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This is another statistics problem that I have, which I cannot make sense of:

An IQ-test is normal to $ \mu = 100$ and $\sigma = 10$. What is the standard deviation of the sample mean of a sample with size $n = 50$?

We know the mean is $\mu = n \times p$ which we solve for $n=\frac{100}{p}$ and substitute in $Var(X) = n \times p(1-p)$ which is $\sigma^2$ but yet I don't seem to be able to obtain the value of the standard deviation $\sigma$ using this way. The supposed solution of this problem is $1.4142$.

Thanks for your help!

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The question says normal, as in the normal distribution. The fact that you are applying a formula that relates to the binomial distribution indicates that you do not understand what you are doing or what the formulas you are using actually mean.

Go back and review your textbook and notes pertaining to the following topics:

  • Normal distribution
  • Sample mean
  • Sampling distribution

Then show how you have modified your computation accordingly.

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  • $\begingroup$ I did not come up with this by myself, my textbook actually does apply this formula $\sqrt{n} \times \frac{\overline{X}-\mu}{\sigma}$ to this problem. One can obtain $\mu$ with $n \times p$ if it is not given, but here it is given so that is not relevant. And the standard deviation $\sigma$ can be gotten by $\sqrt{np \times (1-p)}$ which is what I am trying to follow here to find it. $\endgroup$ – Khaled Apr 6 '18 at 2:48

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