Answer:
12.1, 12.3,12.4,12.5,12.3,12.1,12.2

And for the standard deviation we can use the following formula:

And after replace we got:

And as we can ee we got a small value for the deviation <1 on this case.
Step-by-step explanation:
For example if we have the following data:
12.1, 12.3,12.4,12.5,12.3,12.1,12.2
We see that the data are similar for all the observations so we would expect a small standard deviation
If we calculate the sample mean we can use the following formula:

And replacing we got:

And for the standard deviation we can use the following formula:

And after replace we got:

And as we can ee we got a small value for the deviation <1 on this case.