Question

Math

Posted 3 months ago

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

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Answer from Sia

Posted 3 months ago

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 $n$.

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