• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Standard deviation question ...

E

edwardraff

Senior member
I have a statistics question ?
If you have average sales of 10 cars/week with a standard deviation of 2
The average of three weeks would be 30 and the standard deviation for three weeks would also just triple to 6? Or do you have to do a more complicated calculation to find the standard deviation using something like variance? If you have to do another calculation could you show/explain it?

Thanks in advance ?

 
StdD(1 week) = 2 = sqrt(sum(i=1 to n of (X-Xi)^2)/(n-1))
4 = sum(i=1 to n of (X-Xi)^2)/(n-1)
4*(n-1) = sum(i=1 to n of (X-Xi)^2)

now expand that to three weeks
StdD(3 weeks) = sqrt(sum(i=1 to 3n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*sum(i=1 to n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*4*(n-1)/(3n-1))
StdD(3 weeks) ~= sqrt(4) ~= 2

Edit: the equation for standard deviation can also be written as this:
sqrt(sum(i=1 to n of (X-Xi)^2)/n)
which makes that derivation come out to exactly 2 for three weeks.

 
Originally posted by: Kyteland
StdD(1 week) = 2 = sqrt(sum(i=1 to n of (X-Xi)^2)/(n-1))
4 = sum(i=1 to n of (X-Xi)^2)/(n-1)
4*(n-1) = sum(i=1 to n of (X-Xi)^2)

now expand that to three weeks
StdD(3 weeks) = sqrt(sum(i=1 to 3n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*sum(i=1 to n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*4*(n-1)/(3n-1))
StdD(3 weeks) ~= sqrt(4) ~= 2

Edit: the equation for standard deviation can also be written as this:
sqrt(sum(i=1 to n of (X-Xi)^2)/n)
which makes that derivation come out to exactly 2 for three weeks.
Click to expand...

that's wrong, son. see my 2 posts above for correct derivation.
 
I think the answer is that you cannot ascertain what the standard dev would be from the information the OP gave. It would depend on the data.

For example. If I had a weekly sales average of 10 and a std dev of 2 for a year long period, you could not compute the monthly figures from that data alone.

More detalied example: week 1, week 2...etc = 10, 9, 10, 11, 10, 9, 10, 11

Average is 10 std dev is .75

If you put that data into monthly figures you would get 40, 40.

Average is 40, std dev is 0
 
Originally posted by: welst10
that's wrong, son. see my 2 posts above for correct derivation.
Click to expand...
I believe the equation you provided is incorrect.

Edit: I must say yours isn't really a derivation as much as it is an equation. Where did that come from? Please point out the mistake in my derivation. I used the definition of of StdDev and algebraic manipulation to show it is still 2.
 
right... does anyone have commonly accepted answer ... there seem to be at least three different theories (school of thought) going on in this thread at the moment ... anyone?
 
Originally posted by: ggnl
I think the answer is that you cannot ascertain what the standard dev would be from the information the OP gave. It would depend on the data.

For example. If I had a weekly sales average of 10 and a std dev of 2 for a year long period, you could not compute the monthly figures from that data alone.

More detalied example: week 1, week 2...etc = 10, 9, 10, 11, 10, 9, 10, 11

Average is 10 std dev is .75

If you put that data into monthly figures you would get 40, 40.

Average is 40, std dev is 0
Click to expand...

 
GOOOOOOOOODDAMMIT. Variance is additive assuming independant variables. So if your standard deviation is 2, variance is 4. Variance over three trials is additive so total variance is 12. Total standard deviation then becomes sqrt(12)
 
Originally posted by: Kyteland
Originally posted by: welst10
that's wrong, son. see my 2 posts above for correct derivation.
Click to expand...
I believe the equation you provided is incorrect.
Click to expand...

Keep in mind that there's more than one way to compute standard deviation.
 
Originally posted by: Kyteland
StdD(1 week) = 2 = sqrt(sum(i=1 to n of (X-Xi)^2)/(n-1))
4 = sum(i=1 to n of (X-Xi)^2)/(n-1)
4*(n-1) = sum(i=1 to n of (X-Xi)^2)

now expand that to three weeks
StdD(3 weeks) = sqrt(sum(i=1 to 3n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*sum(i=1 to n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*4*(n-1)/(3n-1))
StdD(3 weeks) ~= sqrt(4) ~= 2

Edit: the equation for standard deviation can also be written as this:
sqrt(sum(i=1 to n of (X-Xi)^2)/n)
which makes that derivation come out to exactly 2 for three weeks.
Click to expand...

I bolded the corresponding error. You cannot factor out the 3 because Xi is a random variable that I believe is independant of Xj where j != i.
 
Originally posted by: ggnl
I think the answer is that you cannot ascertain what the standard dev would be from the information the OP gave. It would depend on the data.

For example. If I had a weekly sales average of 10 and a std dev of 2 for a year long period, you could not compute the monthly figures from that data alone.

More detalied example: week 1, week 2...etc = 10, 9, 10, 11, 10, 9, 10, 11

Average is 10 std dev is .75

If you put that data into monthly figures you would get 40, 40.

Average is 40, std dev is 0
Click to expand...

That's just bad. That's like flipping a FAIR coin 4x and getting HHHT and concluding that the propability of a head is definitely 75%.
 
Originally posted by: TuxDave
Originally posted by: Kyteland
StdD(1 week) = 2 = sqrt(sum(i=1 to n of (X-Xi)^2)/(n-1))
4 = sum(i=1 to n of (X-Xi)^2)/(n-1)
4*(n-1) = sum(i=1 to n of (X-Xi)^2)

now expand that to three weeks
StdD(3 weeks) = sqrt(sum(i=1 to 3n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*sum(i=1 to n of (X-Xi)^2)/(3n-1))
StdD(3 weeks) = sqrt(3*4*(n-1)/(3n-1))
StdD(3 weeks) ~= sqrt(4) ~= 2

Edit: the equation for standard deviation can also be written as this:
sqrt(sum(i=1 to n of (X-Xi)^2)/n)
which makes that derivation come out to exactly 2 for three weeks.
Click to expand...

I bolded the corresponding error. You cannot factor out the 3 because Xi is a random variable that I believe is independant of Xj where j != i.
Click to expand...
That is not an error. The assumption is that the average on any given week is 10 and the standard deviation is 2. For every week. That means that a 3 week period can be broken down in 3 1 week periods in that manner.

Take a concrete example using these sets.

{8,8,8,8,8,12,12,12,12,12}
{8,8,10,10,10,10,10,10,12,14}
{7,9,9,9,9,9,10,11,13,14}

The average is 10 and stdDev is 2 for all three individually.
The average is 10 and stdDev is 2 also for all three sets taken as a whole.

Please provide a similar example that works for StdDev = 6.
 
Originally posted by: Kyteland

That is not an error. The assumption is that the average on any given week is 10 and the standard deviation is 2. For every week. That means that a 3 week period can be broken down in 3 1 week periods in that manner.

Take a concrete example using these sets.

{8,8,8,8,8,12,12,12,12,12}
{8,8,10,10,10,10,10,10,12,14}
{7,9,9,9,9,9,10,11,13,14}

The average is 10 and stdDev is 2 for all three individually.
The average is 10 and stdDev is 2 also for all three sets taken as a whole.

Please provide a similar example that works for StdDev = 6.
Click to expand...

First of all, I made a mistake, I got StDev and Variance backwards. It's variance that is additive, if Stdev=sigma and Variance = sigma^2, then you add the variance. The correct answer for 3 weeks is:

sqrt(12)... gimme a sec to address your post after I correct all my previous posts.
 
Originally posted by: TuxDave
Originally posted by: ggnl
I think the answer is that you cannot ascertain what the standard dev would be from the information the OP gave. It would depend on the data.

For example. If I had a weekly sales average of 10 and a std dev of 2 for a year long period, you could not compute the monthly figures from that data alone.

More detalied example: week 1, week 2...etc = 10, 9, 10, 11, 10, 9, 10, 11

Average is 10 std dev is .75

If you put that data into monthly figures you would get 40, 40.

Average is 40, std dev is 0
Click to expand...

That's just bad. That's like flipping a FAIR coin 4x and getting HHHT and concluding that the propability of a head is definitely 75%.
Click to expand...

Bad or not, it demonstrates my point. If you are working with large samples, then yes, the SD probably would qradruple in my example. But, there is no way to ascertain exactly what the SD would be given the information in the OP. The only way to compute it would be to go back to the original data.
 
Originally posted by: TuxDave
Ok dammit... I'm too lazy to argue with you, so just read this:

http://www.stat.yale.edu/Cours...997-98/101/rvmnvar.htm

Scroll down to

For independent random variables X and Y, the variance of their sum or difference is the sum of their variances:

QED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!ONE
Click to expand...
welst10 and TuxDave,

I don't disagree with your equation. I just disagree that the variables are independant at that point. If you force the condtraints of Avg=10 StdDev=2 upon the sets X and Y then the standard deviation of the two sets combined will still be 2.

If the variable were really independant and random then it yould be possible to have individual sets with higher/lower Avg/StdDev, as long as the overall data fit with avg=10 StdDev=2. For instance, a week of 8 +-4 is possible, although it's probability of occuring is different than 9.5 +- 2.2 week.

In this case StdDev=6, but once you set constraints on the data that every set must be Avg=1- StdDev=2 then the resulting StdDev will be 2 when the sets are combined.
 
Originally posted by: Kyteland
welst10 and TuxDave,

I don't disagree with your equation. I just disagree that the variables are independant at that point. If you force the condtraints of Avg=10 StdDev=2 upon the sets X and Y then the standard deviation of the two sets combined will still be 2.

If the variable were really independant and random then it yould be possible to have individual sets with higher/lower Avg/StdDev, as long as the overall data fit with avg=10 StdDev=2. For instance, a week of 8 +-4 is possible, although it's probability of occuring is different than 9.5 +- 2.2 week.

In this case StdDev=6, but once you set constraints on the data that every set must be Avg=1- StdDev=2 then the resulting StdDev will be 2 when the sets are combined.
Click to expand...

Simplified to clear up all confustion

I bolded the parts where you are correct. He didn't state that every set (an arbitrary sample of sales per week) has to have a mean of 10 and stdev of 2. Just overall with infinite samples, it has a mean of 10 and stdev of 2. Therefore, they are independant by the phrasing of the question.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top