Why isn't standard deviation influenced by outliers?

Good morning Sophie 9I know it's Sunday morning fairly early (sorry to wake you - I don't run a B & B, but you can share my full English if you like! (I could do with reducing my weight by 10 g!)

Ok - down to business:-
When you take range, it gives the minimum value TO maximum value (the difference between the two OR the actual min and max, yeah? So, if you had results like these:

11, 14, 2, 13, 13, 12, 51 (e.g.)
then range would be: 2 to 51 (= massive in comparison to actual values or mean, yes? (Mean = 116/7 = 16.57 [approx. 17])

But standard deviation, because it is the AVERAGE deviation from the mean as I will explain in a sec, therefore it evens out any large deviations (= outliers), still with me? (I know you are starving, and I would be happy to put an egg or some bacon into my microphone, but we do not have the technology of Star Trek yet! ).

So think with your brain and try to ignore the noise in your stomach for just a few minutes!

OK I'll prove it you in reality now:
SD for this e.g. = (approx..[mean = 17]): [Sq root of{(17-11)squared + (17-14)squared + (17-2)squared, etc etc …..}]/n-1
= [sq root of{36 + 9 + 225 + 16 + 16 + 25 + 1156}]/6

= [sq rt of 1483]/6 = 38.5/6 = 6.4

SO TO SUMMARISE: With Range: Mean 17 +- (up to 34 [51-17] do you see?)
With Standard deviation: 17 +- 6.4

So range is plus/minus 34
SD s plus/minus 6ish - a world of difference, yeah?

Can you now see why SD is hardly affected by outliers (actually, in a well-designed study with a much larger sample (e.g. tens of 1000s of patients in famous Framlingham study on heart disease ages ago - google it), any odd few outliers will have a negligible effect on SD – also our outliers in this e.g. are extreme [2 and 51 are very far from other average teens values, agreed?]).).

You are allowed to GO TO BED AGAIN NOW! Good night!! zzzzzzzzzzzzzzzz---

M