Question

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 1

μx = 80, σx = 0.52

Explanation:

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Answer 2

`\mu_(\bar X) = 90`

`\sigma_(\bar X) =(\sigma)/(√(n))= (9.6)/(√(144))= 0.8`

Explanation:

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

`\mu = 90, \sigma =9.6`

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:

`\bar X \sim N( \mu, (\sigma)/(√(n)))`

And replacing we got:

`\mu_(\bar X) = 90`

And the standard deviation is given by:

`\sigma_(\bar X) =(\sigma)/(√(n))= (9.6)/(√(144))= 0.8`