Should I use n2 or n+12 for the median position?

Use n+12 for raw discrete data, and n2 when interpolating in grouped continuous data.

2.02StatisticsFoundation

Measures of Central Tendency

Q: What does coding do to the median?

The same as to the mean: subtract a and divide by b.

An average summarises a whole data set with one representative value. You already know the mean, median and mode; A-Level adds the notation to calculate them efficiently from frequency tables and grouped data, the technique of interpolation for grouped medians, and — most importantly — the judgement to say which average suits a particular data set and why.

35 min Data Analysis

The big picture

Statistics questions at A-Level are rarely just calculations. OCR expects you to interpret: which average is appropriate, how an outlier distorts the mean but not the median, and what a summary statistic says about a real context such as the large data set. The mean uses every value and has the best mathematical properties, which is why it underpins variance, the normal distribution and hypothesis testing later on. The median is resistant to extreme values. The mode is the only average for categorical data. Sigma notation, $\bar{x} = \dfrac{\sum fx}{\sum f}$ , is the language for all of it, and coding shows how averages respond when data are shifted or rescaled.

What you'll be able to do

Calculate the mean, median and mode from raw data and frequency tables
Estimate the mean and median from grouped data, using interpolation for the median
Use sigma notation and the effect of linear coding on the mean
Choose and justify an appropriate average, including in the presence of outliers

Mean, median and mode

The $\textbf{mean}$ is $\bar{x} = \dfrac{\sum x}{n}$ : the sum of the values divided by how many there are.

The $\textbf{median}$ is the middle value when the data are ordered. For $n$ values it is the $\tfrac{n+1}{2}$ th value; if $n$ is even, average the two middle values.

The $\textbf{mode}$ is the most frequent value. Grouped data have a $\textbf{modal class}$ instead. Data can be bimodal, or have no mode.

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

Mean of raw data.

Example 1Find the mean and median of

12, 7, 15, 9, 7, 22, 11, 13

1Sum

= 96

n = 8

, so

\bar{x} = 12

2Ordered:

7, 7, 9, 11, 12, 13, 15, 22

. The 4th and 5th values are 11 and 12.

AnswerMean

12

; median

11.5

Frequency tables

When values repeat, a frequency table records each value $x$ with its frequency $f$ . The mean is $\bar{x} = \dfrac{\sum fx}{\sum f}$ .

For the median, find its position $\tfrac{n+1}{2}$ (for discrete data) and use cumulative frequencies to locate the value.

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

Mean from a frequency table.

Example 1The number of goals in 20 matches: 0 goals ×4, 1 ×6, 2 ×5, 3 ×3, 4 ×2. Find the mean and median.

\sum fx = 0 + 6 + 10 + 9 + 8 = 33

\sum f = 20

, so

\bar{x} = 1.65

2Median position: between the 10th and 11th values. Cumulative frequencies:

4, 10, 15, \dots

3The 10th value is 1 and the 11th is 2.

AnswerMean

1.65

; median

1.5

Grouped data

With grouped data the exact values are lost, so averages are $\textbf{estimates}$ . For the mean, use the class midpoint as $x$ in $\dfrac{\sum fx}{\sum f}$ .

For the median of continuous grouped data, find the $\tfrac{n}{2}$ th position and use $\textbf{linear interpolation}$ within the class that contains it, assuming values are spread evenly across the class.

Pay attention to class boundaries. Ages "20–29" in completed years run from 20 up to (but not including) 30.

Example 1Times (minutes) for 50 commuters:

0 \le t < 10

: 8;

10 \le t < 20

: 17;

20 \le t < 30

: 15;

30 \le t < 50

: 10. Estimate the mean and median.

1Midpoints

5, 15, 25, 40

\sum fx = 40 + 255 + 375 + 400 = 1070

, so

\bar{x} \approx 21.4

2Median at position

25

. Cumulative:

8, 25, 40, 50

— the 25th value is at the end of the

10

–

20

class.

3Interpolating:

10 + \dfrac{25 - 8}{17} \times 10 = 20

AnswerMean

\approx 21.4

minutes; median

\approx 20

minutes.

Tip — State that grouped values are estimates, and why: "the midpoint is used because the exact values within each class are unknown".

Coding and choosing an average

If data are coded by $y = \dfrac{x - a}{b}$ , then $\bar{y} = \dfrac{\bar{x} - a}{b}$ . Adding a constant shifts the mean; multiplying scales it. The median and mode transform the same way.

An $\textbf{outlier}$ pulls the mean towards it but hardly affects the median. So for skewed data or data with outliers, the median is usually more representative.

The mean is preferred for roughly symmetric data because it uses every value. The mode is the only average possible for categorical (non-numerical) data, such as method of travel.

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

Coding and the mean.

Why this works

Average income is a classic trap: a few very high earners drag the mean well above what a typical person earns. Reported "typical" incomes normally use the median for exactly this reason.

Think like an examiner

State that grouped-data averages are estimates.
Show the interpolation calculation for a grouped median.
Justify the choice of average with a reason tied to the data — outliers, skew, or data type.
Interpret averages in context with units.

Common misconceptions

Dividing

\sum fx

by the number of rows in the table.

Divide by the total frequency

\sum f

Using class widths or upper bounds in place of midpoints.

Use midpoints for estimating the mean.

Taking the median of a frequency table as the middle row.

Use cumulative frequency to find the middle data value.

Choosing the mean for a data set with a large outlier.

The median is less affected by extreme values.

Averages

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

Raw data.

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

Frequency or grouped data.

\bar{y} = \tfrac{\bar{x} - a}{b}

Coding.

Stretch yourself

The mean mass of 12 parcels is 3.5 kg. Two more parcels are added and the mean of all 14 becomes 3.8 kg. One of the new parcels weighs 5.1 kg. Find the mass of the other, and explain whether it is likely to be an outlier.

Hint — Convert each mean into a total mass.

Reveal solution

SolutionOriginal total

= 12 \times 3.5 = 42

kg. New total

= 14 \times 3.8 = 53.2

kg. The two new parcels together weigh

53.2 - 42 = 11.2

kg, so the other has mass

11.2 - 5.1 = 6.1

kg. Whether it is an outlier cannot be decided from averages alone: we would need a measure of spread for the original 12 parcels. If, for instance, their standard deviation were about 0.5 kg, a 6.1 kg parcel would be over five standard deviations above the mean and clearly an outlier; if the spread were large, it might be unremarkable.

\blacksquare

Questions students ask

Why are averages from grouped data only estimates?

Grouping loses the individual values, so you assume each value sits at the class midpoint (for the mean) or is spread evenly (for the median).

Should I use

\tfrac{n}{2}

\tfrac{n+1}{2}

for the median position?

Use

\tfrac{n+1}{2}

for raw discrete data, and

\tfrac{n}{2}

when interpolating in grouped continuous data.

Which average is best?

It depends on the data: mean for symmetric numerical data, median when there are outliers or skew, mode for categorical data.

What does coding do to the median?

The same as to the mean: subtract

a

and divide by

b

Key takeaways

Mean $\bar{x} = \sum x / n$ , or $\sum fx / \sum f$ from a table.
Median is the middle value; interpolate within the median class for grouped data.
Mode is the most common value or class; the only average for categorical data.
Grouped-data averages are estimates.
Median resists outliers; the mean uses every value.