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# Maths! watch

1. Hi
How do you carry out standard deviation?
thanx!
Hi
How do you carry out standard deviation?
thanx!
you work out the mean of all your numbers. you take the mean away from all those numbers, and square these new numbers. you then work out the mean of this new set of numbers, then squareroot it.

i.e. sd²=mean[(x-mean(x))²]
3. (Original post by elpaw)
you work out the mean of all your numbers. you take the mean away from all those numbers, and square these new numbers. you then work out the mean of this new set of numbers, then squareroot it.

i.e. sd²=mean[(x-mean(x))²]

What would you use standard deviation to prove?
What would you use standard deviation to prove?
its not really used to prove anything, the same way the mean doesnt prove anything. its just a way of measuring how spread-out the data is.

for example {-1,0,1} and {-100,0,100} have the same mean, but the second one has a higher sd because it is more spread out
5. (Original post by elpaw)
its not really used to prove anything, the same way the mean doesnt prove anything. its just a way of measuring how spread-out the data is.

for example {-1,0,1} and {-100,0,100} have the same mean, but the second one has a higher sd because it is more spread out

right, so if the answer 4 the standard deviation was 24.7, what would you say?
right, so if the answer 4 the standard deviation was 24.7, what would you say?
what can you say? it depends on what data you have.
7. (Original post by elpaw)
what can you say? it depends on what data you have.

k. so say I had the data for the BMIs for 13 female students in year 7 and the standard deviation thingy answer was 24.7 wot wud I say?
8. The standard deviation figure (in this case 24.7) shows that 68% of the data lies 24.7 units either side of the mean. Which if you're looking at BMIs suggests a wide spread. So wide that I'd try and calculate it again.
k. so say I had the data for the BMIs for 13 female students in year 7 and the standard deviation thingy answer was 24.7 wot wud I say?
as it stands, on its own the number does not mean anything. you have to give it a scale to work with. the only thing you can say so far is that the results are varied. (which is obvious from the outset anyway)
10. (Original post by XTinaA)
The standard deviation figure (in this case 24.7) shows that 68% of the data lies 24.7 units either side of the mean. Which if you're looking at BMIs suggests a wide spread. So wide that I'd try and calculate it again.

I just made the number up. but if it was the number and the mean was say 30, half the data wud be up to 24.7 units below it and the other half of the data would be up to 24.7 units above it? right? So if this were true, I would say that there is an immense variation in the bmi's of female students in year 7?
I just made the number up. but if it was the number and the mean was say 30, half the data wud be up to 24.7 units below it and the other half of the data would be up to 24.7 units above it? right? So if this were true, I would say that there is an immense variation in the bmi's of female students in year 7?
No, 34% lies 24.7 units below, 34% lies above. The smaller the figure, the less the spread.
Hi
How do you carry out standard deviation?
thanx!
dead easy!!!! The square root of the mean of the squares minus the square of the mean
13. (Original post by XTinaA)
No, 34% lies 24.7 units below, 34% lies above. The smaller the figure, the less the spread.
sooooo how do you know it is 34%????
sooooo how do you know it is 34%????
At GCSE you just take it as a given. You may find out at A-Level when you study the normal distribution in S1...
sooooo how do you know it is 34%????
from the normal distribution
16. (Original post by XTinaA)
At GCSE you just take it as a given. You may find out at A-Level when you study the normal distribution in S1...

is it always 32% then?
is it always 32% then?
sorry, meant 34%!
is it always 32% then?
It's always 68% either side for one standard deviation. For 99% of the data you need to go 2.58 standard deviations either side.
is it always 32% then?
yes, always 68% of the data lies within + or - the sd of the mean

(assuming data is normally distributed)
20. (Original post by elpaw)
yes, always 68% of the data lies within + or - the sd of the mean

(assuming data is normally distributed)
Paw, you said you did the remainder theorem by a way other than long division, how?

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