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I'm okay with solving regular normal distribution questions (where $X$ is a normal random variable with mean $\mu$ and standard deviation $\sigma$). However, we're currently dealing with samples within a larger population and I'm struggling to understand how the process of solving a question goes.

Here's an example:

The weight of a car is normally distributed with a mean of $2500$ and a standard deviation of $50$. Next, a random sample of $5$ cars is taken. What is the standard error of the sample mean?

I know that the standard error of the sample is:

$$ \frac{\sigma^2}{n} $$

but a question can't be this easy to answer can it? It's leading me to believe I'm missing a concept. If someone would be able to walk through the procedure of solving a question of this nature it would be greatly appreciated.

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    $\begingroup$ The question is in fact that simple. $\frac{\sigma^2}{n}$ is the standard error of the sample mean. "standard error of the sample" (what you said it is) doesn't make any sense. $\endgroup$ – Jonathan Christensen Nov 11 '12 at 2:19
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As Jonathan wrote, your answer is correct.

In my experience this phenomenon of doubting your own knowledge on the basis of the assumption that you couldn't possibly find it easy to answer a question is quite widespread. My advice would be to question that assumption. A realistic assessment of one's own abilities is important, and that applies just as much to not underestimating them as it does to not overestimating them.

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I hate necro-posting but your answer is incorrect.

The sample mean $M$ of a sample of size $n$ from a normally distributed RV with mean $\mu$ and variance $\sigma^2$ is a RV $\sim N(\mu,\sigma^2/n)$. The standard error of the mean $s_M$ is the standard deviation of the sample mean and so $s_M=\sigma/\sqrt{n}$.

What you have given, and the other answers and comments have given is the variance of the sample mean.

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