Confidence interval calculation: why sometimes the SD is dividted by the sqrt of sample size, and sometimes not? I have trouble understanding the following:
Looking how reference ranges for laboratory values are calculated, I found the following:

In our sample of 72 printers, the standard error of the mean was 0.53
mmHg. The sample mean plus or minus 196 times its standard error gives
the following two figures:
88 + (1.96 x 0.53) = 89.04 mmHg
88 – (1.96
x 0.53) = 86.96 mmHg.
This is called the 95% confidence interval , and we can say that there
is only a 5% chance that the range 86.96 to 89.04 mmHg excludes the
mean of the population.

That is clear to me: I multiply the Z for 2 SD by the value of SD and subtract and add it to the mean to define the limits.
However, I often see another definition of CI, that divides the SD by the square root of the number of samples. The result is therefore quite different, in this case, it would be 88 ± 0.122 (vs 88 ± 0.53 in the calculation above).
What is the difference and meaning of that? In the latter, how would I call the value of 0.122 ( I guess that is Margin of Error)

You're mixed up about standard deviation and standard error. Consider learning the difference to be a rite of passage, as I think every statistician is confused about this early in their training.
When you see the standard error stated as $0.53$, that has already divided by $\sqrt{n}$. Standard error is the standard deviation divided by $\sqrt{n}$ (though more advanced statistics will expand on this). This means that the standard deviation is $0.53\sqrt{72} \approx 4.5$.
Then when you go to calculate the standard error, you divide the standard deviation by $\sqrt{n}$, resulting in the expected $4.5/\sqrt{72} \approx 0.53$.
The value of $0.122$ comes from mistakenly dividing the standard error, rather than the standard deviation, by $\sqrt{n}$.
