ISO 1938 calculation example (plain limit gauges of linear size)

hello Gentelmens,

with regards to ISO 1938...
I have it in front of my eyes and I am not sure if I am doing this correctly

I have a slot (to be checked with bar gauge type F) with envelope requirement, let's say 20mm +/- 0.2

Cause the size tolerance is not given as an ISO code, I am using first IT grade from Table7-11 with a tolerance interval lower than or equal to the tolerance interval in the same nominal size range

So 0.4mm is 400 micrometers, for size 20 it would be IT13 (T=330, z=36), table 10

Now, in the formula for the new state "z" is one of the factors.
There is also H - for the form limit values which are "half the values given in column 2xF in Table 6"...

So now I am going to Table 6.
For IT13 bar gauge:
size H - IT7
form and orientation 2xF - IT5

Going to ISO 286 IT grades table:
Size H for IT7 (20mm) = 21 micrometers
form and orientation 2xF for IT5 (20mm) = 9 micrometers (<- this shall be multiplied by 2?...)

can someone check if my values selection from tables is correct?