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Assume that house prices in an area are normally distributed with a standard deviation of \$ 20,000. A random sample of 16 houses is taken. What is the probability that the sample mean differs from the population mean by more than \$ 5,000.

I can write two equations as follows

For case I - $N(\mu_1, 20000)$

For case II - $N\left(\mu_2, \sqrt{\frac{20000^2}{16}}\right)$

But in both cases $\mu_1 = \mu_2$ right? Can anyone give me a tip?

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The arithmetic average of $K$ equally normal-distributed quantities has the same mean $\mu$; but the standard deviation decreases by the factor of $\sqrt{K}$. In this case, $K=16$ so the standard deviation of the mean will be $20,000/\sqrt{16} = 5,000$ dollars. It just happens to be the value equal to the deviation in the last, main question you asked.

So the question is what's the probability that the quantity (average of 16 prices) differs from the mean by more than 1 standard deviations. This value is well-known numerically. The probability that the quantity is "within one sigma" is 68.3% or so. The probability that the deviation exceeds one standard deviation is 37.3% and that's the answer to your question.

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  • $\begingroup$ Thank you for the answer. Let me check if I got it correctly :) So if the sample size is 20, then the standard deviation decreases by a factor of 4.472 and that means $\frac{5000}{4472} = 1.118$ standard deviations? $\endgroup$ – Blogger Jan 15 '16 at 7:56
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    $\begingroup$ Sorry, the last step is $4472/5000=0.894$ standard deviations. $\endgroup$ – Luboš Motl Jan 15 '16 at 16:25

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