S4 - Averages, range and grouped-data estimates

A single “average” can hide very different stories. Learn to calculate and choose measures of centre and spread, work with frequency tables and grouped estimates, and recover or combine totals when a mean is given.

Mean, median and mode

Can five people have an average of 5 books when nobody has exactly 5? Yes: the values 2, 3, 3, 7 and 10 do exactly that. Different averages describe different features.

You already know

A frequency counts repeated values. To find a middle value, put observations in increasing order first; to find a total, include every observation.

The mean shares a total equally. The median is the middle ordered value, or the mean of the two middle ordered values when there is an even number. The mode is the most frequent value. A tie can give two or more modes; if all values occur equally often, there is no uniquely most frequent value.

Arithmetic mean

For at least one observation,

\text{mean}=\frac{\text{sum of all values}}{\text{number of values}}

Not on the formula sheet: learn it.

For an odd count $n$ , use median position $(n+1)/2$ . For an even count, average the values at positions $n/2$ and $n/2+1$ . These position rules are not on the formula sheet: derive them from the middle of an ordered list.

Worked example 1: three different averages

Question. Find mean, median and mode of $2,3,3,7,10$ .

Plan. Calculate the total, then use order and repetition separately.

\begin{aligned} \text{mean}&=(2+3+3+7+10)\div5=5 &&\text{share total 25}\\ \text{median}&=3 &&\text{third ordered value}\\ \text{mode}&=3 &&\text{only value appearing twice} \end{aligned}

Why separate these? The mean uses every value's size; the median uses position; the mode uses frequency.

Answer. Mean 5, median 3, mode 3.

Check. $5\times5=25$ , the original total. An average need not appear in the list.

The next example changes only the number of observations by adding 11. There are now two middle positions.

Worked example 2: an even number of values

Question. Find mean, median and mode of $2,3,3,7,10,11$ .

Plan. Recalculate the total and average the two middle values.

\begin{aligned} \text{mean}&=(25+11)\div6=6 &&\text{new total and count}\\ \text{median}&=(3+7)\div2=5 &&\text{positions 3 and 4}\\ \text{mode}&=3 &&\text{highest frequency unchanged} \end{aligned}

Why this step? Neither middle position alone splits an even list equally.

Answer. Mean 6, median 5, mode 3.

Check. Three values lie at or below 5 and three at or above it.

An unsorted list such as 8, 1, 4, 2 must first become 1, 2, 4, 8. Its median is $(2+4)/2=3$ , and its mean is $15/4=3.75$ . Neither has to be an observed value. A fractional mean of a count describes an average, not one person's count.

Watch out

Wrong: take the middle two numbers of the unsorted list 8, 1, 4, 2. Their positions depend on how the list was written. Sort first, then use 2 and 4 to obtain median 3.

Choose a mode for the most common travel category: taking a numerical mean of category names makes no sense. The median often gives a more useful typical numerical value when a few extremes pull the mean away from most observations.

Averages from ungrouped frequency tables

A frequency table saves writing repeated observations, but the repeats still count in an average.

If score 2 occurs three times, its contribution to the total score is $2\times3$ , not just 2. A zero value contributes zero to the total but its observations still enter the count.

Mean from a frequency table

\text{mean}=\frac{\sum fx}{\sum f}

Here $x$ is a value, $f$ is its frequency, and $\sum$ means “add all”. The denominator is the total number of observations, which must be positive.

Not on the formula sheet: derive it from the arithmetic mean.

Running frequency totals locate median positions without expanding the whole list.

Worked example 3: weighting repeated values

Question. Book counts 0, 1 and 2 have frequencies 2, 5 and 3. Find mean, median and mode.

Plan. Find weighted total and count, then locate the middle observations.

\begin{aligned} \sum f&=2+5+3=10 &&\text{number of members}\\ \sum fx&=0\times2+1\times5+2\times3=11 &&\text{number of books}\\ \text{mean}&=11\div10=1.1 &&\text{books per member}\\ \text{running totals}&=2,7,10 &&\text{locate positions 5 and 6} \end{aligned}

Why this step? Both middle positions fall among the five members with 1 book.

Answer. Mean 1.1 books; median 1; mode 1, whose frequency is greatest.

Check. The expanded list has ten entries: two zeros, five ones and three twos.

The next example moves two observations from value 1 to value 2. The total count stays ten, but the middle positions now straddle two values.

Worked example 4: a median between values

Question. Counts 0, 1 and 2 now have frequencies 2, 3 and 5. Find the mean and median.

Plan. Reweight the total and locate each middle position separately.

\begin{aligned} \sum fx&=0\times2+1\times3+2\times5=13 &&\text{new weighted total}\\ \text{mean}&=13\div10=1.3 &&\text{same total count}\\ \text{running totals}&=2,5,10 &&\text{position 5 is 1; position 6 is 2}\\ \text{median}&=(1+2)\div2=1.5 &&\text{average the two values} \end{aligned}

Why this step? A cumulative total exactly equal to position 5 does not put position 6 in the same value.

Answer. Mean 1.3; median 1.5.

Check. The ordered list's middle pair is 1 and 2.

The same method works for scores 2, 4 and 7 with frequencies 3, 2 and 5. The count is 10 and weighted total is $6+8+35=49$ , giving mean 4.9. Running totals 3, 5, 10 put middle scores at 4 and 7, so the median is 5.5.

Watch out

Wrong: $(0+1+2)/3=1$ for the first table. That gives each listed value equal weight, ignoring repeats. Use the weighted total 11 divided by all 10 members, giving 1.1.

Apply both weighting and position to a new table.

Estimating from grouped data

Grouping hides individual values, so replace each class by its midpoint to estimate a mean.

The midpoint is halfway between the class endpoints. It represents the class as if its values balanced around that point. The actual values may not balance there, which is why this is an estimate.

Worked example 5: grouped journey times

Question. Estimate the mean of 20 times in minutes: $0<t\leq10$ , frequency 4; $10<t\leq20$ , frequency 10; $20<t\leq30$ , frequency 6.

Plan. Use class midpoints in the weighted-mean calculation.

\begin{aligned} \text{midpoints}&=5,15,25 &&\text{halfway between boundaries}\\ \text{estimated total}&=4\times5+10\times15+6\times25=320 &&\text{weight the midpoints}\\ \text{estimated mean}&=320\div20=16\,\mathrm{min} &&\text{divide by all times} \end{aligned}

Why midpoints? The individual times are unknown; treating every class member as its midpoint supplies a representative value.

Answer. Estimated mean 16 minutes.

Check. It lies between the smallest and largest class midpoints, 5 and 25.

The next example changes the kind of classes to inclusive groups of discrete counts. Use the listed whole-number endpoints for their midpoints.

Worked example 6: grouped discrete counts

Question. Item counts 0–4, 5–9 and 10–14 have frequencies 6, 8 and 2. Estimate the mean.

Plan. Represent each count group by its central value.

\begin{aligned} \text{midpoints}&=2,7,12 &&\text{average each pair of endpoints}\\ \text{estimated total}&=6\times2+8\times7+2\times12=92 &&\text{weighted representatives}\\ \text{estimated mean}&=92\div(6+8+2)=5.75 &&\text{divide by 16 customers} \end{aligned}

Why this step? The first group includes 0, 1, 2, 3 and 4; its centre is 2, not 2.5.

Answer. Estimated mean 5.75 items per customer.

Check. It lies between 2 and 12, and below the most common class midpoint 7 because the lower class has more weight than the upper class.

The modal class has the greatest frequency: $10<t\leq20$ in the journey table, and 5–9 in the item table. It is an interval, not proof that the midpoint is the exact mode. In the journey table, median positions 10 and 11 both fall in $10<t\leq20$ ; in the item table, positions 8 and 9 fall in 5–9. Neither table reveals the exact middle values.

The midpoint relationship and grouped estimate are not on the formula sheet: derive them from halfway values and the weighted mean.

Watch out

Wrong: “The exact mean is 16 minutes, and the mode is 15 minutes.” The first uses assumed midpoint representatives; the second invents an individual mode. Say “estimated mean 16 minutes” and “modal class $10<t\leq20$ ”.

Use midpoints for another measured variable and explain the estimate.

Last Accessed

Papers 1, 2 and 3

S4 - Averages, range and grouped-data estimates

Mean, median and mode

Averages from ungrouped frequency tables

Estimating from grouped data

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