Determine μoverbar(x)and σoverbar(x) from the given parameters of the population and the sample size. Round the answer to the nearest thousandth where appropriate. μ = 25, σ = 12, n = 16

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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