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I saw this somewhere. Is this right?

$ P(z< -4.3333)=0.00003 $

Most z-score table stops at 3.4. If this is right, kindly explain how it was gotten. Thanks

john scott
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2 Answers2

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This $z$-score is calculated as $$P(z<-4.3333) = \int_{-\infty}^{4.333} f(z)\mathrm{d}z$$

where $f(z)$ is the pdf for the standard normal distribution.Every $z$-score is calculated in this way.

G. Snapsmath
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  • Thanks for the answer. Although, I don't really comprehend. This looks advance. Can you explain more? – john scott Oct 09 '16 at 14:42
  • Sure! The $z$-score is nothing more than the area under the standard normal curve. The area under the curve is given by the integral, which is a calculus notion. If you can imagine approximating the area by rectangles, then do it again with smaller rectangles you can begin to understand the intuition of an integral. The tables in your book stop at 3.4 and -3.4 because at that point you will have about 99.99% of all data accounted for. – G. Snapsmath Oct 09 '16 at 14:48
  • Check out this page for a nice intuitive explanation of integral http://math.stackexchange.com/questions/916569/intuitive-explanation-for-integration – G. Snapsmath Oct 09 '16 at 14:50
  • Thank you very much. I just need one more thing. Can you kindly expatiate on this using this particular question as an example so I can understand well. P(z< -4.3333)=0.00003. Can you solve here to show how it was gotten – john scott Oct 09 '16 at 14:56
  • We can't. The integral cannot be evaluated by elementary methods. We can only approximate it using power series or other tools. Hence, why we have tables and computers to this. – Sean Roberson Oct 09 '16 at 14:59
  • @SeanRoberson I tried using Excel to get it. It displayed an error message. What software can I use to solve it then? – john scott Oct 09 '16 at 15:03
  • You likely typed it in wrong then. Try =NORM.DIST(-4.3333, 0, 1, TRUE). Read the documentation or do a search for other tools online. – Sean Roberson Oct 09 '16 at 15:07
  • @SeanRoberson Thanks a lot. I really appreciate it – john scott Oct 09 '16 at 15:33
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It was most likely obtained by the integral representation:

$$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{t} e^{-\frac{z^2}{2}} \ dz $$

Of course any computer can do this. Many tables stop at a score of 3 because it is less than anything practical.

Sean Roberson
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  • Thanks for the answer. this looks kinda advanced. How can I find it without using a computer – john scott Oct 09 '16 at 14:46
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    If you don't have a computer or a calculator capable of doing these calculations, then for all purposes, assume that for $z < -3.5$ we have $P(Z < z) = 0$. You can only use the limitations of your table. – Sean Roberson Oct 09 '16 at 14:50