A machine automatically dispenses a beverage of a desired size. When set to "small," the machine dispenses varying amounts of liquid with a mean of $275 \mathrm{~mL}$ and standard deviation of $10 \mathrm{~mL}$. Suppose that we take random samples of 5 of these drinks and calculate the mean amount of liquid $\bar{x}$ in each sample. We can assume that individual drinks are independent.

Calculate the mean and standard deviation of the sampling distribution of $\bar{x}$.
You may round to one decimal place.
$$
\begin{array}{l}
\mu_{\bar{x}}=\square \mathrm{mL} \
\sigma_{\bar{x}}=\square \mathrm{mL}
\end{array}
$$

Question

A machine automatically dispenses a beverage of a desired size. When set to "small," the machine dispenses varying amounts of liquid with a mean of $275 \mathrm{~mL}$ and standard deviation of $10 \mathrm{~mL}$. Suppose that we take random samples of 5 of these drinks and calculate the mean amount of liquid $\bar{x}$ in each sample. We can assume that individual drinks are independent.

Calculate the mean and standard deviation of the sampling distribution of $\bar{x}$.
You may round to one decimal place.
$$
\begin{array}{l}
\mu_{\bar{x}}=\square \mathrm{mL} \
\sigma_{\bar{x}}=\square \mathrm{mL}
\end{array}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the mean of the sampling distribution of the sample mean $\bar{x}$, we use the population mean $\mu$. ‖ ‖ step 2 ⋮ Given the population mean $\mu = 275$ mL, the mean of the sampling distribution of the sample mean is also $\mu_{\bar{x}} = 275$ mL. ‖ ‖ step 3 ⋮ To find the standard deviation of the sampling distribution of the sample mean $\sigma_{\bar{x}}$, we use the population standard deviation $\sigma$ and the sample size $n$. ‖ ‖ step 4 ⋮ The formula for the standard deviation of the sampling distribution of the sample mean is $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. ‖ ‖ step 5 ⋮ Given the population standard deviation $\sigma = 10$ mL and the sample size $n = 5$, we calculate $\sigma_{\bar{x}} = \frac{10}{\sqrt{5}}$ mL. ‖ ‖ step 6 ⋮ Simplifying the expression, we find $\sigma_{\bar{x}} = \frac{10}{\sqrt{5}} \approx 4.5$ mL (rounded to one decimal place). ‖ ∻[1] Answer∻ ⚹ $\mu_{\bar{x}} = 275$ mL ⚹ ⚹ $\sigma_{\bar{x}} \approx 4.5$ mL ⚹ ∻Key Concept∻ ⚹Central Limit Theorem and Sampling Distributions⚹ ∻Explanation∻ ⚹The mean of the sampling distribution of the sample mean is equal to the population mean, and the standard deviation of the sampling distribution of the sample mean is the population standard deviation divided by the square root of the sample size. This is derived from the Central Limit Theorem, which applies because the sample size is sufficiently large and the population standard deviation is known.⚹◊What is the formula for calculating the standard deviation of the sampling distribution of the mean ($\sigma_{\bar{x}}$) in this scenario?⍭ Generate me a similar question◊

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