MD10, MD11 - Averages, percentiles and spread | OCR A-Level Mathematics B (MEI)

Choose and interpret means, medians, modes, midranges and weighted means in MD10, then use percentiles, quartiles, range and interquartile range in MD11. Real pulse readings from the current health large data set show how a summary can change while the middle observation stays put.

Mean, median, mode and midrange

Can a changed extreme reading move an average without changing the middle reading? In the first ten LDS records, nine pulse readings are available: $74,92,70,68,92,84,82,62,64$ beats in 60 seconds.

You already know

A missing observation is not zero. The pulse column has nine available readings in these ten records, so a mean of those readings uses denominator 9.

These records belong to people aged at least 16 in the American NHANES survey of 2003–04, supplied as LDS9 for June 2027. The pulse unit is beats in 60 seconds. The summaries below describe this small extract, without establishing a person's health or today's population pattern.

Four measures of central tendency

The arithmetic mean is total divided by count. The median is the middle ordered value, or the mean of the middle pair for an even count. A mode is a most frequent value or category. The midrange is the midpoint between minimum and maximum; it need not equal the median.

A distribution may have several tied modes or no uniquely most frequent value. The mode of marital status is meaningful, while the mean of arbitrary marital-status codes is not. For grouped data the modal class need not identify an exact modal observation; unequal-width histogram height indicates concentration, rather than necessarily the largest class count.

Mean and midrange

For $n>0$ numerical observations:

\bar x=\frac{\sum x}{n},\qquad \text{midrange}=\frac{x_{\min}+x_{\max}}{2}.

Not in the formula sheet: learn it.

The midrange uses only two observations; the mean uses all of them. Calculate both before deciding which serves a statistical purpose.

Worked example 1: four summaries of real pulse readings

Question. Find the mean, median, mode and midrange of $74,92,70,68,92,84,82,62,64$ . Give the mean to one decimal place.

Plan. Order the readings, use their total for the mean, and use positions, repeats and extremes for the other summaries.

\begin{aligned} \text{ordered readings}&=62,64,68,70,74,82,84,92,92\\ \bar x&=\frac{688}{9}=76.444\ldots &&\text{all nine readings}\\ \text{median}&=74,\qquad \text{mode}=92 &&\text{middle and most frequent}\\ \text{midrange}&=\frac{62+92}{2}=77 &&\text{extreme endpoints} \end{aligned}

Why use nine, not ten? Only nine pulse readings were supplied in this extract.

Answer. Mean $76.4$ , median 74, mode 92 and midrange 77 beats in 60 seconds.

Check. $9(688/9)=688$ ; all four summaries lie between 62 and 92.

Change only one reading: replace one of the 92s by 110 in a what-if calculation. This is not a correction to the real LDS.

Worked example 2: one changed extreme reading

Question. Find the four summaries after that replacement, again giving the mean to one decimal place.

Plan. Update the total and ordering, then check the middle, repeated values and new extremes.

\begin{aligned} \text{new total}&=688+(110-92)=706 &&\text{one replacement}\\ \bar x&=\frac{706}{9}=78.444\ldots\\ \text{ordered readings}&=62,64,68,70,74,82,84,92,110\\ \text{median}&=74,\qquad \text{midrange}=\frac{62+110}{2}=86 \end{aligned}

Why check the mode again? The two tied 92s are no longer repeated, and all readings now occur once.

Answer. Mean $78.4$ , median 74 and midrange 86 beats in 60 seconds; there is no uniquely most frequent value.

Check. The mean rises by $18/9=2$ , while the median stays fixed. The midrange rises by $18/2=9$ .

This small change shows why choice matters more than routine arithmetic. For skewed delivery times or wages, a median often describes a typical observation more usefully than a mean pulled towards a tail. The mean is useful when totals matter; multiplying it by the count recovers the total. The midrange can move sharply because a single extreme controls half its value.

For example, constructed times $2,3,4,4,5,6,8,8,23$ have mean 7, median 5, modes 4 and 8, and midrange 12.5 minutes. Replacing 23 by 41 changes mean to 9 and midrange to 21.5, while median and modes stay fixed.

Watch out

Wrong: “The midrange is the middle observation.” It is the midpoint of the two extremes. The median is found by ordered position, so the two measures can differ greatly.

Choose a typical-value summary for prices with one exceptional maximum.

Weights preserve the sizes of groups

Combining group means requires knowing how many observations each group represents.

Weighted mean

A weighted mean gives each value influence in proportion to an appropriate weight. Frequencies weight repeated values, group sizes weight group means, and relevant population counts can weight area summaries.

The quantities must describe the same variable in the same units, with non-overlapping groups. To find an overall percentage, weights must match the percentages' denominators; simply averaging area percentages gives areas equal influence.

Weighted mean

For non-negative weights with positive total:

\bar x=\frac{\sum w_i x_i}{\sum w_i}.

When $x_i$ is a group mean, $w_i x_i$ is that group's total.

Not in the formula sheet: learn it.

Begin with two constructed school groups so that the effect of unequal sizes is clear.

Worked example 3: unequal group sizes

Question. School A has 80 students with mean journey time 12 minutes. School B has 20 with mean 22 minutes. Find the combined mean.

Plan. Recover each total time, add the totals, and divide by the combined number of students.

\begin{aligned} \text{total time}&=80(12)+20(22)=1400 &&\text{weight each mean}\\ \bar x&=\frac{1400}{80+20}=14 &&\text{divide by all 100 students} \end{aligned}

Why not $(12+22)/2$ ? Equal weighting would give the small school the same influence as the large one.

Answer. $14\,\mathrm{min}$ .

Check. The answer lies between 12 and 22, closer to 12 because A has more students.

Change only the size of school A from 80 to 20. The two group means stay fixed.

Worked example 4: when equal weighting becomes appropriate

Question. A now has 20 students with mean 12 minutes; B still has 20 with mean 22 minutes. Find the combined mean.

Plan. Use the same totals method with the new group size.

\begin{aligned} \text{total time}&=20(12)+20(22)=680 &&\text{new first weight}\\ \bar x&=\frac{680}{20+20}=17 &&\text{divide by 40} \end{aligned}

Why does the simple average work now? Each group represents the same number of students.

Answer. $17\,\mathrm{min}$ .

Check. $(12+22)/2=17$ , matching the weighted calculation.

The real LDS has 103 female records with total age 4,559 years and 97 male records with total age 4,399 years. The group mean ages are $4559/103$ and $4399/97$ , so the overall recorded mean age is

\frac{103(4559/103)+97(4399/97)}{200} =\frac{8958}{200}=44.79\text{ years}.

The weights count records, not survey population weights: this is the mean of the 200 LDS records, not a weighted national NHANES population estimate. Published rounded group means would introduce a rounding limitation.

Watch out

Wrong: “The overall mean is the average of the two group means.” That works only when the groups have equal weights. Reconstruct the totals when their sizes differ.

Carry the same method into a fresh unequal pair.

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MD10, MD11 - Averages, percentiles and spread

Mean, median, mode and midrange

Weights preserve the sizes of groups

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