I'm having difficulty in determining what exactly the difference is between the 2, especially when given an exercise and I have to choose which of the 2 to use. These is how my text book describes them:

Sum standard deviation

Given is a population with a normally distributed random variable $X$. When you have a sample $n$ from this population the population is:

$X_{sum} = X_1 + X_2 ... + X_n$ with

$\mu_{Xsum} = n \times \mu_x$ and $\sigma_{Xsum} = \sqrt{n} \times \sigma_x$.

Standard error

When you have a normally distributed random variable $X$ with mean $\mu_X$ and standard deviation $\sigma_X$ and sample length $n$, the sample mean $\bar{X}$ is normally distributed with $\mu_{\bar{x}} = \mu_X$ and $\sigma_{\bar{x}} = \dfrac{\sigma_X}{\sqrt{n}}$

These 2 are awefully similair to me to the point I can't at all decide which to use where. Here are the problems where I discovered I couldn't:

Problem 1

A filling machine fills bottles of lemonade. The amount is normally distributed with $\mu = 102 \space cl$.

$\sigma$ = $1.93\space cl$.

  • Calculate the chance that out of 12 bottles the average volume is $100 \space cl$.

The problem itself is easy, however the troublesome part is what to choose for the standard deviation of the sample. Here they use $\dfrac{1.93}{\sqrt{12}} $ which I can live with, until I encountered the second problem.

Problem 2

A tea company puts 20 teabags in one package. The weight of a teabag is normally distributed with $\mu = 5.3 \space g$ and $\sigma = 0.5 \space g.$

  • Calculate the chance that a package weighs less than 100 grams.

Here I thought they'd also use $\dfrac{0.5}{\sqrt{20}}$, but instead they use $\sqrt{20} \times 0.5$.

Can someone clear up the confusion?

  • $\begingroup$ You should tag this as "homework" as well, since it seems to be a homework question. $\endgroup$
    – Placidia
    Jan 20, 2013 at 17:33
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    $\begingroup$ @Placidia Are you kidding me?! This isn't homework, this is about understanding and differentiating 2 general concepts in statistics, which then could be implemented in homework questions.. like every other mathematical concept.. $\endgroup$ Jan 20, 2013 at 17:36
  • $\begingroup$ My textbook confused me; (Freedman, Pisani, Purves, Statistics, Fourth Edition). The chapter is titled "The Standard Error", uses $\sqrt{n} \times \sigma_x$ but the acronym is "SE", which must be a hint that my textbook is referring to "sum standard deviation"; indeed the textbook describes the experiment: "When drawing at random with replacement from a box of numbered tickets; the standard error for the sum of the draws is..." $\endgroup$ Nov 7, 2019 at 7:26
  • $\begingroup$ ... more from the confusing part of Statistics textbook Freedman, Pisani, Purves, Fourth Edition: "In this book, we use SD for data and SE for chance quantities (random variables). This distinction is not standard and the term SD is often used in both situations" Indeed, as in this SE question, we refer to the "SD of X" ($\sigma_{X}$) and the "SD of sum of X" ($\sigma_{X_{sum}}$) $\endgroup$ Nov 7, 2019 at 7:37

2 Answers 2


The sum standard deviation is, as the name suggests, the standard deviation of the sum of $n$ random variables. The standard error you're talking about is just another name for the standard deviation of the mean of $n$ random variables. As you noted, the two formulas are closely related; since the sum of $n$ random variables is $n$ times the mean of $n$ random variables, the standard deviation of the sum is also $n$ times the standard deviation of the mean:

$\sigma_{X_{sum}} = \sqrt n\sigma_X = n \times \frac{\sigma_X}{\sqrt n} = n\times \sigma_\bar{X}$.

In the first problem you are dealing with a mean, the average of twelve bottles, so you use the standard deviation of the mean, which is called standard error. In the second problem you are dealing with a sum, the total weight of 20 packages, so you use the standard deviation of the sum.

Summary: use standard error when dealing with the mean (averages); use sum standard deviation when dealing with the sum (totals).

  • $\begingroup$ But what I think is that theyask one about the sum of 12 bottles, and the mean of that sum? In other words, they're too similair to me.. $\endgroup$ Jan 20, 2013 at 18:37
  • $\begingroup$ There's no sum in question one. Each bottle is filled with an amount given by a normal distribution with mean 102, the question asks about the mean of twelve bottles. Where do you see a sum? $\endgroup$ Jan 20, 2013 at 18:45
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    $\begingroup$ Oh wait nevermind, I was being a little bit blind! In the first one they ask about the MEAN (i.e. average) out of a sample, in the second they transform a sample of 6 into '1' object (>namely, the box of teabags), with its own SD and M! $\endgroup$ Jan 20, 2013 at 18:45
  • $\begingroup$ I think I was writing my response the same time you were doing yours. Nice answer. $\endgroup$
    – Placidia
    Jan 20, 2013 at 18:46
  • 1
    $\begingroup$ I meant 20, my brain is random, no idea how I got to 6 $\endgroup$ Jan 20, 2013 at 18:48

The first standard deviation formula you gave is the SD for a sum. The standard error is the SD of the sample mean. Remember that: $\text{Var}(aX)=a^2 \text{Var}(X)$ and the variance of the sum is the sum of the variances (First formula). So

$\text{Var}(\bar{X})=\frac{n\sigma^2}{n^2}=\sigma^2/n$. Taking the square root gives the result.


$\text{Var}(\sum X_i)=\sum (\text{Var}(X_i)=n \sigma^2.$ The Variance of the sums.

Problem 1 is looking for a statement about the sample mean; Problem 2 is about the sum, since the weight of the package is the sum of the weights of individual tea bags.

  • $\begingroup$ Nice answer, +1, but I gave the other one a best answer since I read it first and it answered my question first. $\endgroup$ Jan 20, 2013 at 18:48

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