# Basic Standard Deviation

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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?

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?

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(Original post by

Multiply your answer by the square root of 6/5. Why?

**BabyMaths**)Multiply your answer by the square root of 6/5. Why?

sorry im confused

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#4

(Original post by

sorry im confused

**cera ess six**)sorry im confused

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(Original post by

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.

**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.

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#6

Standard deviation = square root of [(the square of the means) minus (the mean of the squares)].

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#7

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

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#8

(Original post by

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

**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

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#9

(Original post by

That just gives you a maths error

**Super199**)That just gives you a maths error

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

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#11

(Original post by

fx^2 = 1816.15

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

isnt giving me 1.42

**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

standard deviation =

or .

This gives the unbiased estimate of the population standard deviation.

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(Original post by

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

standard deviation =

or .

This gives the unbiased estimate of the population standard deviation.

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

standard deviation =

or .

This gives the unbiased estimate of the population standard deviation.

why do you have to do the (n-1)

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#13

(Original post by

why do you have to do the (n-1)

**cera ess six**)why do you have to do the (n-1)

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#14

(Original post by

why do you have to do the (n-1)

**cera ess six**)why do you have to do the (n-1)

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#15

(Original post by

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

**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

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#16

(Original post by

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

**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

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.

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#17

**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

**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!

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**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

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#19

(Original post by

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

**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

(This does not stop books, authors who should know better, and even exam questions, from getting it wrong )

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#20

(Original post by

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.

**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.

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