Mathematics 9709 · Probability & Statistics 1

Working from totals & combining sets

Variance only needs three totals: $n,\ \sum x,\ \sum x^2$ .
That means you can run the formula backwards to find a missing total.
It also means you can combine two groups by pooling totals.
Totals add. Variances and standard deviations do not.

Run it backwards, and combine

Totals add; variances don't. To combine two groups, pool Σx, Σx², n and re-run the ONE formula — never average the SDs.

Worked example

Backwards: find a missing total

Given $n=25,\ \sum x=275,\ \operatorname{Var}=7$ , find $\sum x^2$ .

$\tfrac{\sum x^2}{25}-11^2=7\Rightarrow \tfrac{\sum x^2}{25}=128$

Answer

\sum x^2=25\times128=3200

To combine sets $x$ and $y$ , add the totals and re-run the one formula:

$\bar X=\dfrac{\sum x+\sum y}{n_x+n_y},\qquad \operatorname{Var}=\dfrac{\sum x^2+\sum y^2}{n_x+n_y}-\bar X^2$

Pool: $N=n_x+n_y,\ \Sigma X=\sum x+\sum y,\ \Sigma X^2=\sum x^2+\sum y^2$ .
$\bar X=\Sigma X/N$ — keep it exact (don't round before squaring).
$\operatorname{Var}=\Sigma X^2/N-\bar X^2$ ; $\text{SD}=\sqrt{\operatorname{Var}}$ .

Worked example

Combine two datasets (the canonical case)

10 boys: $\sum x=1650,\ \sum x^2=275490$ ; 15 girls: $\sum y=2370,\ \sum y^2=377835$ . Pool over all 25: $\Sigma X=4020,\ \Sigma X^2=653325$ , so $\bar X=\tfrac{4020}{25}=160.8$ .

$\operatorname{Var}=\tfrac{653325}{25}-160.8^2=26133-25856.6=276.4$

Answer

\text{SD}=\sqrt{276.4}\approx 16.6\text{ cm}

Worked example

Combine when the mean is not clean

$x: n=15,\ \sum x=605,\ \sum x^2=25000$ ; $y: n=15,\ \sum y=525,\ \sum y^2=19200$ . Pool $N=30,\ \Sigma X=1130,\ \Sigma X^2=44200$ ; $\bar X=\tfrac{1130}{30}=37.6\overline{6}$ — carry it exact. (9709/52/O/N/24 Q6)

$\operatorname{Var}=\tfrac{44200}{30}-\big(\tfrac{1130}{30}\big)^2=1473.33-1418.78=54.56$

Answer

\text{SD}=\sqrt{54.56}\approx 7.39

Worked example

Reverse-combine: find a group's mean from the combined SD

120 boys: $\sum x=7020,\ \sum x^2=424320$ ; 80 girls: $\sum y^2=352130$ . The SD of all 200 is 10. Find the girls' mean.

Pool into the variance formula and set it to $10^2$ : $\tfrac{424320+352130}{200}-\big(\tfrac{7020+\sum y}{200}\big)^2=100$ , i.e. $3882.25-\big(\tfrac{7020+\sum y}{200}\big)^2=100$ .

So $\tfrac{7020+\sum y}{200}=\sqrt{3782.25}=61.5\Rightarrow \sum y=12300-7020=5280$ .

Answer

\text{girls' mean}=\tfrac{5280}{80}=66

Given	Want	Move
$n,\ \sum x,\ \operatorname{Var}$	$\sum x^2$	$n(\operatorname{Var}+\bar x^2)$
$n,\ \sum x,\ \sum x^2$	Var / SD	$\sum x^2/n-\bar x^2$
Two groups' totals	combined SD	pool, then re-run
Combined SD + one group	other group's mean	set pooled Var = SD², solve for the total

Examiner note

The method mark is the pooled variance formula with pooled totals substituted,

(\sum x^2+\sum y^2)/N-((\sum x+\sum y)/N)^2

, shown unsimplified; the exact combined mean must appear inside it (don't round, e.g. 37.67, then square). Final SD identified.

Common mistake

Never average the two variances or SDs — that ignores how far apart the two means are. If given a mean and SD per group, recover the totals first with

\sum x=n\bar x

and

\sum x^2=n(\operatorname{Var}+\bar x^2)

Why you can't average the two variances

Two groups with the same SD but different means form a combined set that is more spread than either — the gap between the means adds extra variance. So $\operatorname{Var}(\text{combined})\ne\tfrac12(\operatorname{Var}_x+\operatorname{Var}_y)$ ; you must pool the squared totals to capture that gap.

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Working from totals & combining sets

Variance only needs three totals: $n,\ \sum x,\ \sum x^2$ .
That means you can run the formula backwards to find a missing total.
It also means you can combine two groups by pooling totals.
Totals add. Variances and standard deviations do not.

Run it backwards, and combine

Totals add; variances don't. To combine two groups, pool Σx, Σx², n and re-run the ONE formula — never average the SDs.

Worked example

Backwards: find a missing total

Given $n=25,\ \sum x=275,\ \operatorname{Var}=7$ , find $\sum x^2$ .

$\tfrac{\sum x^2}{25}-11^2=7\Rightarrow \tfrac{\sum x^2}{25}=128$

Answer

\sum x^2=25\times128=3200

To combine sets $x$ and $y$ , add the totals and re-run the one formula:

$\bar X=\dfrac{\sum x+\sum y}{n_x+n_y},\qquad \operatorname{Var}=\dfrac{\sum x^2+\sum y^2}{n_x+n_y}-\bar X^2$

Pool: $N=n_x+n_y,\ \Sigma X=\sum x+\sum y,\ \Sigma X^2=\sum x^2+\sum y^2$ .
$\bar X=\Sigma X/N$ — keep it exact (don't round before squaring).
$\operatorname{Var}=\Sigma X^2/N-\bar X^2$ ; $\text{SD}=\sqrt{\operatorname{Var}}$ .

Worked example

Combine two datasets (the canonical case)

10 boys: $\sum x=1650,\ \sum x^2=275490$ ; 15 girls: $\sum y=2370,\ \sum y^2=377835$ . Pool over all 25: $\Sigma X=4020,\ \Sigma X^2=653325$ , so $\bar X=\tfrac{4020}{25}=160.8$ .

$\operatorname{Var}=\tfrac{653325}{25}-160.8^2=26133-25856.6=276.4$

Answer

\text{SD}=\sqrt{276.4}\approx 16.6\text{ cm}

Worked example

Combine when the mean is not clean

$\operatorname{Var}=\tfrac{44200}{30}-\big(\tfrac{1130}{30}\big)^2=1473.33-1418.78=54.56$

Answer

\text{SD}=\sqrt{54.56}\approx 7.39

Worked example

Reverse-combine: find a group's mean from the combined SD

120 boys: $\sum x=7020,\ \sum x^2=424320$ ; 80 girls: $\sum y^2=352130$ . The SD of all 200 is 10. Find the girls' mean.

Pool into the variance formula and set it to $10^2$ : $\tfrac{424320+352130}{200}-\big(\tfrac{7020+\sum y}{200}\big)^2=100$ , i.e. $3882.25-\big(\tfrac{7020+\sum y}{200}\big)^2=100$ .

So $\tfrac{7020+\sum y}{200}=\sqrt{3782.25}=61.5\Rightarrow \sum y=12300-7020=5280$ .

Answer

\text{girls' mean}=\tfrac{5280}{80}=66

Given	Want	Move
$n,\ \sum x,\ \operatorname{Var}$	$\sum x^2$	$n(\operatorname{Var}+\bar x^2)$
$n,\ \sum x,\ \sum x^2$	Var / SD	$\sum x^2/n-\bar x^2$
Two groups' totals	combined SD	pool, then re-run
Combined SD + one group	other group's mean	set pooled Var = SD², solve for the total

Examiner note

The method mark is the pooled variance formula with pooled totals substituted,

(\sum x^2+\sum y^2)/N-((\sum x+\sum y)/N)^2

, shown unsimplified; the exact combined mean must appear inside it (don't round, e.g. 37.67, then square). Final SD identified.

Common mistake

Never average the two variances or SDs — that ignores how far apart the two means are. If given a mean and SD per group, recover the totals first with

\sum x=n\bar x

and

\sum x^2=n(\operatorname{Var}+\bar x^2)

Why you can't average the two variances