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Gaussian Distribution Question.

  1. Oct 21, 2011 #1
    I am studying the Gaussian distribution and am doing one of the problems for practice. The problem states that the standard deviation is equal to 15 and the actual value recorded in the experiment is 385.0. It then asks what is the probability that a single measurement lies in the range of the following..
    a.) 385.0-385.1
    b.) 400.0-400.1
    c.) 451.0-415.1
    d.) 370.0-400.0
    e.) 355.0-415.0
    f.) 340.0-430.0

    The following equations are to be used, based on the Gaussian distribution.
    [PLAIN]http://img62.imageshack.us/img62/8594/what2c.gif [Broken]
    I couldn't get the same answer as the book. This is the books work for a.)
    [PLAIN]http://img585.imageshack.us/img585/1271/whati.gif [Broken]

    There is no explaination as to where 0.399 came from, and the section on Gaussian distribution is very short (As this isn't a statistics course). It says for problems d e and f use the integral equation. Why can't I use the integral equation for all of them? Can anyone help?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 21, 2011 #2

    Stephen Tashi

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    The problem would make sense if it said that 385.0 was the actual population mean.

    Taking [itex] z= \frac{(385.1 - 385.0)}{15} [/itex] and substituting this in the expression [itex] \frac{1}{\sqrt{2 \pi} } e^{\frac{-z^2}{2}} [/itex] is probably where it came from. Or they might have used 385.05 as being more representative of the interval from 385 to 385.1 than the endpoint 385.1.

    I think you coud use the integral expression for all of them if you have numerical table of that integral.
  4. Oct 21, 2011 #3
    Yup that was it! But what confuses me is, wouldnt that mean that a and b have the same answer? Since the differences are all 0.1? The book as the the answer to b as, 0.0161.
  5. Oct 21, 2011 #4

    Stephen Tashi

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    The z for a value v is calculated by (v - 385.0)/15. It just happens that in a) the interval begins at 385. In the other examples, the intervals don't begin there.
  6. Oct 21, 2011 #5
    Ohh! Thank you very much! I understand now :]
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