# Sum standard deviation vs standard error

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 litrebottles of lemonade. The amount is normally distributed with $\mu = 102 \space cl$. Standard deviation is $1.93\space cl$.

• 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 an average of 5.3 grams and a standard deviation of 0.5 grams.

• 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?

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In second problem you need to calculate standard deviation of a package weights which is total of 20 teabags not the average standard deviation. So $\sqrt{20} \times 0.5$ is correct.