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Averages and Measures of Spread

Introduction:

Averages and measures of spread help us to summarise large sets of data and compare different data sets. The most common types of averages are mean, median, and mode. Each type has its own strengths and is used in different situations.

1. The Mean

The mean is what most people commonly refer to as the "average." It is calculated by adding up all the values in a data set and then dividing by the total number of values.

How to Work Out the Mean:

  1. Add up all the values in the data set.
  2. Divide the total by the number of data values.
Mean=Sum of all data valuesNumber of data valuesMean=\frac{\text{Sum of all data values}}{\text{Number of data values}}
infoNote

Example 1: Finding the Mean

Question: Find the mean of the following set of numbers: 4,6,8,10,12.4, 6, 8, 10, 12.

Solution:

  1. Add up all the values:
4+6+8+10+12=404+6+8+10+12=40
  1. Divide the total by the number of values:
405=8\frac{40}5=8

The mean is 88.


2. The Median

The median is the middle value in a data set when all the values are arranged in ascending order (from smallest to largest). If there is an even number of values, the median is the average of the two middle numbers.

How to Work Out the Median:

  1. Arrange the data values in ascending order.
  2. Find the middle value. If there are two middle values, find their mean.
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Example 2: Finding the Median

Question: Find the median of the following set of numbers: 3,1,9,7,5.3, 1, 9, 7, 5.

Solution:

  1. Arrange the numbers in ascending order: 1,3,5,7,91,3,5,7,9

  2. The middle value is 5, so the median is 55.

Note: If the data set has an even number of values, you find the mean of the two middle numbers.

infoNote

Example 3: Median with Even Number of Values

Question: Find the median of the following set of numbers: 2,4,6,8.2, 4, 6, 8.

Solution:

  1. Arrange the numbers in ascending order: 2,4,6,82,4,6,8

  2. The middle values are 44 and 66. Find the mean of these two values: 4+62=5\frac{4+6}{2}=5

The median is 55.


3. The Mode

The mode is the value that appears the most frequently in a data set. A data set may have more than one mode if multiple values occur with the same highest frequency. If no value repeats, there is no mode.

How to Work Out the Mode:

  1. Identify the most frequent value(s) in the data set.
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Example 4: Finding the Mode

Question: Find the mode of the following set of numbers: 2,4,4,5,6,6,6,72, 4, 4, 5, 6, 6, 6, 7.

Solution:

  • The number 66 appears three times, more frequently than any other number. The mode is 66.

4. The Range

The range measures the spread of the data by finding the difference between the largest and smallest values. A large range indicates that the data varies greatly, while a small range suggests that the data is more consistent

infoNote

Example: Finding the Range

Question: Find the range of the following set of numbers: 12,7,5,15,912, 7, 5, 15, 9.

Solution:

  1. The largest value is 1515.
  2. The smallest value is 55.
  3. Subtract the smallest value from the largest value:
155=1015−5=10

The range is 1010.

Why Is the Range Important?

The range gives us an idea of how spread out the data is. A large range indicates that the values in the data set vary significantly, while a small range means the values are clustered closer together.


Worked Example (The Big Cricket Example):

infoNote

Example

Andrew Flintoff and Michael Vaughan are comparing their cricket performances. Their scores are as follows:

  • Andrew Flintoff's Scores:
20,35,22,55,60,10,17,32,64,86,14,32,50,24,3020, 35, 22, 55, 60, 10, 17, 32, 64, 86, 14, 32, 50, 24, 30
  • Michael Vaughan's Scores:
0,0,0,2,3,15,5,370,250,14,0,3,5,10, 0, 0, 2, 3, 15, 5, 370, 250, 14, 0, 3, 5, 1

We can calculate the Mean for both cricketers.


1. The Mean

Step 1: Add up all their scores.

  • Andrew Flintoff's Total Runs:
20+35+22+55+60+10+17+32+64+86+14+32+50+24+30=551 runs.20 + 35 + 22 + 55 + 60 + 10 + 17 + 32 + 64 + 86 + 14 + 32 + 50 + 24 + 30 = 551 \ runs.
  • Michael Vaughan's Total Runs:
0+0+0+2+3+15+5+370+250+14+0+3+5+1=651 runs.0 + 0 + 0 + 2 + 3 + 15 + 5 + 370 + 250 + 14 + 0 + 3 + 5 + 1 = 651 \ runs.

Step 2: Divide the total by the number of games.

  • Andrew Flintoff's Games Played: 1515 games.

  • Michael Vaughan's Games Played: 1414 games.

Andrew Flintoff’s Mean=55115=36.7runs\text{Andrew Flintoff's Mean}=\frac{551}{15}=36.7 runsMichael Vaughan’s Mean=65114=46.5runs\text{Michael Vaughan's Mean}=\frac{651}{14}=46.5 runs

What Does This Tell Us?

  • On average, Michael Vaughan has scored 46.546.5 runs per game, while Andrew Flintoff has scored 36.736.7 runs.
  • It looks like Michael Vaughan has had the better season, based on the mean alone.

2. The Median

How to Work Out the Median:

  1. Arrange the data values in ascending order.
  2. Identify the middle value. If there are two middle values (even number of data points), find the mean of those two values.

Example: Comparing Cricketers Using the Median

Let's find the median for each cricketer's scores:

Andrew Flintoff's Scores:

10,14,17,20,22,24,30,32,32,35,50,55,60,64,8610, 14, 17, 20, 22, 24, 30, 32, 32, 35, 50, 55, 60, 64, 86
  • The median is the middle value. Since there are 1515 scores, the median is the 8th8th value:
Median=32Median = 32

Michael Vaughan's Scores:

0,0,0,0,1,2,3,3,5,5,14,15,250,3700, 0, 0, 0, 1, 2, 3, 3, 5, 5, 14, 15, 250, 370
  • The median is the mean of the 7th7th and 8th8th values:
3+32=3\frac{3+3}{2}=3

What does this tell us?

The median shows a different picture. Michael Vaughan's median score is 33, which is much lower than Andrew Flintoff's median of 3232. This suggests that Michael Vaughan was much more inconsistent, scoring very low in most games but achieving a couple of exceptionally high scores.

  • Good thing about the median: It is not affected by outliers, so it gives a better sense of typical performance.
  • Bad thing about the median: It doesn't consider all the data points, only focusing on the middle, which can ignore important information about the rest of the data.

3. The Mode

How to Work Out the Mode:

  1. Identify the most common value in the data set.
  2. This value is the mode.

Example: Comparing Cricketers Using the Mode

Andrew Flintoff's Mode:

  • The most frequent score is 32, which appears multiple times in his data set.

  • Mode = 3232 Michael Vaughan's Mode:

  • The most frequent score is 00, which appears the most often in his data set.

  • Mode = 00 What does this tell us?

Once again, using the mode suggests that Andrew Flintoff had the better season. His most frequent score was 3232, which shows that he consistently scored decent runs. Michael Vaughan, however, had 0 as his mode, indicating that he often scored no runs in many games.

  • Good thing about the mode: It's very quick and easy to find.
  • Bad thing about the mode: The mode can sometimes be misleading, especially when data is skewed or has outliers. In Michael Vaughan's case, his mode is 00 because of his many low scores, but this doesn't reflect the two very high scores of 370370 and 250250.

4. The Range

How to Work Out the Range:

  1. Subtract the smallest value from the largest value.
Range=LargestvalueSmallestvalueRange=Largest value−Smallest value

Example: Comparing Cricketers Using the Range

Andrew Flintoff's Range:

  • Largest score: 8686
  • Smallest score: 1010
  • Range:
8610=7686−10=76

Michael Vaughan's Range:

  • Largest score: 370370
  • Smallest score: 00
  • Range:
3700=370370−0=370

What does this tell us?

At first glance, someone might think that Michael Vaughan had the better season because he has the larger range of 370370, but that's not entirely correct. A larger range means that his scores varied significantly, showing inconsistency. Andrew Flintoff, with a smaller range of 7676, had a more consistent performance.

  • Good thing about the range: It gives a quick idea of how spread out the data is.
  • Bad thing about the range: The range is heavily affected by outliers. Michael Vaughan's very high scores of 370370 and 250250 make his range large, but they don't reflect his many low scores of 00.

Who Had the Better Season?

When comparing Andrew Flintoff and Michael Vaughan using these statistics, it becomes clear that:

  • Andrew Flintoff had a more consistent season, with a higher mode of 3232 and a smaller range of 7676. This suggests he performed well in most matches.
  • Michael Vaughan, although he had some brilliant high scores (like 370370 and 250250), had many low scores, leading to an inconsistent performance. His mean was distorted by the outliers, and his range was very large, showing great variability in his results.

Worked Example: Estimating the Mean Attendance at Football Matches

infoNote

Example:

The table below shows the attendance at Preston North End football matches over a season:

Attendance RangeFrequency (Number of Matches)
0x<5,0000≤x<5,0005
5,000x<10,0005,000≤x<10,0001212
10,000x<15,00010,000≤x<15,0002424
15,000x20,00015,000≤x≤20,00088

Step-by-Step Guide:

  1. Work out the mid-point for each group.
  • The mid-point is the average of the lower and upper boundaries of each class.
  • For the first group 0x<5,000,0≤x<5,000, the mid-point is:
Midpoint=0+5,0002=2,500Mid-point=\frac{0+5,000}2=2,500
  • Continue for the remaining groups.
  1. Multiply the mid-point by the frequency for each group.
  • This gives an estimate of the total attendance for each group of matches.
  1. Calculate the estimated mean using the formula:
Mean=(Midpoint×Frequency)TotalFrequencyMean=\frac{∑(Mid-point×Frequency)}{Total Frequency}

Step-by-Step Solution:

Attendance RangeMid-pointFrequencyMid-point × Frequency
0x<5,0000≤x<5,0002,5002,5005512,50012,500
5,000x<10,0005,000≤x<10,0007,5007,500121290,00090,000
10,000x<15,00010,000≤x<15,00012,50012,5002424300,000300,000
15,000x20,00015,000≤x≤20,00017,50017,50088140,000140,000

Totals:

  • Total frequency (number of matches) = 4949
  • Total of mid-point × frequency = 542,500542,500 Now, calculate the estimated mean:
Mean=542,50049=11,071(nearest whole number)Mean=\frac{542,500}{49}=11,071 (\text{nearest whole number})
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