Confusing Labels for Function Generators and Oscilloscopes in Tinkercad

In Tinkercad, amplitude definition for Function Generators and scale definition for Oscilloscopes are quite confusing. Here is an ss from Tinkercad's function generator:

On the device 6.20 V is represented as peak-to-peak voltage, look at the red-lines I've marked. But on the panel right-hand-side, we input it as the amplitude, look at the green line I've marked. Which one is true?

And I cannot deduce the answer using an oscilloscope, because there is not enough info about oscilloscope. (At least, I couldn't find enough info.) Here is the input signal from the function generator above:

Answer is not obvious, because the meaning of 10 V placed on y_axis is ambiguous. Is it +/- 10 V as in 20 V in total, i.e. the voltage-per-division is 2 V (first explanation)? Or, is it +/- 5 V as in 10 V in total, i.e. voltage-per-division is 1 V (second explanation)? In some Youtube lectures the explanation is first one. But, I'm not quite sure. Because, if 6.2 V is amplitude and voltage-per-division is 2 V, then this is noncontradictory. But if 6.2 V is peak-to-peak voltage and voltage-per-division is 1 V, then this, too, is noncontradictory. Again, which one is true?

And also, while studying, I've realise that a real life experiment indicates that the second explanation should be true. Let me explain the experiment step by step.

Theory: Full Wave Rectifier Circuits

Assume we apply V_in as the amplitude, the peak-peak voltage is, V_peaktopeak = 2 * V_in. And for output signal we have,

V_out = (V_in - n * V_diode) * R_L / (R_L + r_d),

where n is the number of diode in conduction, V_diode is bias of a diode and R_L is load resistor. Load resistor is choosen big enough so that R_L >> r_d and we get,

V_out = V_in - n * V_diode.

In a real experiment r_d is in between 1 \ohm and 25 \ohm, and we choose R_L on the order of kilo \ohm. Therefore, we can ignore R_L / (R_L + r_d) part, safely.

And for DC voltage corresponding to the output signal we have,

V_DC = 2 * V_out / \pi = 0.637 * V_out.

Sheme of Circuit in an Experiment

Here is circuit scheme,

As you may see, for positive half-periode only two of four diode is in conduction. And for negative half-periode, the other two is in conduction. Thus n is 2 for this circuit. Let's construct this experiment on Tinkercad. I didn't use breadboard to show more similarity between the scheme of circuit and the circuit built in Tinkercad.

Scenerio #1 - Theoretical Expectations

Let's assume 6.2 V to be the amplitude. Then, V_in=6.2 V. And V_peaktopeak is 12.4 V. As output signal we calculate,

V_out = V_in - n * V_diode = 6.2 V - 2 * 0.7 V = 4.8 V.

And for DC equivalent we theoretically get,

V_DC = 0.637 * V_out = 3.06 V.

But in multimeter, we see 1.06 V. This indicates nearly %60 percantage error.

Scenerio #2 - Theoretical Expectations

Let's assume 6.2 V to be the peak-to-peak voltage. Then, V_in=3.1 V. And V_peaktopeak is 6.2 V. As output signal we calculate,

V_out = V_in - n * V_diode = 3.1 V - 2 * 0.7 V = 1.7 V.

And for DC equivalent we theoretically get,

V_DC = 0.637 * V_out = 1.08 V.

And in multimeter, we see 1.06 V. There values are pretty close to each other.

Conclusion

Based on these results, we may conclude that 6.2 V is peak-to-peak voltage, scheme on the function generator is true, the tag "Amplitude" in the description of function generator is wrong and the y-scale of an oscilloscope represents the total voltage half of which is positive and the other half is negative.

BUT

I cannot be sure, and since I'll teach this material in my electronic laboratory class, I really need to be sure about this conclusion. Therefore, here I'm asking you about your opinions, conclusions or maybe other references that I've missed.

TinkerCAD refers to peak-to-peak voltage as amplitude for some reason. I believe the second explanation (+/- 5V, 10 V total) is correct, based on the x axis and frequency value.