A researcher wants to take a random sample of size 40 of a certain population with a mean of 23.5 and a variance of 18.49. What is the approximate standard deviation of sample means?
σXˉ=σn=18.4940≈0.68.\sigma_{\bar X}=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{18.49}}{\sqrt{40}}\approx 0.68.σXˉ=nσ=4018.49≈0.68.
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