2.02f - Averages, percentiles and spread | OCR A-Level Mathematics A

Use 2.02f to calculate and interpret averages, percentiles, quartiles and measures of variation. You will choose summaries for the question being asked, and compare distributions without pretending that one number describes every observation.

Mean, median and mode describe different centres

Can one delayed delivery make the “average” delivery slower while leaving the typical middle delivery unchanged? Start with nine constructed delivery times, in minutes: $2,3,4,4,5,6,8,8,23$ .

You already know

Order observations before finding a middle value. Repeated values represent repeated observations, not values to discard.

The total is 63 minutes and the fifth value is 5. Those facts describe different features: the total determines the mean, while the position determines the median.

Mean, median and mode

The arithmetic mean is total divided by number of observations. The median is the middle ordered value, or the mean of the two middle values for an even-sized dataset. The mode is a most frequent value or category; there can be more than one mode or no uniquely most frequent value.

The mode of travel method is meaningful; the mean of arbitrary category codes is not. For grouped data, a class does not reveal an exact modal observation. In an unequal-width histogram the tallest bar represents greatest concentration per unit, not necessarily the greatest class frequency.

Arithmetic mean

For $n>0$ observations:

\bar x=\frac{\sum x}{n},\qquad \sum x=n\bar x.

Not in the formula sheet: learn it.

Use the mean when totals matter, and consider the median when a few extreme values distort the typical observation.

Worked example 1: three different averages

Question. Find the mean, median and modes of $2,3,4,4,5,6,8,8,23$ minutes.

Plan. Total all values for the mean, locate the middle ordered value, and count repetitions.

\begin{aligned} \bar x&=\frac{63}{9}=7 &&\text{use every observation}\\ \text{median}&=x_5=5 &&\text{fifth of nine ordered values}\\ \text{modes}&=4,\ 8 &&\text{each occurs twice} \end{aligned}

Why can both 4 and 8 be modes? They tie for the greatest frequency.

Answer. Mean $7\,\mathrm{min}$ , median $5\,\mathrm{min}$ , modes $4$ and $8\,\mathrm{min}$ .

Check. $9(7)=63$ , recovering the total.

Change only the last delivery, from 23 to 41 minutes. The number of deliveries stays fixed.

Worked example 2: changing one extreme observation

Question. Find the mean, median and modes after that change, and describe its effect.

Plan. Update the total, then check whether the middle position or frequencies change.

\begin{aligned} \text{new total}&=63+(41-23)=81 &&\text{replace one observation}\\ \bar x&=\frac{81}{9}=9 &&\text{same denominator}\\ \text{median}&=5,\qquad \text{modes}=4,\ 8 &&\text{positions and repeats unchanged} \end{aligned}

Why does the mean rise by 2? The extra 18 minutes is shared over nine observations.

Answer. Mean $9\,\mathrm{min}$ ; median and modes stay unchanged. One delay raised the mean without moving the middle delivery.

Check. $(41-23)/9=2=9-7$ .

The real LDS makes the observational unit equally important. In the 2011 census, County Durham's mean age was 41 years and median age 42; Darlington's were 40.2 and 41. These summarise residents, whereas averaging the two authorities' published means gives equal weight to authorities, not to their different numbers of residents.

Watch out

Wrong: “The mean always describes a typical person best.” A high tail can pull the mean away from most values. State why your choice suits the purpose; the median's resistance is useful for skewed wages or delivery times.

Compare an even-sized dataset with an exceptional high value.

Percentiles locate proportions of the data

A percentile is a threshold position, not the value taken by that percentage of the observations.

Percentiles and quartiles

The percentile of rank $p$ is a threshold below which approximately $p\%$ of observations lie. The lower quartile $Q_1$ , median and upper quartile $Q_3$ are the 25th, 50th and 75th percentiles. Ties and small discrete datasets can prevent exact percentage divisions.

A 90th-percentile delivery time of 18 minutes means about $90\%$ take no more than 18 minutes, not that $90\%$ take exactly 18. Small-sample quartile conventions differ. Follow a question's stated convention and know which one your calculator uses. Here, take medians of the lower and upper halves, excluding a single overall median when $n$ is odd.

Worked example 3: quartiles for an odd number of values

Question. Find the quartiles of $2,3,4,4,5,6,8,8,23$ , using the stated median-of-halves convention.

Plan. Remove the overall median from the two halves, then find each half's median.

\begin{aligned} \text{lower half}&:2,3,4,4 &&\text{exclude the middle 5}\\ Q_1&=\frac{3+4}{2}=3.5 &&\text{middle pair of lower half}\\ \text{upper half}&:6,8,8,23\\ Q_3&=\frac{8+8}{2}=8 &&\text{middle pair of upper half} \end{aligned}

Why exclude the 5? This is the explicitly chosen convention for an odd-sized set.

Answer. $Q_1=3.5\,\mathrm{min}$ , $Q_3=8\,\mathrm{min}$ .

Check. The median 5 lies between the two quartiles.

Change only the sample size by adding one observation, 41 minutes. Each half now has five observations.

Worked example 4: quartiles for an even number of values

Question. Find the quartiles of $2,3,4,4,5,6,8,8,23,41$ , using the same convention.

Plan. Split the ten ordered observations into equal halves and take their middle values.

\begin{aligned} \text{lower half}&:2,3,4,4,5,\qquad Q_1=4 &&\text{third of five}\\ \text{upper half}&:6,8,8,23,41,\qquad Q_3=8 &&\text{third of five} \end{aligned}

Why is no observation excluded? An even-sized dataset splits into two equal halves without a single middle observation.

Answer. $Q_1=4\,\mathrm{min}$ , $Q_3=8\,\mathrm{min}$ .

Check. The full median is $(5+6)/2=5.5$ , between those quartiles.

On a cumulative-frequency diagram with $n$ observations, read $Q_1$ and $Q_3$ at cumulative frequencies $n/4$ and $3n/4$ ; read percentile $p$ at $pn/100$ . Thus the 90th percentile of 120 observations uses cumulative frequency 108, and the 85th percentile of 200 uses 170. Move horizontally to the curve and then down to the measurement axis. These readings or interpolations are estimates, and need not equal a small raw dataset's half-median quartiles.

Watch out

Wrong: “The 85th percentile is 32 minutes, so 85% take exactly 32 minutes.” Correct: about $85\%$ are at or below that threshold; about $15\%$ are above it, allowing for ties and estimation.

Use the same raw-data convention with a fresh ordered list.

Range and interquartile range measure spread

A measure of spread tells you how widely values are distributed; it answers a different question from an average.

Range and interquartile range

The range is maximum minus minimum. The interquartile range, $\mathrm{IQR}$ , is $Q_3-Q_1$ , the width of the middle approximately $50\%$ of observations. Both use the data's units.

The range uses only two observations. IQR focuses on central spread and resists changes confined to the extreme tails, though it is not immune to every change. A wider box means a larger IQR, not more observations; a sample of 20 and one of 200 each have roughly half their observations inside the box.

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2.02f - Averages, percentiles and spread

Mean, median and mode describe different centres

Percentiles locate proportions of the data

Range and interquartile range measure spread

Continue this lesson.

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