Mathematics 9709 · Probability & Statistics 1

The mean

The mean is the fair share: put every value into one total and split it equally.
That is why the total $\sum x$ matters so much.
Most exam twists are just rearrangements of $\bar x=\sum x/n$ .
If a group changes, rebuild the total first, then divide again.

Three forms of the same idea

$\bar x=\dfrac{\sum x}{n}\qquad \bar x=\dfrac{\sum xf}{\sum f}\qquad \Sigma x=\bar x\,n$

For a frequency table, build the $xf$ row and divide by $\sum f$ . For grouped data you never see the raw values, so you estimate by putting everyone at the class mid-value, $\tfrac12(\text{lower}+\text{upper boundary})$ .

The grouped mean is an estimate. The most used tool is Σx = x̄n: recover the total first, then add, drop, or combine values.

Worked example

Start simple: a frequency-table mean

$x=31,32,33,34,35$ with $f=5,7,9,8,11$ . $\sum f=40$ , $\sum xf=155+224+297+272+385=1333$ .

Answer

\bar x=\tfrac{1333}{40}=33.325

The key skill: back out a total with Σx = x̄n

Worked example

Back out the total, then use it

Five labourers have a mean mass of 70.2 kg and want to share a lift with a 500 kg limit. The mean alone hides the total — recover it first: $\sum x=\bar x\,n=70.2\times5=351$ kg.

So the cement can mass at most $500-351$ kg.

Answer

\text{greatest cement }=149\text{ kg}

Combine two groups by totals, never by averaging averages:

Get each group's total via $\sum x=\bar x\,n$ (or read it off).
Add the totals; add the counts.
Divide: $\dfrac{\sum x_1+\sum x_2}{n_1+n_2}$ .
Averaging the two means only works if $n_1=n_2$ .

Worked example

Combine two sets given as totals

A large bag has 72 sweets massing 852.4 g; a small bag has 24 sweets massing 282.8 g. Pool both totals and both counts: $\dfrac{852.4+282.8}{72+24}=\dfrac{1135.2}{96}$ .

Answer

\text{mean mass}=11.825\text{ g}

Worked example

Combine two teams, then adjust one value

Team P: 12 runners, $\sum x=540$ . Team Q: 18 runners, $\sum y=882$ . Mean of all 30: $\dfrac{540+882}{30}=\dfrac{1422}{30}$ . (9709/52/O/N/24 Q6)

Add/drop twist: the mean of 12 salaries is $1950; a 13th is hired and the mean drops by $8. New total $=1942\times13=25246$ ; old total $=1950\times12=23400$ ; the 13th salary is $25246-23400=1846$ .

Answer

\text{mean}=47.4\text{ min};\quad \text{13th salary }=\$1846

Combine by TOTALS, not by averaging averages — the bigger group pulls harder.

Grouped data: mid-values give an estimate

Worked example

Grouped mean — mind the age trap

50 students: sixteen 18-year-olds, twenty 19-year-olds, fourteen aged 20–21. “18” means $18\le A<19$ , so the mid-value is 18.5, not 18. Mid-values $18.5,\ 19.5,\ 21$ :

$\sum xf=18.5(16)+19.5(20)+21(14)=980$

Answer

\bar x=\tfrac{980}{50}=19.6\text{ years (an estimate)}

Using 18, 19, 20.5 gives the trap answer 19.1 — half a year off.

Worked example

Grouped mean, then a per-item follow-up

75 crates of coconuts (40 per crate): 46 crates at $20\le m<25$ , 22 at $25\le m<40$ , 7 at $40\le m<54$ kg. Mid-values $22.5,\ 32.5,\ 47$ :

$\sum xf=22.5(46)+32.5(22)+47(7)=1035+715+329=2079$ .
Mean per crate $=\tfrac{2079}{75}=27.72$ kg.
Per coconut: divide by 40, $\tfrac{27.72}{40}=0.693$ kg.

Answer

27.72\text{ kg per crate},\ \ 0.693\text{ kg per coconut}

You are given…	Move
A raw list	$\sum x / n$
A frequency table	$\sum xf / \sum f$
Grouped / histogram / CF	mid-values → an estimate
A total Σx	$\bar x=\sum x/n$
Two groups	combine totals
Add/drop one value	adjust Σx, then re-divide

Examiner note

For grouped data, write the mid-values row explicitly — the M1 hinges on correct mid-values — and call the result an estimate. For a combine, the M1 is the

\sum=\bar x\,n

line for each set before you add.

Common mistake

Grouped mid-value off by a boundary (18 instead of 18.5 — the half-year trap), and treating an estimated grouped mean as exact (forgetting the word “estimate”).

Common mistake

Combining two groups by averaging the two means. That only works if the groups are the same size — otherwise use

\dfrac{\sum x_1+\sum x_2}{n_1+n_2}

Your turn— tap to reveal the worked answer (combined mean)

22 boys average 71%, 28 girls average 76%. Mean of all 50?

$\dfrac{71(22)+76(28)}{50}=\dfrac{1562+2128}{50}$ , pulled toward the larger group.

Answer

73.8\%\ (\text{not } 73.5)

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The mean

The mean is the fair share: put every value into one total and split it equally.
That is why the total $\sum x$ matters so much.
Most exam twists are just rearrangements of $\bar x=\sum x/n$ .
If a group changes, rebuild the total first, then divide again.

Three forms of the same idea

$\bar x=\dfrac{\sum x}{n}\qquad \bar x=\dfrac{\sum xf}{\sum f}\qquad \Sigma x=\bar x\,n$

The grouped mean is an estimate. The most used tool is Σx = x̄n: recover the total first, then add, drop, or combine values.

Worked example

Start simple: a frequency-table mean

$x=31,32,33,34,35$ with $f=5,7,9,8,11$ . $\sum f=40$ , $\sum xf=155+224+297+272+385=1333$ .

Answer

\bar x=\tfrac{1333}{40}=33.325

The key skill: back out a total with Σx = x̄n

Worked example

Back out the total, then use it

Five labourers have a mean mass of 70.2 kg and want to share a lift with a 500 kg limit. The mean alone hides the total — recover it first: $\sum x=\bar x\,n=70.2\times5=351$ kg.

So the cement can mass at most $500-351$ kg.

Answer

\text{greatest cement }=149\text{ kg}

Combine two groups by totals, never by averaging averages:

Get each group's total via $\sum x=\bar x\,n$ (or read it off).
Add the totals; add the counts.
Divide: $\dfrac{\sum x_1+\sum x_2}{n_1+n_2}$ .
Averaging the two means only works if $n_1=n_2$ .

Worked example

Combine two sets given as totals

A large bag has 72 sweets massing 852.4 g; a small bag has 24 sweets massing 282.8 g. Pool both totals and both counts: $\dfrac{852.4+282.8}{72+24}=\dfrac{1135.2}{96}$ .

Answer

\text{mean mass}=11.825\text{ g}

Worked example

Combine two teams, then adjust one value

Team P: 12 runners, $\sum x=540$ . Team Q: 18 runners, $\sum y=882$ . Mean of all 30: $\dfrac{540+882}{30}=\dfrac{1422}{30}$ . (9709/52/O/N/24 Q6)

Add/drop twist: the mean of 12 salaries is $1950; a 13th is hired and the mean drops by $8. New total $=1942\times13=25246$ ; old total $=1950\times12=23400$ ; the 13th salary is $25246-23400=1846$ .

Answer

\text{mean}=47.4\text{ min};\quad \text{13th salary }=\$1846

Combine by TOTALS, not by averaging averages — the bigger group pulls harder.

Grouped data: mid-values give an estimate

Worked example

Grouped mean — mind the age trap

50 students: sixteen 18-year-olds, twenty 19-year-olds, fourteen aged 20–21. “18” means $18\le A<19$ , so the mid-value is 18.5, not 18. Mid-values $18.5,\ 19.5,\ 21$ :

$\sum xf=18.5(16)+19.5(20)+21(14)=980$

Answer

\bar x=\tfrac{980}{50}=19.6\text{ years (an estimate)}

Using 18, 19, 20.5 gives the trap answer 19.1 — half a year off.

Worked example

Grouped mean, then a per-item follow-up

75 crates of coconuts (40 per crate): 46 crates at $20\le m<25$ , 22 at $25\le m<40$ , 7 at $40\le m<54$ kg. Mid-values $22.5,\ 32.5,\ 47$ :

$\sum xf=22.5(46)+32.5(22)+47(7)=1035+715+329=2079$ .
Mean per crate $=\tfrac{2079}{75}=27.72$ kg.
Per coconut: divide by 40, $\tfrac{27.72}{40}=0.693$ kg.

Answer

27.72\text{ kg per crate},\ \ 0.693\text{ kg per coconut}

You are given…	Move
A raw list	$\sum x / n$
A frequency table	$\sum xf / \sum f$
Grouped / histogram / CF	mid-values → an estimate
A total Σx	$\bar x=\sum x/n$
Two groups	combine totals
Add/drop one value	adjust Σx, then re-divide

Examiner note

For grouped data, write the mid-values row explicitly — the M1 hinges on correct mid-values — and call the result an estimate. For a combine, the M1 is the

\sum=\bar x\,n

line for each set before you add.

Common mistake

Grouped mid-value off by a boundary (18 instead of 18.5 — the half-year trap), and treating an estimated grouped mean as exact (forgetting the word “estimate”).

Common mistake

Combining two groups by averaging the two means. That only works if the groups are the same size — otherwise use

\dfrac{\sum x_1+\sum x_2}{n_1+n_2}

Your turn— tap to reveal the worked answer (combined mean)

22 boys average 71%, 28 girls average 76%. Mean of all 50?

$\dfrac{71(22)+76(28)}{50}=\dfrac{1562+2128}{50}$ , pulled toward the larger group.

Answer

73.8\%\ (\text{not } 73.5)