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<font color='#000080'>Right, first correct answer wins on this one. God I hate uni work. I'm not a Mathematician, so I want answers.

If I have data for the height of men, and the mean and 5th and 95th percentiles, can I work out what percentile will be excluded by a cut-off at a certain height?

To give you my data, the 5th percentile is 1625mm, mean 1740mm, and 95th percentile 1855mm. Now I want to now what percentage is above 1800mm, and what percentage is above 2000mm.

Answers on a postcard. And no, this is not cheating. It's using my networking skills to save me half an hour.

And any working must be included, or you won't get all the marks.

Digs.
 

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Jonah
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Not accurately, you'd need the standard deviation as well as the mean.

What you could do - assuming the distribution is pretty close to normal (which it should be from height) is work out what the standard deviation is from the percentiles, then do the calculation to work out what proportion is beyond the cutoff points you  mentioned.

If you have no idea what I'm talking about then forget it, but that's the approach I'd probably use to do the calculation if I had time (busy right now).
 

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The Bridget Jones of Diving
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Yes you can.

I can't remember exactly how to do it though  


I've been taught statistics about three times now, and each time I manage to remember how to do it for the exam, and then it just disappears from my brain!

iirc 96% is three standard deviations from the mean, and there are figs for what % falls at the 1st and 2nd. Can't remember much more than that unfortunately


Pretty Flowers.... proud to work in Market Research!
 

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<font color='#000080'>OK, we now have the SD for everyone that couldn't answer that. Standard Deviation for the data is 70.

Now who gets the beer, eh?
 

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<font color='#8D38C9'>if I had a clue about anything I did at uni I could help you but I don't
to busy diving
 

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<font color='#8D38C9'>oo no i guess at 3%

I'm going home now if your still stuck at 7pm pm me and I'll get the book out
 

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I have suggested Lou (Fluice) has a look to relieve her boredom at work. I knew the units question is the first thing she would ask. You never know, PhD in maths and all that!!

Paul
 

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Jonah
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OK, based on the figures you supplied (mean 1740 and SD 70), 1800 is 0.857 SD above the mean. 2000 is 3.714 SD above the mean.

According to normal distribution tables, the percentage beyond 1800 is 19.49%. You might want to round that to 20%. The proportion beyond 2000 is 0.01%.

How much beer is that?
 

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<font color='#000080'>Sorry Tom, but the beer goes to my maths monkey, who I have conveniently employed as a girlfriend.

She has the answer of 99.98% for 2000mm, and 80.23% for the 1800mm. And the numbers work! They work I tell you!

Accuracy is the key. Oh, and you've not put your working down, so you only get a mark for each answer. Half a pint of cheap lager to you at our next gig. Nice try.
 

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<font color='#000080'>Just a sec. Right answer, but still no working. I give you one point. You get one pint. Pints mean prizes!
 

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Even my biochemical knowledge of maths tells me that if the mean is 1740mm, 99.98% cannot be over 2000mm. I would also say that even the figure over 1800 is looking a bit dodgy?

Paul
 

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Jonah
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Better spank your monkey, she's wrong. What she's given you are the percentages BELOW the cutoff points, not above. Think about it.
 

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<font color='#0000FF'>Is th answer 42 or is that a different question


Seriously though, good luck. Its beyond me nowadays.
 

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<font color='#000080'>that is undeniably cheating Digger, which I approve of. Who's the lawyer on here...I might try the same approach!
 

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Diggs...

If the 5th and 95th percentile values are acuurate then the standard deviation should come to 69.909 (3dp)

Using that value of S.D....
Below 1800 is 80.511%
Above 2000 is 0.01%

I can post working to back up my answer if you want it!!!


Mr Bez.
A-Level Statistics Teacher.
 
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