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Basic Standard Deviation

Find the standard deviation

15.6, 17.3, 19.4, 16, 18.3, 17.5


fx = 104.1
fx^2 = 1816.15

(1815.15 / 6) - (104.1 / 6)^2 = 1.5025 >>>>> sq1.5025 = 1.23

isnt giving me 1.42

where am i going wrong?
(edited 10 years ago)

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Reply 1
Multiply your answer by the square root of 6/5. Why?
Reply 2
Original post by BabyMaths
Multiply your answer by the square root of 6/5. Why?



sorry im confused
Reply 3
Original post by cera ess six
sorry im confused

I get 1.31 as my standard deviation? In your workings you put 1815.15 when it should be 1816.5. Regardless I don't get 1.42 either.
Reply 4
Original post by Super199
I get 1.31 as my standard deviation? In your workings you put 1815.15 when it should be 1816.5. Regardless I don't get 1.42 either.


the calculator gives 1.42 as well the answer sheet so i guess we both doing something wrong
Standard deviation = square root of [(the square of the means) minus (the mean of the squares)].
Reply 6
Your calculator and the answer uses Bessel's correction. Which is using (N-1) instead of N. If you change 6 (N) to 5 (N-1) you'll get 1.42
Reply 7
Original post by MuzzyYT
Your calculator and the answer uses Bessel's correction. Which is using (N-1) instead of N. If you change 6 (N) to 5 (N-1) you'll get 1.42

That just gives you a maths error :confused:
Reply 8
Original post by Super199
That just gives you a maths error :confused:



15.6, 17.3, 19.4, 16, 18.3, 17.5


fx = 104.1 , mean = 17.35
fx^2 = 1816.15

Altered to correction.

Formula SQRT( (Fx^2/n) - (mean)^2)
SQRT ( 1816.5/6 - (17.35^2) = SQRT(1.669) = 1.29

Bessel's Correction. If it's an old question, maybe this is why this is used, otherwise it may be the context. I can only remember the correction for the alternate formula of SD, being the

SQRT( Sigma(X-mean)^2) / n)
SQRT(10.015/6) = SQRT(1.669) = 1.29

With Bessel's correction

SQRT(10.015/5) = SQRT(2.0164) = 1.42
Square root the variance
Reply 10
Original post by cera ess six

fx^2 = 1816.15

(1815.15 / 6) - (104.1 / 6)^2 = 1.5025 >>>>> sq1.5025 = 1.23

isnt giving me 1.42



The error above is part of the problem. The other problem is that you should apparently be using

standard deviation =Σx2n1(Σx)2n(n1)\sqrt{\frac{\Sigma x^2}{n-1}-\frac{(\Sigma x)^2}{n(n-1)}}

or Σ(xxˉ)2n1 \sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}}.

This gives the unbiased estimate of the population standard deviation.
Reply 11
Original post by BabyMaths
The error above is part of the problem. The other problem is that you should apparently be using

standard deviation =Σx2n1(Σx)2n(n1)\sqrt{\frac{\Sigma x^2}{n-1}-\frac{(\Sigma x)^2}{n(n-1)}}

or Σ(xxˉ)2n1 \sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}}.

This gives the unbiased estimate of the population standard deviation.



why do you have to do the (n-1)
Original post by cera ess six
why do you have to do the (n-1)


It provides an unbiased estimate.
Reply 13
Original post by cera ess six
why do you have to do the (n-1)


This may not mean much at the moment but it does answer your question.
Reply 14
Original post by MuzzyYT
15.6, 17.3, 19.4, 16, 18.3, 17.5


fx = 104.1 , mean = 17.35
fx^2 = 1816.15

Altered to correction.

Formula SQRT( (Fx^2/n) - (mean)^2)
SQRT ( 1816.5/6 - (17.35^2) = SQRT(1.669) = 1.29

Bessel's Correction. If it's an old question, maybe this is why this is used, otherwise it may be the context. I can only remember the correction for the alternate formula of SD, being the

SQRT( Sigma(X-mean)^2) / n)
SQRT(10.015/6) = SQRT(1.669) = 1.29

With Bessel's correction

SQRT(10.015/5) = SQRT(2.0164) = 1.42

Ah I see... So when do we use Bessel's correction? We were never actually taught this so it's a new thing. But thanks :smile:
Reply 15
Original post by Super199
Ah I see... So when do we use Bessel's correction? We were never actually taught this so it's a new thing. But thanks :smile:


It was never even mentioned last year when I did stats in AS maths. The only reason I know about it is because we were taught SD in biology too, and there was a little fued as our Teacher and teacher's calculator used N-1, but this calculator was so old you had to practically shovel coal into it to make it work.

We were told it's usually used to remove bias and provide a better value from a sample which is not a totality.
For example if you were to measure the height of a sample of 20 giraffes, you'd probably use Bessel's Correction(N-1), but if you were to measure the height of every giraffe in existence, then it would be N.
(edited 10 years ago)
Reply 16
Original post by Super199
Ah I see... So when do we use Bessel's correction? We were never actually taught this so it's a new thing. But thanks :smile:


The only time you need to use it is if you have taken a sample of data from a population and you want to estimate the variance of the population from the sample data.

Unfortunately some statistics courses seems to include it as a "standard" alternative to the variance which is obtained by dividing by n, and this leads to a lot of confusion!
Reply 17
Original post by BabyMaths
This may not mean much at the moment but it does answer your question.



Original post by MuzzyYT
15.6, 17.3, 19.4, 16, 18.3, 17.5


fx = 104.1 , mean = 17.35
fx^2 = 1816.15

Altered to correction.

Formula SQRT( (Fx^2/n) - (mean)^2)
SQRT ( 1816.5/6 - (17.35^2) = SQRT(1.669) = 1.29

Bessel's Correction. If it's an old question, maybe this is why this is used, otherwise it may be the context. I can only remember the correction for the alternate formula of SD, being the

SQRT( Sigma(X-mean)^2) / n)
SQRT(10.015/6) = SQRT(1.669) = 1.29

With Bessel's correction

SQRT(10.015/5) = SQRT(2.0164) = 1.42




do you have to do the whole (n-1) all the time when working out SD im sure in the past ive done it with just using N and got the correct answer
Reply 18
Original post by cera ess six
do you have to do the whole (n-1) all the time when working out SD im sure in the past ive done it with just using N and got the correct answer


See my previous answer! The n-1 version is only for estimating population variance from a sample of data - ordinarily n should be used.
(This does not stop books, authors who should know better, and even exam questions, from getting it wrong :smile: )
Reply 19
Original post by MuzzyYT
It was never even mentioned last year when I did stats in AS maths. The only reason I know about it is because we were taught SD in biology too, and there was a little fued as our Teacher and teacher's calculator used N-1, but this calculator was so old you had to practically shovel coal into it to make it work.

We were told it's usually used to remove bias and provide a better value from a sample which is not a totality.
For example if you were to measure the height of a sample of 20 giraffes, you'd probably use Bessel's Correction(N-1), but if you were to measure the height of every giraffe in existence, then it would be N.

I take it that most question in AS maths tend to be on just n and not n-1?

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