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Stats help please

  1. Mar 2, 2006 #1
    Hi Everyone,

    I am new to the forum and really need so help before I have a test tomorrow. I have been going over my book and working through the problems but I am stuck on this problem. Would someone please help me? thank you very much for your time

    From a study
    average height of men =68in sd=2.7inch
    average forearm length=18in sd 1inch r=0.80

    a) what percentage of men have forearms which are 18 inches long to the nearest inch? for this problem I figured out 38%

    b) of the men who are 68inches tall, what percentage have forearms which are 18 inches long to the nearest inch?

    I know that the nearest inch refers to 17.5 and 18.5 and the answer in the back of the book is 60% however, I do not know how they solve the problem to get 60%
  2. jcsd
  3. Mar 3, 2006 #2


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    Staff Emeritus
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    Since I don't believe that "forearm length" and "height" are independent, I don't see any way to do this with knowing the correlation.
  4. Mar 11, 2006 #3


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    Homework Helper

    I wonder whether "r=0.80" in the OP stands for corr.

    For indep. normal distributions, since sd = 1 for the forearm length, the question is, what is the probability around the mean plus/minus half a sd? You can look it up from a standard normal table (e.g. here). I believe 38.3%.

    For correlated distributions you need to think about multivariate normal.
  5. Mar 13, 2006 #4
    found the answer

    Thank you for the replies, I already found a way to solve the problem.

    r stands for coefficient correlation
    and the answer as I included is 60%
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