dnd 5e – When performing a group check, is it better to have an odd number of creatures? And more creatures?

The rules for group checks say:

To make a group ability check, everyone in the group makes the ability check. If at least half the group succeeds, the whole group succeeds.

So, for groups of 2 and 3, 2 need to succeed. For groups of 4 and 5, 3 need to succeed. And so on.

Running this anydice:

loop N over {1..8} {
  output ((N+2)/2)@Nd20


enter image description here

Each of the odd-numbered groups has a symmetric distribution centred on 10.5 – this makes sense because we are looking at the middle dice in the group. That is, the median and the mean are always 10.5. As you add more creatures, the variance decreases which makes you less likely to succeed at hard things but more likely to succeed at easy things.

With the even-numbered groups, we are looking at the $nover2$ dice which will always be lower. So even-numbered groups are at a disadvantage than the next-larger odd-numbered group. That means that 3 creatures can, say, Hide more easily than 2, 5 more easily than 4 and so on.

Further, the more creatures there are in even-numbered groups the higher the mean and median but the lower the variance. So 2 are likelier to do worse than 4 who are more likely to do worse than 6 and so on unless the check is really hard.

Wierd practical outcomes

If the DC is such that the target for the d20 roll is 10, say, a group trying to use Dexterity (Stealth) of $x$ against an observer with an equal Wisdom (Perception) modifier then it’s easier to hide the more people you throw at it. For odd-numbered groups the more the merrier; it’s always better than 1. For even-numbered groups, you want to get at least 8 people together; 8 are better than 1, 10 are better than 8, and so on.

enter image description here

However, when the target is 11, odd-numbered groups always perform the same but even-number groups start bad but get better. For 12, odd-numbers get worse and even-numbers get better. This is for sizes up to 8, eventually, the shrinking variance will mean that you are always getting 10s or 11s.

Is this analysis correct?