OK first of all you're doing some mass averaging here. Thats NOT the same thing as a median!
your using median.. thats so much fail on you
4,4,5,8,16
your median is 5
average is 7.4
so your under estimating.. (your pink fluffy clouding numbers)
EG
if there was only 5 seats on a bus and the bus had a 'bid for seat' system where by 10people offered
4,4,4,5,5,5,5,5,8,16
these people will get to sit down
5,5,5,8,16
these wont
4,4,4,5,5
so if you looked at a median and seen so far median is 5 and you paid
5. you would have been at the back of the queue of the people that paid 5 so you as an 11th person would be part of the 4,4,4,5,5,
5 lot that wont get a seat
where as by going for average 7.4 i would have jumped the queue your stuck in and i would have got a seat
5,5,7.4,8,16
leaving you and another 5 payer waiting for the next bus, HOPING to get on
4,4,4,5,5,
5,5
then 10 further fresh people see the average is now 8.28
4,4,4,5,5,
5,5 + 8,8,8,9,9,9,9,9,16,16,32,32
guess what
9,16,16,32,32
get on the bus... and your still waiting
franky1, if your data doesn't have any outliers, then by any means, use the average. But in reality, there are outliers. It does not always come out in the set of numbers that you posted.
4, 4, 5, 8, 500 can happen too. Would you say that the average of that would be right at the estimated middle?