Is this a valid way to calculate statistically probable value?

[30osk]

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.

[HallsofIvy]

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.

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
\int_{-\infty}^\infty f(x)g(x)dx
If g(x)= 0 for x< a then that is
\int_a^\infty f(x)g(x)dx

[30osk]

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!