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Standard deviation

I don’t know what SD formula to use...

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Bump please help
Original post by Reece.W.J
Bump please help


No need to bump threads a few moments after creation :smile: Are you talking about,

σ=∑fx2∑f−(x‾)2\displaystyle \sigma = \sqrt{\frac{\sum fx^2} {\sum f} - (\overline x)^2}

vs.

σ=∑f(x−x‾)2∑f\displaystyle \sigma = \sqrt{\frac{\sum f(x-\overline x)^2} {\sum f}}
Reply 3
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
I don’t know what SD formula to use...

It is nearly always best to use this formula:

σ=∑x2n−xˉ2\displaystyle \sigma = \sqrt{\frac{\sum x^2}{n}-\bar{x}^2}

(Or an equivalent formula with frequency).

You may also be able to find it using your calculator. Are you doing new spec and do you have an example question?

Also, please do not bump your threads since then they will appear answered and people may not look at your question. Be patient.
Original post by Notnek
It is nearly always best to use this formula:

σ=∑x2n−xˉ2\displaystyle \sigma = \sqrt{\frac{\sum x^2}{n}-\bar{x}^2}

(Or an equivalent formula with frequency).

You may also be able to find it using your calculator. Are you doing new spec and do you have an example question?

Also, please do not bump your threads since then they will appear answered and people may not look at your question. Be patient.


New spec. I’ve done the whole table but is the sigma 4 (4 rows down) or 220 (total f)?
Reply 5
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
New spec. I’ve done the whole table but is the sigma 4 (4 rows down) or 220 (total f)?

I don't know what you're talking about (but maybe someone else will). You haven't posted an example question so I don't know where these numbers are coming from.
Original post by Notnek
I don't know what you're talking about (but maybe someone else will). You haven't posted an example question so I don't know where these numbers are coming from.


48EFCB92-CA93-47BC-BEA3-6B7AF3A298CC.jpg.jpeg
Original post by Reece.W.J
48EFCB92-CA93-47BC-BEA3-6B7AF3A298CC.jpg.jpeg


Might help using the formulae to label your sums at the bottom as ∑f\sum f, ∑fx\sum fx etc.
Original post by _gcx
Might help using the formulae to label your sums at the bottom as ∑f\sum f, ∑fx\sum fx etc.


Is midpoint Xi or X?
Reply 9
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
48EFCB92-CA93-47BC-BEA3-6B7AF3A298CC.jpg.jpeg

It looks like you are dealing with a grouped frequency table here (you should have made your question more clear initially) so you can only estimate the standard deviation.

So for a group like e.g. 10<x≤2010<x\leq 20 you assume that all the data points are equal to 15 i.e. the midpoint. The formula to use is

σ2=∑fx2∑f−xˉ2=∑fx2∑f−(∑fx∑f)2\displaystyle \sigma^2 = \frac{\sum fx^2} {\sum f}-\bar{x}^2 = \frac{\sum fx^2} {\sum f}- \left(\frac{\sum fx}{\sum f}\right)^2

So something like ∑fx\sum fx is the sum of all the midpoints multiplied by the corresponding frequencies so that's the sum of your f x MP column.
Original post by Reece.W.J
I don’t know what SD formula to use...


Remember: there's 2 types... Sample and Population.

Are you doing WJEC spec by any chance? :smile:
Original post by Notnek
It looks like you are dealing with a grouped frequency table here (you should have made your question more clear initially) so you can only estimate the standard deviation.

So for a group like e.g. 10<x≤2010<x\leq 20 you assume that all the data points are equal to 15 i.e. the midpoint. The formula to use is

σ2=∑fx2∑f−xˉ2=∑fx2∑f−(∑fx∑f)2\displaystyle \sigma^2 = \frac{\sum fx^2} {\sum f}-\bar{x}^2 = \frac{\sum fx^2} {\sum f}- \left(\frac{\sum fx}{\sum f}\right)^2

So something like ∑fx\sum fx is the sum of all the midpoints multiplied by the corresponding frequencies so that's the sum of your f x MP column.


Ok I don’t really get it when it’s typed out like that. Should I do root 220(3590-16.32)^2/5-1?
Original post by RoyalSheepy
Remember: there's 2 types... Sample and Population.

Are you doing WJEC spec by any chance? :smile:


Yes WJEC but we’ve done 3 types (small easy data, one with f, one w/o f)
Reply 13
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
Ok I don’t really get it when it’s typed out like that. Should I do root 220(3590-16.32)^2/5-1?

It will be worth you getting used to this notation. I'll rewrite it in words

σ2=sum of frequency x midpoints2sum of frequencies−(sum of frequency x midpointssum of frequencies)2\displaystyle \sigma^2 = \frac{\text{sum of frequency x midpoints}^2}{\text{sum of frequencies}}- \left(\frac{\text{sum of frequency x midpoints}}{\text{sum of frequencies}}\right)^2
(edited 6 years ago)
Reply 14
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
Ok I don’t really get it when it’s typed out like that. Should I do root 220(3590-16.32)^2/5-1?

It's not clear where all of these numbers came from. Please post your working in full.
Original post by Notnek
It's not clear where all of these numbers came from. Please post your working in full.


3944BF3C-BD48-4485-B062-7C7EE9A56AC3.jpg.jpeg
Reply 16
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Notnek
21
Original post by Reece.W.J
3944BF3C-BD48-4485-B062-7C7EE9A56AC3.jpg.jpeg

What was your final answer. I get 35.62.

No I don't, sorry! 2 secs...
(edited 6 years ago)
Original post by Notnek
What was your final answer. I get 35.62.


Your answer seems correct but I got 26503
mean of the squares take away the square of the mean is a nice way to remember variance formula
Reply 19
Avatar for Notnek
Notnek
21
Original post by Reece.W.J
Your answer seems correct but I got 26503

Sorry I read your table wrong - I actually get 12.58. Your answer is not right. Can you please post your working in full?

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