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Normal distribution, find mean and SD

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose X is a normally distributed random variable. Suppose also that P ( X > 44.7 ) = 0.33 and P ( X < 46 ) = 0.7123. What is the mean and standard deviation of X ?

    2. Relevant equations

    3. The attempt at a solution
    P(X<44.7) = 1-P(X>44.7) = 1-.33 = .67
    P(X<44.7) = P((x-μ)/σ < (44.7-μ)/σ) = .67
    P(Z < (44.7-μ)/σ) = .67

    P(X<46) = P((x-μ)/σ < (46-μ)/σ) = P(Z < (46-μ)/σ) = .712

    Looked up corresponding z-scores:
    Z1 = .4
    Z2 = .5
    It's possible these are wrong

    Setting up system of equations.

    (44.7-μ)/σ = .4
    (46-μ)/σ = .5

    Solved using matrix -> rref

    μ = 39.5
    σ = 13

    These answers aren't correct. What am I doing wrong here?

  2. jcsd
  3. Mar 14, 2013 #2

    Ray Vickson

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    The problem is premature rounding. Just because data is given to a small number of significant figures does NOT mean that you should work through the problem with a small number of figures. The accurate values of z1 and z2 are
    z1 = .4399131657 and z2 = .5601164657. If you use these you will get
    μ = 39.94233436 ≈ 39.9 and σ = 10.81501090 ≈ 10.8 . Note that we round off at the end, after doing all the calculations!
  4. Mar 14, 2013 #3
    Thank you!

    I guess I'm not really sure how to use my z-score table correctly. I don't know how you get z1 and z2 to that many decimal places of accuracy
  5. Mar 14, 2013 #4

    Ray Vickson

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    I just used a computer package (Maple in my case). You can use the EXCEL spreadsheet (Solver tool) or the on-line program Wolfram Alpha. Some scientific hand-held calculators give similar accuracy.
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