RMS value of clipped sine
As Hans mentioned in the comments, your scope might be ignoring the DC offset of your signal.

You described that your signal has a peak-to-peak voltage of \$3.88~V\$. This is an amplitude (from the center) of
$$
A = \frac{3.88~V}{2} = 1.94~V 
$$
so if your signal was a perfect sine wave, it would have an RMS amplitude of
$$
A_{RMS} = \frac{A}{\sqrt{2}} = \frac{1.94~V}{\sqrt{2}} = 1.38~V
$$
This is almost exactly what your scope has measured, so I think it's right.

Note: to add a DC bias to an RMS level, you can use the relationship
$$
A_{shifted} = \sqrt{A_{RMS}^2 + A_{DC}^2}
$$
In your case, this would be
$$
A_{shifted} = \sqrt{(1.39~V)^2 + (1.94~V)^2} = 2.38~V
$$