Measures of Location and Spread – A Level Maths Year 1

1 Measures of Location

A measure of location (or average, or measure of central tendency) is a single value that represents a typical member of the data set.

The mean

$\bar{x} = \dfrac{\sum x}{n}$ or, from a frequency table, $\bar{x} = \dfrac{\sum fx}{\sum f}$

Measure	How to find it	Notes
Mean $\bar{x}$	Total $\div$ number of values	Uses every value; affected by extreme values.
Median $Q_2$	Middle value when ordered: the $\tfrac{n+1}{2}$th value	Not affected by extreme values.
Mode	Most common value (or modal class)	The only average for qualitative data; may not be unique.

Worked Example 1 — Raw data

Find the mean, median and mode of $12,\; 15,\; 9,\; 18,\; 15,\; 11,\; 20,\; 14$.

①$\sum x = 114$, $n = 8$, so $\bar{x} = \dfrac{114}{8} = 14.25$.

②Ordered: $9, 11, 12, 14, 15, 15, 18, 20$. The median is halfway between the $4$th and $5$th values: $\dfrac{14 + 15}{2} = 14.5$.

③The mode is $15$.

Worked Example 2 — A frequency table

The number of goals $x$ scored by a team in $25$ matches is shown. Find the mean, median and mode.

$x$	$0$	$1$	$2$	$3$	$4$
$f$	$5$	$9$	$7$	$3$	$1$

①$\sum fx = 0 + 9 + 14 + 9 + 4 = 36$, so $\bar{x} = \dfrac{36}{25} = 1.44$ goals.

②Median is the $13$th value. Cumulative frequencies $5, 14, \ldots$: the $6$th to $14$th values are all $1$, so the median is $1$.

③Mode: $1$ (frequency $9$).

2 Grouped Data and Interpolation

When data are grouped, the original values are lost, so every measure is an estimate. For the mean, use each class midpoint as $x$. For the median and quartiles, use linear interpolation: assume the values in a class are spread evenly across it.

Linear interpolation for grouped data

value $\approx$ lower boundary $+ \dfrac{\text{position} - \text{cumulative frequency before class}}{\text{frequency of class}} \times \text{class width}$
with position $\tfrac{n}{4}$, $\tfrac{n}{2}$, $\tfrac{3n}{4}$ for $Q_1$, median, $Q_3$

Worked Example 3 — Estimating the mean

Estimate the mean commuting time from this table.

Time $t$ (min)	$f$	Midpoint $x$	$fx$
$0 \leqslant t < 10$	$8$	$5$	$40$
$10 \leqslant t < 20$	$22$	$15$	$330$
$20 \leqslant t < 30$	$26$	$25$	$650$
$30 \leqslant t < 50$	$18$	$40$	$720$
$50 \leqslant t < 80$	$6$	$65$	$390$

①$\sum f = 80$, $\sum fx = 2130$.

②$\bar{x} \approx \dfrac{2130}{80} = 26.6$ minutes (to 3 s.f.).

Worked Example 4 — Median and percentiles by interpolation

For the same data, estimate the median and the $10$th percentile.

①Cumulative frequencies: $8, 30, 56, 74, 80$.

②Median: position $40$, in the $20$–$30$ class: $20 + \dfrac{40 - 30}{26} \times 10 = 23.8$ minutes.

③$10$th percentile: position $0.1 \times 80 = 8$, which is exactly the end of the first class: $P_{10} = 10$ minutes.

The mean ($26.6$) is greater than the median ($23.8$): a few long journeys pull the mean up. This is a sign of positive skew.

Quartiles of raw (ungrouped) data. Conventions vary slightly. A common one: find the median, then $Q_1$ is the median of the lower half and $Q_3$ the median of the upper half (excluding the median itself when $n$ is odd). Calculators and software may give slightly different values. Any sensible method is normally accepted, and questions are usually set so the methods agree.

3 Measures of Spread

Measure	Definition	Notes
Range	Largest $-$ smallest	Very sensitive to extreme values.
Interquartile range	$Q_3 - Q_1$	Spread of the middle $50\%$; not affected by extremes.
Interpercentile range	e.g. $P_{90} - P_{10}$	Spread of the middle $80\%$.
Variance $\sigma^2$	Mean of the squared deviations from the mean	Uses every value; units are squared.
Standard deviation $\sigma$	$\sqrt{\text{variance}}$	Same units as the data; pairs with the mean.

Variance and standard deviation

$\sigma^2 = \dfrac{\sum (x - \bar{x})^2}{n} = \dfrac{\sum x^2}{n} - \bar{x}^2$ ("mean of the squares minus square of the mean")
From a frequency table: $\sigma^2 = \dfrac{\sum fx^2}{\sum f} - \bar{x}^2$ $\sigma = \sqrt{\sigma^2}$

You may also see $S_{xx} = \sum (x - \bar{x})^2 = \sum x^2 - \dfrac{\left(\sum x\right)^2}{n}$, so that $\sigma^2 = \dfrac{S_{xx}}{n}$. Some courses also use the sample variance $s^2 = \dfrac{S_{xx}}{n - 1}$, which gives a slightly larger value. Your calculator shows both ($\sigma_x$ or $\sigma_n$, and $s_x$ or $\sigma_{n-1}$); check which your exam board expects. In this page we divide by $n$.

Worked Example 5 — From summary statistics

For $20$ values, $\sum x = 360$ and $\sum x^2 = 7100$. Find the mean and standard deviation.

①$\bar{x} = \dfrac{360}{20} = 18$.

②$\sigma^2 = \dfrac{7100}{20} - 18^2 = 355 - 324 = 31$.

③$\sigma = \sqrt{31} = 5.57$ (to 3 s.f.).

Worked Example 6 — Standard deviation from a frequency table

Find the standard deviation of the goals data in Worked Example 2.

①$\sum fx^2 = 5(0) + 9(1) + 7(4) + 3(9) + 1(16) = 0 + 9 + 28 + 27 + 16 = 80$.

②$\sigma^2 = \dfrac{80}{25} - 1.44^2 = 3.2 - 2.0736 = 1.1264$.

③$\sigma = \sqrt{1.1264} = 1.06$ goals (to 3 s.f.).

Square then multiply by $f$: $fx^2$ means $f \times x^2$, not $(fx)^2$.

Worked Example 7 — Grouped data

Estimate the standard deviation of the commuting times in Worked Example 3.

①$\sum fx^2 = 8(25) + 22(225) + 26(625) + 18(1600) + 6(4225)$

②$= 200 + 4950 + 16\,250 + 28\,800 + 25\,350 = 75\,550$

③$\sigma^2 \approx \dfrac{75\,550}{80} - 26.625^2 = 944.375 - 708.890625 = 235.48\ldots$

④$\sigma \approx 15.3$ minutes.

Keep the unrounded mean ($26.625$) when squaring it — rounding first can noticeably change the answer.

Worked Example 8 — Adding a value to a data set

Using the data in Worked Example 5, a $21$st value $x = 18$ is added. Find the new mean and standard deviation, and explain the change.

①New totals: $\sum x = 378$, $\sum x^2 = 7100 + 324 = 7424$, $n = 21$.

②$\bar{x} = \dfrac{378}{21} = 18$ — unchanged.

③$\sigma^2 = \dfrac{7424}{21} - 324 = 29.52\ldots$, so $\sigma = 5.43$.

④The new value equals the mean, so it adds nothing to $\sum (x - \bar{x})^2$ but increases $n$: the standard deviation falls.

4 Changing the Data, and Choosing Measures

Effect of a linear change $y = ax + b$

$\bar{y} = a\bar{x} + b$ $\sigma_y = |a|\,\sigma_x$
Adding $b$ shifts every value, so the average moves but the spread does not.

Situation	Best average	Best spread
Roughly symmetrical, no outliers	Mean	Standard deviation
Skewed, or with outliers	Median	Interquartile range
Qualitative data, or "most popular"	Mode	—

Justify your choice. "The median, because the data contain an outlier ($95$) which would distort the mean" earns the mark; "the median, because it is more accurate" does not.

5 Quick Reference

Mean

$\bar{x} = \dfrac{\sum fx}{\sum f}$; midpoints for grouped data.

Median (raw)

$\tfrac{n+1}{2}$th ordered value.

Median (grouped)

Interpolate at $\tfrac{n}{2}$.

Percentile

$P_k$ at position $\tfrac{k}{100}n$.

Variance

$\dfrac{\sum x^2}{n} - \bar{x}^2$

$S_{xx}$

$\sum x^2 - \dfrac{(\sum x)^2}{n}$; $\sigma^2 = \dfrac{S_{xx}}{n}$

Linear change

$y = ax + b$: mean $a\bar{x} + b$, sd $|a|\sigma$.

Outliers present

Use median and IQR.

Grouped data

All measures are estimates.

6 Practice Questions

Question 1

Find the mean, median and mode of $4,\; 9,\; 6,\; 4,\; 11,\; 8,\; 7$.

▶ Show solution

$\sum x = 49$, so $\bar{x} = 7$.

Ordered: $4, 4, 6, 7, 8, 9, 11$; median (4th value) $= 7$.

Mode $= 4$.

Question 2

Find the mean and median of the data: $x$: $1, 2, 3, 4, 5$ with frequencies $3, 7, 10, 6, 4$.

▶ Show solution

$\sum f = 30$, $\sum fx = 3 + 14 + 30 + 24 + 20 = 91$, so $\bar{x} = \dfrac{91}{30} = 3.03$ (3 s.f.).

Median: between the $15$th and $16$th values. Cumulative frequencies $3, 10, 20$, so both are $3$. Median $= 3$.

Question 3

For $8$ values, $\sum x = 84$ and $\sum x^2 = 1020$. Find the mean and standard deviation.

▶ Show solution

$\bar{x} = \dfrac{84}{8} = 10.5$

$\sigma^2 = \dfrac{1020}{8} - 10.5^2 = 127.5 - 110.25 = 17.25$

$\sigma = \sqrt{17.25} = 4.15$ (3 s.f.)

Question 4

The masses $m$ g of $40$ apples: $0 \leqslant m < 20$: $5$; $20 \leqslant m < 40$: $12$; $40 \leqslant m < 60$: $15$; $60 \leqslant m < 100$: $8$. Estimate the mean.

▶ Show solution

Midpoints $10, 30, 50, 80$.

$\sum fx = 50 + 360 + 750 + 640 = 1800$

$\bar{x} \approx \dfrac{1800}{40} = 45$ g

Question 5

For the data in Question 4, use interpolation to estimate the median and the interquartile range.

▶ Show solution

Cumulative frequencies: $5, 17, 32, 40$.

Median ($20$th): $40 + \dfrac{20 - 17}{15} \times 20 = 44$ g.

$Q_1$ ($10$th): $20 + \dfrac{10 - 5}{12} \times 20 = 28.3$ g. $Q_3$ ($30$th): $40 + \dfrac{30 - 17}{15} \times 20 = 57.3$ g.

IQR $\approx 57.3 - 28.3 = 29$ g.

Question 6

Find the standard deviation of the data in Question 2.

▶ Show solution

$\sum fx^2 = 3(1) + 7(4) + 10(9) + 6(16) + 4(25) = 3 + 28 + 90 + 96 + 100 = 317$

$\sigma^2 = \dfrac{317}{30} - \left(\dfrac{91}{30}\right)^2 = 10.5667 - 9.2011 = 1.3656$

$\sigma = 1.17$ (3 s.f.)

Question 7

Twelve values are $2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 17, 40$. Find the median and interquartile range, and explain why these are better summaries than the mean and standard deviation here.

▶ Show solution

Median: mean of the $6$th and $7$th values $= \dfrac{9 + 11}{2} = 10$.

Lower half $2, 4, 5, 7, 8, 9$: $Q_1 = \dfrac{5 + 7}{2} = 6$. Upper half $11, 12, 14, 15, 17, 40$: $Q_3 = \dfrac{14 + 15}{2} = 14.5$.

IQR $= 8.5$.

The value $40$ is far from the rest; it would pull the mean up and greatly inflate the standard deviation. The median and IQR are not affected by it.

Question 8

Daily temperatures have mean $12\,^\circ$C and standard deviation $4\,^\circ$C. Using $F = 1.8C + 32$, find the mean and standard deviation in $^\circ$F.

▶ Show solution

Mean: $1.8 \times 12 + 32 = 53.6\,^\circ$F.

Standard deviation: $1.8 \times 4 = 7.2\,^\circ$F (adding $32$ does not change the spread).

Question 9

Group A: $n = 10$, $\sum x = 150$, $\sum x^2 = 2350$. Group B: $n = 15$, $\sum x = 300$, $\sum x^2 = 6150$. Find the mean and standard deviation of the two groups combined.

▶ Show solution

Combine the totals: $n = 25$, $\sum x = 450$, $\sum x^2 = 8500$.

$\bar{x} = \dfrac{450}{25} = 18$

$\sigma^2 = \dfrac{8500}{25} - 18^2 = 340 - 324 = 16$, so $\sigma = 4$.

(Each group on its own has $\sigma^2 = 10$; the combined spread is larger because the group means, $15$ and $20$, differ.)

Question 10

The lengths $t$ minutes of $40$ phone calls are recorded.

$t$	$0 \leqslant t < 5$	$5 \leqslant t < 10$	$10 \leqslant t < 15$	$15 \leqslant t < 30$	$30 \leqslant t < 60$
Frequency	$14$	$11$	$7$	$6$	$2$

(a) Estimate the mean and standard deviation. (b) Estimate the median and quartiles. (c) Explain why your answers are estimates. (d) Which average better represents a typical call? Give a reason. (e) The phone company charges $5$p per call plus $2$p per minute. Estimate the mean and standard deviation of the cost of a call.

▶ Show solution

(a) Midpoints $2.5, 7.5, 12.5, 22.5, 45$.

$\sum fx = 35 + 82.5 + 87.5 + 135 + 90 = 430$, so $\bar{x} \approx \dfrac{430}{40} = 10.75$ min.

$\sum fx^2 = 87.5 + 618.75 + 1093.75 + 3037.5 + 4050 = 8887.5$

$\sigma^2 \approx \dfrac{8887.5}{40} - 10.75^2 = 222.1875 - 115.5625 = 106.625$, so $\sigma \approx 10.3$ min.

(b) Cumulative frequencies $14, 25, 32, 38, 40$.

Median ($20$th): $5 + \dfrac{20 - 14}{11} \times 5 = 7.73$ min.

$Q_1$ ($10$th): $0 + \dfrac{10}{14} \times 5 = 3.57$ min. $Q_3$ ($30$th): $10 + \dfrac{30 - 25}{7} \times 5 = 13.6$ min.

(c) The exact lengths are unknown: the mean assumes every value is at its class midpoint, and interpolation assumes values are spread evenly within each class.

(d) The median. The data are positively skewed (mean $10.75 >$ median $7.73$, and $Q_3 - Q_2 > Q_2 - Q_1$): a few long calls pull the mean up.

(e) Cost $= 2t + 5$ pence. Mean $\approx 2(10.75) + 5 = 26.5$p; standard deviation $\approx 2 \times 10.3 = 20.7$p.

Measures of Location and Spread · A Level Maths Year 1 · Created with MathJax