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Inequalities in Normal Distributions

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data

    How does P(-1<Z<1) equal to 1-2P(Z>1)?
    (So you can find the values on the Normal Distribution Table)
    2. Relevant equations


    3. The attempt at a solution
    I tried P(-1+1<Z+1<1+1) but ended up with P(1<Z+1<2).
     
  2. jcsd
  3. Sep 21, 2015 #2

    ehild

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    The normal distribution function is symmetric, f(z)=f(-z). The table shows the values of Cumulated Distribution Function F(Z) =P(-∞<z <Z). The probability P(Z<z<∞)=1-F(Z). Because of the symmetry, P(Z<z<∞)=P(-∞<z<-Z), F(-Z)=1-F(Z). The probability that the variable z is between -Z and Z is [tex]P(-Z<z<Z)=F(Z)-F(-Z) = F(Z)-(1-F(Z))[/tex].
     
  4. Sep 21, 2015 #3

    SteamKing

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    The standard normal curve is symmetrical about a mean value μ = 0 like this:

    http://www.spiritsd.ca/curr_content/mathb30/data/les6/images/norm_percent.gif [Broken] ​

    The normal curve is scaled such that the total area under the curve is 1.

    Since you are trying to find P(-1 < Z < 1), don't you see how that's the same probability as 1 - 2P(Z>1)?
     
    Last edited by a moderator: May 7, 2017
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