p = (1/(Sqrt[2 π])) (∫_(-x)^x) 
                Exp[-z^2 /
                  2] \[DifferentialD]z;
Plot[p, {x, 0, 4}, Frame -> True, 
      FrameLabel -> {"Standard Deviation", "Probability"}]

Using the the "replacement operator /. (pronounced “slash-dot”)," we can determine the confidence level of result for 1 standard deviation, when x = 1, is

p /. {x ->  1. }

In In[3], we have replaced x in the expression for p in,  In[1] above, with 1.  Confidence level of result for 2 standard deviations is

p /. {x ->  2. }

Confidence level of result for 3 standard deviations is

p /. {x ->  3. }