Determine $μ_{\bar{X}}$ and $σ_{\bar{X}}$ from the given parameters of the population and sample size.
$μ = 87, σ = 22, n = 21$
$\begin{array}{l} μ_{\bar{x}} = ◻ \\ σ_{\bar{x}} = ◻ \end{array}$
(Round to three decimal places as needed.)
Time Remaining: 01:40:37
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So, the final answer is $μ_{\bar{x}} = 87$ and $σ_{\bar{x}} = 4.801$ .

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Step 1 ：The problem is asking for the mean and standard deviation of the sample mean. The mean of the sample mean, denoted as $μ_{\bar{X}}$ , is equal to the population mean, $μ$ . The standard deviation of the sample mean, denoted as $σ_{\bar{X}}$ , is equal to the population standard deviation, $σ$ , divided by the square root of the sample size, $n$ .

Step 2 ：Given that $μ = 87$ , $σ = 22$ , and $n = 21$ , we can calculate $μ_{\bar{X}}$ and $σ_{\bar{X}}$ .

Step 3 ：Since $μ_{\bar{X}}$ is equal to $μ$ , we have $μ_{\bar{X}} = 87$ .

Step 4 ： $σ_{\bar{X}}$ is equal to $σ$ divided by the square root of $n$ , so $σ_{\bar{X}} = \frac{σ}{\sqrt{n}} = \frac{22}{\sqrt{21}} \approx 4.801$ .

Step 5 ：So, the final answer is $μ_{\bar{x}} = 87$ and $σ_{\bar{x}} = 4.801$ .

Determine μX¯ and σX¯ from the given parameters of the population and sample size.μ=87,σ=22,n=21μx¯=◻σx¯=◻(Round to three decimal places as needed.)Time Remaining: 01:40:37Next

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Determine $μ_{\bar{X}}$ and $σ_{\bar{X}}$ from the given parameters of the population and sample size.
$μ = 87, σ = 22, n = 21$
$\begin{array}{l} μ_{\bar{x}} = ◻ \\ σ_{\bar{x}} = ◻ \end{array}$
(Round to three decimal places as needed.)
Time Remaining: 01:40:37
Next