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Continous Random Variable HELP PLEASE

  1. Nov 11, 2008 #1
    Continous Random Variable HELP PLEASE!!

    Scores on a particular test are normally distributed in the population, with a mean of 100 and a standard deviation of 15. What percentage of the population have scores ....

    a) Between 100 and 125

    b) Between 82 and 106

    c) Between 110 and 132

    d) Above 132

    e) Equal to 132
     
  2. jcsd
  3. Nov 12, 2008 #2
    Re: Continous Random Variable HELP PLEASE!!

    This seems like a homework question so maybe someone can move it there.

    Apochema, have you tried anything? If so, show us what and we can better help you. If not, you certainly don't expect us to do your homework for you.
     
  4. Nov 12, 2008 #3

    cronxeh

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    Gold Member

    Re: Continous Random Variable HELP PLEASE!!

    Here is a hint. Look at the definition of a mean = E(X) and standard deviation sqrt(Var(X))

    which both happen to be integrals. Once you get your p(x) from those 2 equations, you simply take the integral for every problem a) integral(p(x)dx, 100, 125), and so on

    mean = E(X) = http://upload.wikimedia.org/math/5/2/b/52bc687e1475806a8abb8b8252f220cf.png = 100
    standard deviation = http://upload.wikimedia.org/math/f/4/c/f4c7ea85a64ca1819288007e6994e349.png = 15
     
  5. Nov 12, 2008 #4
    Re: Continous Random Variable HELP PLEASE!!

    This is certainly not the way to approach the problem. The biggest hint is basically given to you in the problem that the scores are distributed normally.
     
  6. Nov 13, 2008 #5
    Re: Continous Random Variable HELP PLEASE!!

    Right. And you should know what the probability density function for normally distributed random variable (with given mean and variance) looks like.

    Then you will have to integrate this density over the appropriate intervals or better look up the corresponding values in a table of the cumulative distribution function.

    Maybe you first have to apply a linear transformation to make the distribution standard normal if you only have access to cdf values for this special case.
     
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