Determining Whether Values Within One Standard Deviation Meet Some Condition Suppose I have a set of objects, each of which can take on values within some unknown range, and each of which only has one sample of 50+ values associated with it.  The values of each observation is not known.  In order to determine how likely it is for the value of an object in a set of objects $O$ to have a negative value I opted to use the difference $\sigma(X_i)-|E(X_i)|$ with the assumption that $\forall$ $X_i$ $\in $ $ O$, $X_i$ is i.i.d. Gaussian.  In this way I can find out whether one standard deviation from a known mean value for each $X_i$ is negative.  If it is, I can use normality to determine how likely it might be for this object to be negative.  Is this a good way to go about determining the likelihood, or is it possibly too naïve?  I worry that I may be making some errors.
 A: 
with the assumption that $∀ X_i ∈ O, X_i$ is i.i.d. Gaussian  ...
  ... to determine how likely it is for the value [...] to be negative. 

If I understand correctly, this is asking for $P(X_i<0)$

if $\sigma(X_i)-|E(X_i)| < 0$ then the object has failed

The quantity $\sigma(X_i)-|E(X_i)|$ is not in general very closely related to $P(X_i<0)$.
However, the particular value of 0 does tell us something - if $\sigma(X_i)-|E(X_i)| < 0$ then $P(X_i<0)$ is either less than $\Phi(-1)$ (about 0.16) or greater than $\Phi(1)$ (about 0.84), where $\Phi$ is the standard normal cdf; perhaps that is all you want.
There is a direct calculation in terms of the parameters $\mu=E(X)$ and $\sigma = \sqrt{\text{Var}(X)}$ that tells you more directly about $P(X_i<0)$: 
$$P(X_i<0)=\Phi(-\mu/\sigma)$$
So if you know $\mu$ and $\sigma$ and can calculate $\Phi(x)$, you can find $P(X_i<0)$ readily.

Edit in response to edits to the question: The above is predicated on the Gaussian assumption; whuber's discussion of using the sample proportion is the better way to go if you're not confident it's really very close to Gaussian - the sample proportion works whether or not it's Gaussian.
