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the following is the problem that I am working on.

A random sample of size 16 is to be taken from a normal population having mean 100 and variance 4. What is the 90th percentile of the distribution of the sample mean $\overline{X}$ ?

This is what I understand so far.

Assuming independence of the samples, the mean of $\overline{X}$ is 100 and the variance is $1 \over 4$.

The 90th percentile of the normal curve, according to the table I was provided, was equal to 1.28 standard units above the mean, so my answer was $100+1.28(1/2)=100.64$ .

Would this be the correct answer?

Also, I assumed independence in this problem, but is there any reason why assuming that would be a natural thing to do? Is there a way to solve this problem assuming that the samples are actually dependent?

hyg17
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  • In reply to a now-deleted question in a comment asking why the variance is $1/4$: Because when the population variance is $\sigma^2$ and the sample size is $n$, the the variance of the mean of an independent random sample is $\sigma^2/n$. I could write a derivation of that, but I'm not sure a comment is the most comfortable place for that. ${}\quad{}$ – Michael Hardy Jun 26 '14 at 03:04
  • My understanding is $Var[X_i]=4$ so $Var[\sum X_i /16]= \frac{1}{16^2}Var[\sum X_i]={Var[X_i] \over 16} = {1 \over 4}$ – hyg17 Jun 26 '14 at 03:07

1 Answers1

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That is correct.

The author of the question may have taken the phrase "random sample" to imply independence. In some cases the phrase "simple random sample" means sampling without replacement, which in some contexts would imply dependence, but with a normally distributed population it would not. It implies that only if the population is finite, and normality precludes that.

Michael Hardy
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