Need help, rounding sample values in AD conversion

Hi everyone,
I am a newbie and i need this information,

As far as I understand quantization, the sampled voltage values of an analogue signal can be either truncated to next lower quantization level or rounded to closest quantization level.
But only examples that i can found on internet looks like truncation, here is an example:

Example 1
An analogue signal of between 0 volts to 10 volts to be represented by a 3-bit binary word. A 3-bit code
has 2^3 or 8 combinations. We need to divide the analogue signal into 8 subranges. Each sub-range will
represent a voltage range of 10/8 volts. This gives 1.25 volts per sub-range, the first sub-range starting at
0 volts.
D = 2^n x (Vin/Vref) = 2^3 x (Vin/10)

Range Voltage Range Binary Code
0 0.0 –> 1.25 000
1 1.25 –> 2.5 001
2 2.5 –> 3.75 010
3 3.75 –> 5.0 011
4 5.0 –> 6.25 100
5 6.25 –> 7.5 101
6 7.5 –> 8.75 110
7 8.75 –> 10.0 111


What i understand from above is that if Vinput is 1.24 the output will be 000 which is truncation.
So my question is, how do we round sample values?

You need a shift of 0.5LSB so would have to add that before the ADC i.e. an analogue shift.

Keith

Thank you very much, this makes sense,
So this shift should be applied between sample/hold and input of comparator, increasing the input voltage by an amount of 1/2 LSB, am I right?

Is this the common practice for rounding sample values, because although many books about digital audio states that the sample values are rounded to the closest quantization interval, i couldn't find an example or block diagram showing that kind of shift, so i started to think that the reference values coming to the comparator might be shifted accordingly; but again couldn't find any info

Forgive my ignorance in electronics, i just want to understand how this issue is handled in a within a commercial audio AD converter
Thanks again