Graduate students applying for entrance to many universities must take a Miller Analogies Test. It is known that the test scores have a mean of 75 and a variance of 16. In 1990, 50 students applied for entrance into graduate school in physics.

Find the probability that the sample mean deviates from the population mean by more than 1.5.

So I'm trying to understand my teacher's solution for this but I'm having trouble seeing where some values are coming from.

z = 1.5/(4/√50) = 2.6517

P(z>2.6517) = 0.0040

P(z<-2.6517) = 0.0040

P = 0.0040+0.0040 = 0.0080

I get that the teacher is using the central limit theorem, but where did the values 0.0040 come from? I don't see it on the Z-score table. And I was also wondering why the teacher adds 0.0040+0.0040 at the end to get 0.0080.

  • $\begingroup$ you are right that Z-score table has no 0.004 or 2.6517 generally, but it is easy to get that value by computer. Checking the normal distribution pdf curve you would be able to understand why 0.004+0.004. $\endgroup$
    – user158565
    Jul 23, 2019 at 1:02
  • $\begingroup$ As @user158565 said, using a computer does it. In Excel, 0.004 = 1 - NORM.DIST(2.6517,0,1,1) = one tail probability. But there was no specified direction (just deviation by more than 1.5 in magnitude), so both the upper tail and lower tail are 0.004, summing to 0.008. $\endgroup$
    – Ed V
    Jul 23, 2019 at 1:10
  • $\begingroup$ Sorry, I'm a little confused. Is there a specific table with values I should be looking at? $\endgroup$
    – Tessa Gray
    Jul 23, 2019 at 1:19
  • $\begingroup$ The z tables were something from the old days (my days, actually). Now we have computers and spreadsheets. So, if you have a desired single tail probability, such as 0.004, and want to know the corresponding critical z, just use (in Microsoft Excel) critical z = -NORMSINV(0.004). The result is 2.6520698079022. To go the other way, single tail probability = 1 - NORM.DIST(critical z, 0, 1, 1), as per the Excel Help file info. The tables are still useful as a quick check that software is correct and tables may be used in in-class exams. Other than that, tables are (thankfully) history! $\endgroup$
    – Ed V
    Jul 23, 2019 at 2:21

1 Answer 1


The question asks for $P(|\bar{X}-75|>1.5$), and since $\bar{X}$ has mean $\mu=75$ and deviation, $\sigma=4/\sqrt{50}$. First of all, $$\begin{align}P(|\bar{X}-75|>1.5)&=P(\bar{X}-75>1.5)+P(\bar{X}-75<-1.5)\\&=2P(\bar{X}-75<-1.5)\end{align}$$

We can write this since the distribution is symmetric around $\mu=75$. To convert the inner expression into $z$ value, we just divide it by the deviation, i.e. $$P\left(\frac{\bar{X}-75}{4/\sqrt{50}}<\frac{-1.5}{4/\sqrt{50}}\right)\approx P(Z<-2.6517)\approx0.004$$ And, you can find this value in many Z tables. So, the final result is $2\times0.004=0.008$.

Some Z tables just show the positive side of the normal PDF, but even in that case, you'll find $P(Z<2.6517)$ and subtract from $1$.


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