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.
$$
\begin{aligned}
\mu_{\bar{x}} & =\square \mathrm{mL} \
\sigma_{\bar{x}} & =\square \mathrm{mL}
\end{aligned}
$$

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.
$$
\begin{aligned}
\mu_{\bar{x}} & =\square \mathrm{mL} \
\sigma_{\bar{x}} & =\square \mathrm{mL}
\end{aligned}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the mean of the sampling distribution of $\bar{x}$, we use the population mean $\mu$. ‖ ‖ step 2 ⋮ The population mean is given as $\mu = 275$ mL. ‖ ‖ step 3 ⋮ Therefore, the mean of the sampling distribution of $\bar{x}$ is also $\mu_{\bar{x}} = 275$ mL. ‖ ‖ step 4 ⋮ To find the standard deviation of the sampling distribution of $\bar{x}$, we use the formula $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. ‖ ‖ step 5 ⋮ The population standard deviation is given as $\sigma = 10$ mL and the sample size is $n = 5$. ‖ ‖ step 6 ⋮ Plugging the values into the formula gives $\sigma_{\bar{x}} = \frac{10}{\sqrt{5}}$ mL. ‖ ‖ step 7 ⋮ Calculating the standard deviation of the sampling distribution of $\bar{x}$ gives $\sigma_{\bar{x}} = \frac{10}{\sqrt{5}} \approx 4.5$ mL when rounded to one decimal place. ‖ ∻[question 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 $\bar{x}$ is equal to the population mean $\mu$, and the standard deviation of the sampling distribution of $\bar{x}$, known as the standard error, is equal to the population standard deviation $\sigma$ divided by the square root of the sample size $n$.⚹◊What is the formula to calculate the standard deviation of the sampling distribution of the mean amount of liquid ($\bar{x}$) in random samples of 5 drinks?⍭ Generate me a similar question◊

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