# Standard deviation of the mean of sample data

I can't quite understand what this formula means:

$$\sigma_{\overline{x}}=\frac{\sigma}{\sqrt n}$$

I know what standard deviation $\sigma$ is - it's the average distance of my data points (samples) from the mean. But this part is confusing:

For example, suppose the random variable $X$ records a randomly selected student's score on a national test, where the population distribution for the score is normal with mean $70$ and standard deviation $5$ ($N(70,5)$). Given a simple random sample (SRS) of $200$ students, the distribution of the sample mean score has mean $70$ and standard deviation $$\frac{5}{\sqrt{200}} \approx \frac{5}{14.14} \approx 0.35$$

Source

I thought the standard deviation $\sigma = 5$ means that if I take the scores of all students and calculate the mean, then the average distance of a score from that mean will be equal to $5$. The set of all scores is called the 'population', right? But here it says the more students' scores I take, the lower the standard deviation - thus the closer the number of samples gets to the size of population, the lower the standard deviation (and its get further from $5$).

First, the standard deviation is not the average distance to the mean, that is always zero. It is however, a value to measure how far the points are from the mean or not. Assuming the values are normally distributed, we know that 68% of the values are between $\mu-\sigma$ and $\mu+\sigma$, for example.

Suppose we weigh potatoes with average weight 100 g and stadard deviation 5 g. What does hold for the average of the average weight of a group of 4 potatoes? I hope you see that the average of the average weight is still 100 g. But what is the standard deviation of this average weight? That is where you use the formula:

$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{4}} =2.5$$

Feel free to ask if you still don't understand.

Proof that the average distance between the actual data and the mean is 0: $$\frac{\sum^n_{i=1} (x_i-\mu)}{n} = \frac{\sum^n_{i=1} x_i-\mu n}{n} = \frac{\sum^n_{i=1} x_i}{n}-\mu = \mu - \mu = 0$$

• 'The standard deviation is the average distance between the actual data and the mean.' - quote from: controls.engin.umich.edu/wiki/index.php/… – user4205580 May 3 '15 at 14:54
• Then that site is wrong. Note that the average distance between the actual data and the mean is 0. I will write why this in my answer, because I'm getting sick of the fact that pressing enter in a comment saves the comment. – wythagoras May 3 '15 at 14:55

You are computing the standard deviation of the mean, $\sigma_{\bar X}$, not that of the individual samples, $\sigma_X$.

When the variables are independent, the variances do add up. So $$\text{var}_{\sum_i Xi}=n\text{var}_X,$$ and dividing by $n^2$ (the variance is quadratic), $$\text{var}_{\bar X}=\frac1n\text{var}_X.$$

Hence taking the square root

$$\sigma_{\bar X}=\frac1{\sqrt n}\sigma_X.$$

• It's true assuming the variances of $x_i$ are equal. – user4205580 May 3 '15 at 18:25
• @user4205580: homoscedasticity is obvious from the problem statement, there is no need to trouble the OP. – Yves Daoust May 4 '15 at 7:30

Look carefully at the last sentence in the quote: in it 'standard deviation' refers to that of the sample mean. Thus one is essentially looking at all possible samples of 200 students, given that the population's standard deviation is 5. wythagoras' answer provides the formula for the sample mean's standard deviation.

• How to understand 'standard deviation of the mean of sample data'? That we take all possible samples of 200 students, calculate mean of every such set ($x_i$), calculate the mean $\overline{x}$ of the means $x_i$ and then calculate standard deviation using $x_i$ and $\overline{x}$? – user4205580 May 3 '15 at 15:00
• Yes, but in actually computing the formula you can extract the denominator of 200 from the sum of each sample mean, which saves you from having to deal with the mean of every such set (see the step involving dividing by $n^2$ in Yves Daoust's answer), which leads to a nice simple relation between the sample mean's standard deviation and the standard deviation of an individual's score. – Nigel May 3 '15 at 15:12