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Normal Random Variable Probability

  1. Aug 25, 2009 #1
    If X is a normal rv with mean 80 and standard deviation 10, compute the following probabilities by standardizing:

    P(|X-80| <= 10)

    I know how to determine the probability without absolute value, but this confuses me. Any help?
     
  2. jcsd
  3. Aug 25, 2009 #2
    From |X-80| <= 10, you can get -10 <= (X-80) <= 10, -1 <= Z <= 1
    We also know that P(Z<-1) = P(Z>1) due to symmetry. Hope that helps.
     
  4. Aug 26, 2009 #3
    I do understand partially. I remember this from just basic algebra. Afterwards, would I solve for P(70 <= X <= 90) where I added 80 to both sides?
     
  5. Aug 26, 2009 #4
    Yup.
     
  6. Aug 27, 2009 #5
    Ok I am not getting this to work out correctly:

    P(|X-80|) [tex]\leq[/tex] 10)

    P(-10 [tex]\leq[/tex] X-80 [tex]\leq[/tex] 10)

    P( [tex]\frac{-10-80}{10}[/tex] [tex]\leq[/tex] Z [tex]\leq[/tex] [tex]\frac{10-80}{10}[/tex])

    -9 [tex]\leq[/tex] Z [tex]\leq[/tex] -7

    Ok, so these aren't on the Z table, so I am definitely not doing something right. What am I now missing? The 80 is my mean and 10 is my s.d.
     
  7. Aug 27, 2009 #6
    [tex]Z = \frac{X-\mu}{\sigma}[/tex]
    Check your standardisation.
     
  8. Aug 27, 2009 #7
    P(-10 [tex]\leq[/tex] X-80 [tex]\leq[/tex] 10)

    P (70 [tex]\leq[/tex] X [tex]\leq[/tex] 90)

    P ([tex]\frac{70-80}{10}[/tex] [tex]\leq[/tex] Z [tex]\leq[/tex] [tex]\frac{90-80}{10}[/tex])

    .1587-.8413 = -0.6826

    I am funny! What a silly mistake. Should be right now!
     
  9. Aug 27, 2009 #8

    statdad

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    Homework Helper

    Check your final calculation - you have a glaring error.
     
  10. Aug 27, 2009 #9
    Wow, this problem kicked my butt!

    ANSWER IS:
    0.8413 - 0.1587 =0.6826

    Thanks for the help guys!
     
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