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Normal Distribution Question. Need help!

  1. May 13, 2016 #1
    [MENTOR note] Post moved from General Math forum hence no template.

    Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
    My approach is to calculate the Probability for (mean - 2*σ < X < mean + 1.5*σ), not sure how to solve this further.
     
    Last edited by a moderator: May 13, 2016
  2. jcsd
  3. May 13, 2016 #2

    fresh_42

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  4. May 13, 2016 #3

    Ray Vickson

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    Use ##P(a < X < b) = P(X < b) - P(X < a)##, and find both ##P(X < a)## and ##P(X < b)## from standard normal tables.
     
  5. May 13, 2016 #4
    The solution I have with me is:
    Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668)
    My question is how we got (0.5 - 0.0228) + (0.5 - 0.0668) ?
     
  6. May 13, 2016 #5

    NascentOxygen

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  7. May 19, 2016 #6
    Try to break the problem into smaller pieces. What probability do you believe 0.5 - 0.0228 is specifying? (i.e. Probability that X is in what range?).
     
  8. May 19, 2016 #7

    RUber

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    Try both your method and the one suggested by Ray. They should give the same result. Drawing a picture might help as well.
    It might also be interesting to note that (0.5 - 0.0228) + (0.5 - 0.0668) = 1 - 0.0228 - 0.0668. Do you know why those two numbers (0.0228 and 0.0668) are significant in this problem?
     
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