A population has μ = 90 and a standard deviation of σ = 9.6. What are the mean and standard deviation of the sampling distribution x-bar if a sample of 144 were taken?

Answer :

Answer:

[tex] \mu_{\bar X} = 90[/tex]

[tex] \sigma_{\bar X} =\frac{\sigma}{\sqrt{n}}= \frac{9.6}{\sqrt{144}}= 0.8[/tex]

Step-by-step explanation:

For this case we have a population with the following parameters:

[tex]\mu = 90, \sigma =9.6[/tex]

For this case we have a sample of n= 144 and this sample size is larger (>30) then we can apply the central limit theorem and the distirbution for the sample mean is given by:

[tex] \bar X \sim N( \mu, \frac{\sigma}{\sqrt{n}}) [/tex]

And replacing we got:

[tex] \mu_{\bar X} = 90[/tex]

And the standard deviation is given by:

[tex] \sigma_{\bar X} =\frac{\sigma}{\sqrt{n}}= \frac{9.6}{\sqrt{144}}= 0.8[/tex]

Answer:

μx = 80, σx = 0.52

Step-by-step explanation:

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