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Error analysis

  1. Dec 19, 2006 #1
    A student makes 10 measurements of length x and gets the results all in mm


    Using cahuvenenet's criterion should he accept or reject the measurement of 58??

    2. Relevant equations
    [itex] \overline{x} [/itex] = average
    [itex] sigma_{x} [/itex] = standard deviation
    [itex] x_{sus} [/itex] = the measurement we want to reject or accept
    [tex] t_{sus} = \frac{x_{sus}-\overline{x}}{\sigma_{x}} [/tex]
    the number of standard deviationsfrom which x sus differes from x bar
    n(worse than [itex]x_{sus} [/itex]) = N P(outside [itex] t_{sus} \sigma_{x} [/itex])

    is n < 0.5 then 58 is rejected

    if n > 0.5 then 58 is accepted
    3. The attempt at a solution
    well the average
    x bar = 45.8
    standrad deviation = 5.1
    [tex] t_{sus} = 2.4 [/tex] standard deviations

    P(putside 2.4[itex] \sigma [/itex]) = 1 - P(within 2.4 [itex] \sigma [/itex])
    = 1 - 0.984
    the 0.984 is taken from a table which shows the percent probability
    P(within [itex] t\sigma[/itex])= \int_{X-t\sigma}^{X+t\sigma} f_{X,\sigma} (x) dx [/tex], as a function of t

    but why is .984?? Why is it that the probability should be chonse to be 0.01 and not 0.00 ... or 0.02??

    thanks for your help in advance!!
  2. jcsd
  3. Dec 19, 2006 #2


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    Science Advisor
    Homework Helper

    I'm not sure exactly what you are asking, but the .984 is the probablility that a measurement made on normally distributed data will be within 2.4 standard deviations of the mean. I've never heard of cahuvenenet's criterion, but then I'm hardly an expert in things statistical. .984 is the area under the normal distribution function from -2.4 to +2.4 standard deviations. There are several calculators online for finding and graphing these areas. Here is one of them:


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