0

Possible Duplicate:
Evaluate definite interval of normal distribution

Title was changed and question edited bellow.

How is possible that a probability density function defined as following can return some values that are higher than 1 ?! Is there any error in my definition of this pdf ?

float gauss_pdf = exp( -pow(x-mean, 2.0) / (2.0 * pow(sigma, 2.0)) )   /   ( sigma * sqrt(2*M_PI) );

For example, for x = 0.0908182, mean = 0.0552096 and sigma = 0.0241953, it will return 5.583

EDIT:

Well, what I need is the corresponding probability, thus I need to compute the CDF (i.e. P(X < x)) for that mean and sigma. How can I compute the CDF ?

Community
  • 1
shn
  • 2,479
  • 9
  • 31
  • 38
  • 1
    Voting to close. There are several questions that address related matters. Here are some: [Evaluate definite interval of normal distribution](http://stats.stackexchange.com/questions/7200), [Is it possible to analytically integrate x multiplied by the lognormal probability density function?](http://stats.stackexchange.com/questions/9501), [Probability distribution value exceeding 1 is OK?](http://stats.stackexchange.com/questions/4220) and [How can a probability density be greater than one and integrate to one](http://math.stackexchange.com/questions/105455), to give you a start. – cardinal Jun 06 '12 at 13:38

0 Answers0