Statistical Measures - Extended Maths

Averages for discrete data

An average is a single number that summarises a whole list. There are three of them, and each answers a slightly different question. Alongside them sits the range, which measures how spread out the numbers are.

Mean: Add every value, then divide by how many values there are. This is the "fair share" average.
Median: Put the values in order and take the middle one. With an even count, take the mean of the two middle values.
Mode: The value that appears most often. A list can have one mode, several, or none.
Range: The largest value minus the smallest. It measures spread, not a central value.

Written as a formula, the mean of \(n\) values is:

\[ \bar{x} = \frac{\sum x}{n} \]

The symbol \(\bar{x}\) is read "x-bar" and \(\sum x\) means "the sum of all the values".

Worked example

Seven students scored: 4, 7, 7, 2, 9, 5, 8. Find the mean, median, mode and range.

Mean: add them up: \(4+7+7+2+9+5+8 = 42\). Divide by \(n = 7\): \(\bar{x} = \dfrac{42}{7} = 6\).

Median: order the list: \(2,\,4,\,5,\,\mathbf{7},\,7,\,8,\,9\). With 7 values the middle is the 4th, so the median is \(7\).

Mode: \(7\) appears twice, more than any other value, so the mode is \(7\).

Range: largest minus smallest: \(9 - 2 = 7\).

Mean \(= 6\), median \(= 7\), mode \(= 7\), range \(= 7\).

Order first, always

The most common median slip is reading the middle of the unordered list. Sorting the values takes a few seconds and removes the mistake entirely.

The estimated mean from a grouped table

When data comes grouped into class intervals, you no longer know each exact value, only the interval it fell in. So you cannot find the true mean, but you can estimate it. The trick is to assume every value in a class sits at the midpoint of that class.

The midpoint is the mean of the two class boundaries. For the class \(0 \le t < 10\), the midpoint is \(\dfrac{0 + 10}{2} = 5\). Multiply each midpoint \(x\) by its frequency \(f\), total those products, and divide by the total frequency:

\[ \text{estimated mean} = \frac{\sum f x}{\sum f} \]

Worked example

The table shows how long 20 pupils spent on homework one night. Estimate the mean time.

Add a midpoint column \(x\) and an \(f x\) column:

Time \(t\) (min)	Frequency \(f\)	Midpoint \(x\)	\(f x\)
\(0 \le t < 10\)	4	5	20
\(10 \le t < 20\)	7	15	105
\(20 \le t < 30\)	6	25	150
\(30 \le t < 40\)	3	35	105
Total	20		380

Now divide the totals: \(\text{estimated mean} = \dfrac{\sum f x}{\sum f} = \dfrac{380}{20} = 19\) minutes.

Where this is assessed

Building and reading a frequency table sits in Criterion A (knowing and applying the method) and Criterion C (communicating with a clear, labelled table). Choosing which average best describes a real data set, and saying why, is Criterion D.

Quartiles and the interquartile range

The median splits ordered data in half. Quartiles go further and split it into quarters. The lower quartile \(Q_1\) is a quarter of the way in, the median \(Q_2\) is halfway, and the upper quartile \(Q_3\) is three-quarters of the way in.

The interquartile range (IQR) is the distance between the upper and lower quartiles:

\[ \text{IQR} = Q_3 - Q_1 \]

Because it ignores the smallest and largest quarter of the data, the IQR measures the spread of the middle half and is not thrown off by a single extreme value.

Worked example

Find the median, quartiles and IQR of: 3, 5, 6, 8, 8, 9, 11, 12, 14, 15, 18.

The list is already ordered and has \(n = 11\) values. The median is the middle (6th) value: \(Q_2 = 9\).

Take the lower half, the 5 values below the median: \(3, 5, 6, 8, 8\). Its middle is \(Q_1 = 6\).

Take the upper half, the 5 values above the median: \(11, 12, 14, 15, 18\). Its middle is \(Q_3 = 14\).

Interquartile range: \(\text{IQR} = Q_3 - Q_1 = 14 - 6 = 8\).

\(Q_1 = 6\), median \(= 9\), \(Q_3 = 14\), IQR \(= 8\).

Odd counts: leave the median out

With an odd number of values, do not include the median itself in either half when finding the quartiles. Here the median \(9\) is set aside, leaving a clean five in each half. Different textbooks use slightly different rules, so stick to the one your class teaches.

Which measure, and when

Part of the skill is choosing the right summary for the data in front of you, not just calculating all of them.

Mean: uses every value, so it is the most informative, but a single very large or very small value drags it off.
Median: ignores the exact size of extreme values, so it is the safer average when the data has outliers, for example house prices or salaries.
Mode: the only average that works for non-numerical data, such as the most common shoe size or favourite colour.
Range vs IQR: the range is quick but sensitive to one freak value; the IQR describes the reliable middle half.

Justify your choice

In an investigation (Criterion B and D), a sentence such as "the median is more suitable here because the one very high value would distort the mean" earns marks that the number alone never will.

Check yourself

Work each one through before opening the answer.

1. Find the mean, median and range of: 12, 15, 11, 20, 12. +

Sum \(= 12+15+11+20+12 = 70\), so the mean \(= \dfrac{70}{5} = 14\). Ordered: \(11, 12, 12, 15, 20\), so the median (middle of five) is \(12\). Range \(= 20 - 11 = 9\). So mean \(= 14\), median \(= 12\), range \(= 9\).

2. Estimate the mean of this grouped data: \(0\le x<20\) with \(f=3\), \(20\le x<40\) with \(f=5\), \(40\le x<60\) with \(f=2\). +

Midpoints are \(10, 30, 50\). Then \(\sum f x = 3(10) + 5(30) + 2(50) = 30 + 150 + 100 = 280\) and \(\sum f = 3 + 5 + 2 = 10\). Estimated mean \(= \dfrac{280}{10} = \) 28.

3. For 2, 4, 5, 7, 8, 10, 13, find the quartiles and the interquartile range. +

There are \(n = 7\) values, so the median is the 4th value, \(7\). Set the median aside. Lower half \(2, 4, 5\) gives \(Q_1 = 4\); upper half \(8, 10, 13\) gives \(Q_3 = 10\). So \(\text{IQR} = 10 - 4 = \) 6 (with \(Q_1 = 4\), \(Q_3 = 10\)).

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