Is observation a subset of Gaussian distibution

Hi everyone,

I'm tackling with a problem in data analysis. It goes as follows:

I have a random variable X that follows gaussian distribution with mean, width and integral. This (target) distribution has been constructed from measurements of property X from large number of items.

Now, I have a new item from which I measured the property X. I did the measurement in triplicate, so this measurement has also a distribution with mean, width and integral.

The target distribution has mean 85, width 12 and integral 183 (because the distribution was constructed from 183 measurements). Property X from the newest item has mean 99, width 8 and integral 3 (because measurement was done in triplicate). Here it is assumed that both distributions are gaussians, this seems to be reasonable assumption based on the data available. The nature of the property is such that measurements even from single item have large variability.

I want to know the probability of the newest item belonging in fact to this target distribution.

How do I proceed? I only know how to calculate the probability of the new item belonging to the target distribution from single measurement, by using the distance (in standard deviations) of the measurement from the center of the target distribution, and looking up the probability from table. I can't get around the fact that instead of having a distribution and a single value, I have two distributions to compare. Because the new measurement has some uncertainty in it, it must lower the probability compared to single measurement.

EDIT: I just realized that because both distributions have a finite width, the probability of the "overlap" is maybe 1. So, let me modify the question. If I have two target distributions, first with mean 85, width 12 and integral 183, and the second target distribution with mean 130, width 15 and integral 282, what are the probabilities that my new item belongs to either of these distributions. Because there are finite and known number (2) of possibilities, the two probabilities must add to 1.

Cheers,