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Normalcdf question.

  1. Nov 9, 2009 #1
    One summer night at Bellair Lanes, a group of math professors went bowling. In true form, they decided to calculate their mean bowling score and standard deviation to use for the statistics problems. Here are the results: u(mean) 88, o(standard dev.)15.
    Find the probability that if a math professor is selected, his or her score will be greater than 100.
    100-88 / 15 = .8 normal cdf(.8,4) =.2118

    Find the probability that, if a math professor is selected, his or her score will be between 50 and 100.
    50-88 / 15 = -2.53 100-88 / 15 =.8 normal cdf(-2.53,.8)= .7824.

    Did I do the problems right? Thanks
  2. jcsd
  3. Nov 10, 2009 #2


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    Science Advisor

    Why "cdf(.8, 4)" for the first one? Where did the "4" come from?
  4. Nov 10, 2009 #3


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    Homework Helper

    Unless the problem states that the scores resemble a normal distribution your calculation isn't justified. If your instructor left that comment out, it's pretty sloppy.
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