% of above average height uk

[lewy2702]

What percentage of people in the uk are above average height?

[jenstar]

Let's say 50%.

[jills]

jenstar.....LOL!!

[bernardo]

If by "average" you mean "median", then 50%. If by "average" you mean "mean", then I guess about 80% (i.e the 80% are the adults and the 20% are the children). If you mean excluding the children, I guess it's about50%.

[Quizmonster]

Excuse my homersimpsonism again - doh! - but...being totally unmathematical, I've often wondered about the word 'average' in a context such as this. Jenstar's "50%" answer seems the obvious one and, by the same token, the percentage below average would also be 50%.

But - with half above and half below - wouldn't that mean that nobody actually was of average height? That can't be right, surely. Wouldn't there be more people on that height than any other, suggesting the percentage would be quite high rather than 0%? Yeah...I know...I'm as thick as two short planks!

[IndieSinger]

Isn't it 50% minus the percentage of people who are that exact (average) height?

[BenDToy]

There are 3 (to my knowledge) types of average. The "mean" is a theoretical figure which is arrived at by adding all the figures, and dividing by the number of bits of data. ie the mean of 10 and 6 is 8 (10+6 divided by 2 = 8). So QM is sort of right, there may not be anyone of average height, (and the more accurate you are, the fewer people there will be). The "median" is the value which has an equal number of values greater and smaller than it. The "mode" is the most frequently occurring value.

[Quizmonster]

Thanks for that, Indie and Ben. I was vaguely conscious of these other words/ideas, but I have to confess I'm still a bit puzzled. Is the original answer here...50%...actually wrong in this situation? Cheers...sorry for the delay in responding; I was away yesterday.

[Grant]

Well it depends how you define the 'average'. If the arithmetic mean is used ( sum of all the figures divided by number in the sample)then the number above the average would depend on the variance.