Answer:
μoverbar(x) = 25
σoverbar(x) = 3
Explanation:
Given That :
Population mean, μ = 25,
Population standard deviation, σ = 12,
Sample size, n = 16
The sampling distribution is approximately normal from the central. Limit theorem and hence, μoverbar(x) = μ = 25
σoverbar(x) is the standard error and is obtained thus :
σoverbar(x) = σ / sqrt(n)
σoverbar(x) = 12 / sqrt(16) = 12 /4 = 3
Following are the calculation to the [tex]\bold{\mu_{\bar{x}} \ and\ \sigma_{\bar{x}}}[/tex]:
Given:
[tex]\mu = 25\\\\\sigma = 12\\\\ n = 16[/tex]
To find:
[tex]\bold{\mu_{\bar{x}} \ and\ \sigma_{\bar{x}}}[/tex]:=?
Solution:
[tex]\to \sigma_{\bar{x}} = \frac{\sigma }{\sqrt{n}} = \frac{12}{\sqrt{16}} = \frac{12}{4}= 3[/tex]
Therefore, the final value of [tex]\bold{\mu_{\bar{x}} \ and\ \sigma_{\bar{x}}}[/tex] is "25 and 3".
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