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A large population has a mean of 150 and a standard deviation of 40. A sample of 100 observations is to be selected at random from the population. What is the standard deviation of the sample mean? The answer they give is 4 although that might be a rounded answer and I don't know how they got to that answer.

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If we sample from $X\sim N(150, 40^2)$ 100 times, then the sample mean is $$\bar X = \frac{X_1+\dotsb+X_{100}}{100}.$$

Then \begin{align*} \text{Var}(\bar X) &= \text{Var}\left(\frac{X_1+\dotsb+X_{100}}{100}\right) \\ &=\frac{1}{100^2}\text{Var}(X_1+\dotsb+X_{100})\\ &= \frac{1}{100^2}\cdot100\cdot\text{Var}(X_1) \\ &= \frac{40^2}{100} \end{align*} Hence $$\text{SD}(\bar X) = \frac{40}{10} = 4.$$

If you use a box model, then $$\text{SD}_{\text{avg}} = \frac{\text{SD}_{\text{box}}}{\sqrt{\text{#draws}}}.$$

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