{\sim}(\mu,\sigma^2)$$ for $$i=1,\ldots,n$$. Now $$\sigma$$ is obviously the standard deviation of the random variables $$X_1,\ldots,X_n$$, whereas $$\frac{\sigma}{\sqrt{n}}$$ denotes the standard deviation of the sample mean $$\overline{X}=\frac{1}{n}\sum_{i=1}^n X_i$$.