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# S1 Help (GCE- A Levels) watch

1. Hi I'm not too sure how to work out the variance on this question. I tried doing it but come out with a negative number.

https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf

It's question 2 on this link.

In terms of what I done I did this:
(77+(2^2 x 82)+(2^2 x 87)+(4^2 x 92)..../15 )- mean^2 (which I believe to be 92.3
2. (Original post by Xphoenix)
Hi I'm not too sure how to work out the variance on this question. I tried doing it but come out with a negative number.

https://1a388b28f5ce9318f837a64962fc...%20Edexcel.pdf

It's question 2 on this link.

In terms of what I done I did this:
(77+(2^2 x 82)+(2^2 x 87)+(4^2 x 92)..../15 )- mean^2 (which I believe to be 92.3
It should be [(1 x 77^2) + (2 x 82^2) +.... etc] - mean^2

In other words the 'x' terms needs to be squared
3. (Original post by RoseGatz)
It should be [(1 x 77^2) + (2 x 82^2) +.... etc] - mean^2

In other words the 'x' terms needs to be squared
Thanks! Just done it and looked at the mark scheme however I still have a different answer. I got 71.04333333

my sum of all terms is 128855. Then I divided by number of terms (15) to get 8590.3333
then Ans- (92.3^2)

Any clue to where I may have gone wrong?
4. (Original post by Xphoenix)
Thanks! Just done it and looked at the mark scheme however I still have a different answer. I got 71.04333333

my sum of all terms is 128855. Then I divided by number of terms (15) to get 8590.3333
then Ans- (92.3^2)

Any clue to where I may have gone wrong?
It's because you've used the rounded version of the mean i.e 92.3, if you use the original/full value of the mean where the 3 is recurring, you'll get the right answer.

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