The question states "Times spent studying by students the week before exams follows a normal distribution with standard deviation of 8 hours. A random sample of four students was taken in order to estimate the man study time for the population of all students. What is the probability the sample mean exceeds the population mean by more than 2 hours?"

I'm not given any means to work with, so I can't use Z-Score as far as I can tell. I'm clearly not seeing something. Is there a way to back solve for the answer? Is it something along the lines of just 2/standard error and finding Z greater than that?

  • $\begingroup$ The mean will not bring anything here; all that matters is the standard deviation (i.e., "how things are spread"). You could for instance use Chebyshev's inequality, or something more tailored to the Gaussian distribution; but basically the probability is invariant by shifting (i.e., the mean does not matter). $\endgroup$ – Clement C. Oct 13 '15 at 2:22
  • $\begingroup$ @ClementC. So in this case, it would be something like 1-1/k^2, where k is 2? Thus telling us that 75% of the data would lie there, and then take 1 - .75 to get 25% of the data is greater? $\endgroup$ – David Oct 13 '15 at 2:33

The sample mean $\bar{X}$ has normal distribution, mean the population mean $\mu$, and standard deviation $\tau=\frac{8}{\sqrt{4}}$. We want $\Pr(\bar{X}-\mu\gt 2)$, which is $\Pr(Z\gt \frac{2}{\tau})$, where $Z$ is standard normal. This looks like precisely the approach you suggested.


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