Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this a valid way to calculate statistically probable value?

  1. Jun 3, 2006 #1
    Good day,

    Assume an event with a normally distributed numerical outcome. Call the outcome x. Assume that any outcome less than a particular value (called a) has a value of zero and any outcome greater than a has a value of x-a. Call the probability curve of the normal distribution f(x) and x-a = g(x). I am calculating the statistically probable value of any event as

    Integral from a to infinity of [f(x) times g(x)] dx

    Is this valid?

    I hope that I explained this sufficiently. Thank you for your help.
     
  2. jcsd
  3. Jun 3, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Could be clearer! I think that by "statistically probable value" you mean "expected value" (otherwise, I don't know what you mean by "statistically probable"). If so, yes, the "expected value" of any function g, of a random variable x with probability density function f, is given by
    [tex]\int_{-\infty}^\infty f(x)g(x)dx[/tex]
    If g(x)= 0 for x< a then that is
    [tex]\int_a^\infty f(x)g(x)dx[/tex]
     
  4. Jun 3, 2006 #3
    Thanks for your reply. Your presumption is correct, I think. Since I am ignorant of the teminology of statistics, the following is not an argument but an explanation of why I used those words. I didn't use "expected probability" because I don't expect the value of the event's outcome to be the result of the integral, but the value is calculated based on statistical probabilities.

    Thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is this a valid way to calculate statistically probable value?
  1. Probablity &statistics (Replies: 3)

Loading...