L3 - Location, spread and standard deviation | AQA A-Level Mathematics

Calculate measures of centre and variation from raw lists, frequency tables and summary statistics. You will check calculator outputs, recognise grouped estimates and choose summaries that answer a question about the data.

Location from raw data

If one slow journey becomes much slower, must the “average” journey change? The answer depends on which average you mean. For the illustrative ordered times $8,10,10,12,14,16,28$ minutes, the mean is $14$ , the median is $12$ and the mode is $10$ . Each describes a different feature.

You already know

Put observations in order to find positions such as the median. A frequency counts repetitions; an identifier such as ReferenceNumber labels a record and is not a measurement to average.

The mean shares the total equally among all observations. The median is the middle observation when the count is odd, or the mean of the two central observations when it is even. The mode is the most frequent value or category; two values can tie, and a list need not have a useful unique mode. The modal class of grouped data is a most frequent class, rather than a known exact modal observation.

Mean of a list

For $n$ observations $x_1,\ldots,x_n$ ,

\bar x=\frac{\sum x_i}{n}

Here $\sum x_i$ means add the observations, and $\bar x$ denotes their sample mean.
Not in the formula booklet: learn it.

Use the mean if the total matters: total journey time is count multiplied by mean time. Use the median to describe a central position when extremes pull the mean away from most observations.

Worked example 1: three measures of centre

Question. Find the mean, median and mode of $8,10,10,12,14,16,28$ minutes.

Plan. Add for the mean, locate the middle rank, then identify the most frequent value.

\begin{aligned} \sum x&=98 && \text{add all seven times}\\ \bar x&=98/7=14 && \text{equal share of total}\\ (n+1)/2&=4 && \text{middle rank is fourth} \end{aligned}

Why this step? The fourth value is $12$ ; using the middle of the minimum and maximum would instead calculate the midrange.

Answer. Mean $14$ , median $12$ , mode $10$ , all in minutes.

Check. $7\times14=98$ , and $10$ occurs twice while every other value occurs once.

Change only the final observation from $28$ to $98$ . This keeps every rank and repetition pattern unchanged while testing the effect of one extreme value.

Worked example 2: change one extreme observation

Question. Find the mean, median and mode of $8,10,10,12,14,16,98$ minutes. Compare with Worked example 1.

Plan. Update the total by the change in the last observation, then recheck the central rank and repetitions.

\begin{aligned} \sum x&=98+(98-28)=168 && \text{update one observation}\\ \bar x&=168/7=24 && \text{divide by the same count}\\ \text{median}&=x_4=12 && \text{central rank unchanged} \end{aligned}

Why this step? Adding $70$ to one observation adds $70/7=10$ to the mean, but need not move the median or mode.

Answer. Mean $24$ , median $12$ , mode $10$ minutes. Only the mean changes.

Check. The mean rises by $10$ , whereas the middle value and repeated value remain visible in the ordered list.

For a genuine selected list, the first seven AQA workbook masses are $1970,1163,1316,1355,1445,1425,1163\,\mathrm{kg}$ . Their total $9837\,\mathrm{kg}$ gives mean $9837/7\approx1405.29\,\mathrm{kg}$ . Mass includes $75\,\mathrm{kg}$ for an average driver. Being the first rows does not make them a random sample of all cars.

Source: AQA car LDS, $2017$ DfT Stock Vehicle Database, Open Government Licence v3.0.

Watch out

Wrong: median $=(8+98)/2=53$ . This averages the extremes and ignores the central position. Right: order the seven observations and select the fourth, $12$ .

Change the extreme value again to see whether the reasoning, rather than just the answer, carries over.

Quartiles and resistant spread

The range measures the full distance from minimum to maximum, while the IQR focuses on the middle half and is more resistant to one extreme observation.

Range and interquartile range

\text{range}=\text{maximum}-\text{minimum} \qquad \mathrm{IQR}=Q_3-Q_1

Not in the formula booklet: learn it.

Quartile conventions can differ for small samples. Here each question explicitly uses ranks $(n+1)/4$ and $3(n+1)/4$ ; these position rules are not in the formula booklet. Follow a given convention, and record your choice in an investigation. For large datasets, calculator quartiles or graphical readings are useful. You do not need a grouped-data linear-interpolation formula for medians or quartiles.

Worked example 3: locate quartiles

Question. For $8,10,10,12,14,16,28$ minutes, use ranks $(n+1)/4$ and $3(n+1)/4$ to find the quartiles, IQR and range.

Plan. Find the two ranks before selecting the observations.

\begin{aligned} (n+1)/4&=2,\quad 3(n+1)/4=6 && n=7\\ Q_1&=10,\quad Q_3=16 && \text{second and sixth values}\\ \mathrm{IQR}&=16-10=6 && \text{middle-half spread}\\ \text{range}&=28-8=20 && \text{full spread} \end{aligned}

Why this step? A rank tells you where to look; it is not itself the quartile's measured value.

Answer. Quartiles $10,16$ minutes, IQR $6$ minutes, range $20$ minutes.

Check. Both quartiles lie between the minimum and maximum, with the median $12$ between them.

Again change only $28$ to $98$ , keeping the quartile convention fixed. This isolates the effect of a distant maximum.

Worked example 4: resist an extreme value

Question. Find the quartiles, IQR and range for $8,10,10,12,14,16,98$ minutes using the same convention.

Plan. Reuse the ranks, then calculate the two spreads separately.

\begin{aligned} Q_1&=x_2=10,\quad Q_3=x_6=16 && \text{same central ranks}\\ \mathrm{IQR}&=16-10=6 && \text{quartiles unchanged}\\ \text{range}&=98-8=90 && \text{maximum has changed} \end{aligned}

Why this step? IQR ignores how far the largest observation extends once the quartile positions are fixed.

Answer. Quartiles $10,16$ , IQR $6$ , range $90$ , in minutes. The range changes greatly while the IQR stays fixed.

Check. The range rises by $70$ , exactly the increase in the maximum.

“Resistant” does not mean a statistic can never change. Changing enough observations, or moving an observation across a quartile position, can change the IQR.

Watch out

Wrong: “The quartile is $2$ minutes because its rank is $2$ .” A rank is a position. Right: select the second observation; here $Q_1=10$ minutes.

Use a different ordered list for the next check, applying the stated ranks rather than a calculator's potentially different convention.

Variance, standard deviation and the calculator

Standard deviation uses every observation to measure variation around the mean, then expresses the result in the original units.

The deviations from a mean sum to zero: for $8,10,12$ , the mean is $10$ and the deviations are $-2,0,2$ . Averaging signed deviations would therefore hide spread. Squaring prevents cancellation, giving a variance of $(4+0+4)/3=8/3$ and standard deviation $\sqrt{8/3}$ .

Variance and standard deviation

The descriptive variance of a list is the mean squared deviation from its mean, using denominator $n$ . Its standard deviation is the nonnegative square root. Variance has squared units; standard deviation has the original units. Both are zero precisely when every observation is equal.

Expanding the squares explains the convenient summary-statistics form:

\frac{\sum(x_i-\bar x)^2}{n} =\frac{\sum x_i^2-2\bar x\sum x_i+n\bar x^2}{n} =\frac{\sum x_i^2}{n}-\bar x^2

The last equality uses $\sum x_i=n\bar x$ .

Standard deviation of data

\text{standard deviation} =\sqrt{\frac{\sum(x_i-\bar x)^2}{n}} =\sqrt{\frac{\sum x_i^2}{n}-\bar x^2}

In the formula booklet.

Square each observation before adding for $\sum x_i^2$ ; this differs from squaring the total. Take the square root only after finding variance, and keep the mean unrounded during the calculation.

Worked example 5: work from summary statistics

Question. Seven recorded times have $\sum x=98$ and $\sum x^2=1644$ , with $x$ in minutes. Find their descriptive variance and standard deviation.

Plan. Calculate the mean, use the supplied summary-statistics formula, then take the square root.

\begin{aligned} \bar x&=98/7=14 && \text{mean}\\ v&=1644/7-14^2=272/7 && \text{variance, denominator }7\\ \sqrt v&\approx6.23 && \text{restore original units} \end{aligned}

Why this step? $1644/7$ is a mean square; subtracting the square of the mean leaves the variation around that mean.

Answer. Variance $272/7\,\mathrm{min}^2$ ; standard deviation approximately $6.23\,\mathrm{min}$ .

Check. For the original list $8,10,10,12,14,16,28$ , squared deviations from $14$ sum to $272$ .

Change only the form of the supplied summary: the next question gives the squared-deviation total directly. There is no mean square to correct a second time.

Worked example 6: work from squared deviations

Question. Seven times have mean $14$ minutes and $\sum(x-14)^2=272\,\mathrm{min}^2$ . Find the descriptive variance and standard deviation.

Plan. Divide the squared-deviation total by $7$ , then take its square root.

\begin{aligned} v&=272/7 && \text{mean squared deviation}\\ \text{standard deviation}&=\sqrt{272/7}\approx6.23 && \text{square root} \end{aligned}

Why this step? The deviations already measure departures from the mean, so subtracting $14^2$ again would be wrong.

Answer. Variance $272/7\,\mathrm{min}^2$ ; standard deviation approximately $6.23\,\mathrm{min}$ .

Check. Both forms of the same information give the same result as Worked example 5.

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Pure Mathematics

Mechanics

Statistics

L3 - Location, spread and standard deviation

Location from raw data

Quartiles and resistant spread

Variance, standard deviation and the calculator

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