L3StatisticsFoundation

Measures of Location

Q: Why does the median position rule change for grouped data?

For a list you are selecting an actual listed value, so n+12 points at it. For grouped data you are locating a point on a continuous scale, where n2 is the correct halfway position.

A measure of location answers "where is the middle?" — but there are three sensible answers, and they can differ sharply on the same data. Knowing $\textit{which}$ to use, and being able to justify it, is worth more marks than the arithmetic.

30 min Data Presentation

The big picture

The mean uses every value, which is its strength and its weakness: it is efficient, it feeds into everything later (variance, the normal distribution, hypothesis tests on a mean), and a single extreme value can drag it somewhere unrepresentative. The median ignores everything except position, so an outlier cannot move it far — which is exactly why house prices and salaries are reported as medians. Recognising which of these behaviours a data set needs is the judgement being tested, and it recurs the moment you meet skewed distributions later in the chapter.

What you'll be able to do

Calculate the mean, median and mode from a list and from a frequency table
Estimate the mean and median from grouped data using midpoints and interpolation
Choose an appropriate measure of location and justify it in context
Use coding to simplify calculations with awkward numbers

The three averages

The $\textbf{mean}$ is the total divided by how many values there are. It uses every data point, which makes it sensitive to extremes.

The $\textbf{median}$ is the middle value once the data is ordered. With $n$ values it sits at position $\tfrac{n+1}{2}$ ; for even $n$ that lands between two values, so take their mean. Because it depends only on position, extremes barely affect it.

The $\textbf{mode}$ is the most frequent value. It is the only average that works for categorical data, but a data set can have several modes or none at all.

\bar{x} = \dfrac{\Sigma x}{n}

Mean of a list.

\bar{x} = \dfrac{\Sigma fx}{\Sigma f}

Mean from a frequency table.

\text{median position} = \dfrac{n+1}{2}

Position, not value.

Example 1Find the mean and median of

4, 7, 7, 9, 43

1Mean:

\dfrac{4+7+7+9+43}{5} = \dfrac{70}{5} = 14

2Median: five values, so position 3, which is

7

AnswerMean

14

, median

7

Why this works

In that example the mean of 14 is larger than four of the five values. The single 43 has pulled it clear of the data — a compact demonstration of why an outlier makes the mean a poor summary.

Grouped data: estimating the mean

Grouped data tells you how many values fall in each class but not what they were. The standard assumption is that values are evenly spread within a class, so each class is represented by its $\textbf{midpoint}$ .

The result is an $\textbf{estimate}$ , and the word matters — questions ask for "an estimate of the mean" and expect you to know why it cannot be exact.

Class boundaries need care with continuous data. A class recorded as $10$ – $19$ for a quantity measured to the nearest whole number actually spans $9.5$ to $19.5$ , so its midpoint is $14.5$ , not $14.5$ by accident — check the boundaries before halving.

\bar{x} \approx \dfrac{\Sigma f\,x_{\text{mid}}}{\Sigma f}

Midpoints stand in for the raw values.

Example 1Estimate the mean from:

0\!-\!10

(freq 4),

10\!-\!20

(freq 6),

20\!-\!30

(freq 10).

1Midpoints:

5

15

25

\Sigma fx = 4(5) + 6(15) + 10(25) = 20 + 90 + 250 = 360

\Sigma f = 20

AnswerEstimated mean

= \dfrac{360}{20} = 18

Tip — Always call it an estimate and say why: the original values are unknown, so midpoints are assumed to represent each class.

Grouped data: interpolating the median

For grouped data the median is found by linear interpolation. Locate the class containing the $\tfrac{n}{2}$ th value using cumulative frequency, then assume the values are evenly spread across that class and step the appropriate fraction of the way into it.

Note the position rule changes: for grouped data use $\tfrac{n}{2}$ rather than $\tfrac{n+1}{2}$ , because you are locating a position on a continuous scale rather than picking out a listed value.

m = L + \dfrac{\tfrac{n}{2} - F}{f}\times w

L

lower boundary,

F

cumulative frequency before,

f

class frequency,

w

class width.

Example 1Using the table above (

0\!-\!10

: 4,

10\!-\!20

: 6,

20\!-\!30

: 10), estimate the median.

n = 20

, so we want the 10th value.

2Cumulative: 4 after the first class, 10 after the second — so the 10th value is at the top of the

10\!-\!20

class.

m = 10 + \dfrac{10-4}{6}\times10 = 10 + 10

AnswerEstimated median

= 20

Why this works

Interpolation assumes an even spread inside the class, which is the same assumption the midpoint method makes for the mean. Both estimates inherit whatever error that assumption introduces.

Choosing a measure, and coding

Use the $\textbf{mean}$ when the data is roughly symmetric with no extreme values, and when the result feeds into later calculations. Use the $\textbf{median}$ when the data is skewed or contains outliers. Use the $\textbf{mode}$ for categorical data or when the most common value is what matters — a shop ordering shoe sizes wants the mode, not the mean.

$\textbf{Coding}$ simplifies awkward arithmetic. If $y = \dfrac{x - a}{b}$ , then $\bar{x} = b\bar{y} + a$ — the mean transforms exactly as the data does, because it is a linear operation.

y = \dfrac{x-a}{b} \Rightarrow \bar{x} = b\bar{y}+a

Coding the mean.

Example 1The coding

y = \dfrac{x-100}{5}

gives

\bar{y} = 3.2

. Find

\bar{x}

1Reverse the coding:

\bar{x} = 5\bar{y} + 100

= 5(3.2) + 100 = 16 + 100

Answer

\bar{x} = 116

Tip — A justification must name the feature of the data. "The median, because the data contains an outlier at 43 which would distort the mean" scores; "the median, because it is better" does not.

Think like an examiner

Say "estimate" for grouped data and give the reason — the original values are unknown.
Use $\tfrac{n+1}{2}$ for a list and $\tfrac{n}{2}$ for grouped interpolation. Mixing them is a standard error.
Check class boundaries before taking midpoints, especially for data rounded to the nearest unit.
When comparing two data sets, compare a measure of location $\textit{and}$ a measure of spread — one alone rarely scores full marks.

Common misconceptions

Dividing

\Sigma fx

by the number of classes.

Divide by

\Sigma f

, the total frequency, not by how many groups there are.

Finding the median without ordering the data.

The median is the middle of the

\textit{ordered}

list. Sort first.

Quoting a grouped mean as exact.

It is an estimate based on midpoints; the raw values are not recoverable.

Forgetting to reverse the coding at the end.

\bar{y}

is not the answer. Convert back with

\bar{x} = b\bar{y}+a

Location

\bar{x} = \tfrac{\Sigma x}{n}

Mean of a list.

\bar{x} = \tfrac{\Sigma fx}{\Sigma f}

Mean from frequencies.

\tfrac{n+1}{2}

Median position, listed data.

m = L + \tfrac{n/2 - F}{f}w

Median by interpolation.

\bar{x} = b\bar{y}+a

Reversing a coding.

Stretch yourself

Nine employees earn (in £000s): 22, 24, 24, 26, 27, 28, 30, 31, 168. Calculate the mean and median, state which better represents typical pay, and justify your choice.

Hint — Compute both, then look at how many employees actually earn near each value.

Reveal solution

SolutionTotal

= 22+24+24+26+27+28+30+31+168 = 380

, so the mean is

\tfrac{380}{9} = 42.2

(£42,200). With nine values the median is at position

\tfrac{9+1}{2} = 5

, giving £27,000. The median is the better representative: the mean of £42,200 exceeds the pay of eight of the nine employees, because the single £168,000 salary — almost certainly a senior figure and an outlier relative to the rest — pulls it upwards. The median is unaffected by how extreme that one value is, since it depends only on position. This is exactly why national pay statistics are reported as medians rather than means.

\blacksquare

Questions students ask

Why does grouped data give only an estimate of the mean?

Because the individual values are lost. Using midpoints assumes each class is evenly spread around its centre, which is rarely exactly true.

Why does the median position rule change for grouped data?

For a list you are selecting an actual listed value, so

\tfrac{n+1}{2}

points at it. For grouped data you are locating a point on a continuous scale, where

\tfrac{n}{2}

is the correct halfway position.

When is the mode the right choice?

For categorical data, where no mean exists, and when the most common outcome is what matters — stock ordering, most popular option, modal class of a distribution.

Does coding change the answer?

No, provided you reverse it. Coding is a device for keeping the arithmetic manageable; the final value must be converted back to the original scale.

Key takeaways

Mean uses every value and is sensitive to outliers; median depends on position and is not.
Mode is the only average available for categorical data.
Grouped data gives estimates: midpoints for the mean, interpolation for the median.
Justify a choice by naming the feature of the data — skew, outlier, categorical.
Coding simplifies arithmetic but must be reversed with $\bar{x} = b\bar{y}+a$ .