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Averages and spreadIB MYP Maths Standard: Revision notes

Section 1

Mean, median, mode and range

The mean is the total divided by the number of values. The median is the middle value when the data are in order (with an even number of values, halfway between the two middle ones). The mode is the most common value. The range is the largest value minus the smallest, and it measures the spread. Example: 12,15,13,12,40,14,1212, 15, 13, 12, 40, 14, 12. Mean =1187=16.9=\frac{118}{7}=16.9. Ordered: 12,12,12,13,14,15,4012,12,12,13,14,15,40 so the median is 1313, the mode is 1212 and the range is 40−12=2840-12=28.

Key termsmeanmedianmoderangespread
Common mistake

Finding the median without putting the values in order first.

Section 2

Frequency tables

In a frequency table, the total number of values is ∑f\sum f, and the total of all the values is ∑fx\sum fx (multiply each value by its frequency, then add). Mean =∑fx∑f=\frac{\sum fx}{\sum f}. The mode is the value with the highest frequency (not the frequency itself). For the median, use cumulative frequencies to find the middle position. Example: 0,1,2,3,40,1,2,3,4 goals with frequencies 4,6,5,3,24,6,5,3,2 (total 2020). ∑fx=33\sum fx=33, so the mean is 1.651.65. The mode is 11. The 10th value is 11 and the 11th is 22, so the median is 1.51.5.

Key termsfrequency tablecumulative frequency
Common mistake

Adding up the values 0+1+2+3+40+1+2+3+4 and dividing by 55. You must use the frequencies.

Section 3

Grouped data

In a grouped table the individual values are not known, so we can only estimate the mean. Use the midpoint of each class as the value for that class. Estimated mean =∑f×midpoint∑f=\frac{\sum f\times\text{midpoint}}{\sum f}. The modal class is the class with the highest frequency. The median class is the class containing the middle value. Example: classes 00 to <10<10, 1010 to <20<20, 2020 to <30<30, 3030 to <40<40, 4040 to <50<50 with frequencies 4,10,15,8,34,10,15,8,3. Midpoints 5,15,25,35,455,15,25,35,45 give ∑fx=960\sum fx=960 and an estimated mean of 96040=24\frac{960}{40}=24.

Key termsmidpointmodal classestimate
Exam tip

Check the answer is sensible: the estimated mean must lie inside the range of the data.

Section 4

Choosing the best average

  • Mean: uses all the values, but is pulled by outliers (extreme values).
  • Median: not affected by outliers, so it is best for skewed data, such as house prices or one very high value.
  • Mode: the only average for categorical data (such as the most popular colour), and best when the most common value matters, such as the most common shoe size. Example: ages 12,12,12,13,14,15,4012,12,12,13,14,15,40: the mean 16.916.9 is higher than the age of 6 of the 7 members, so the median 1313 is a better typical value.
Key termsoutlierskewed
Common mistake

Choosing the mean every time. Say why the chosen average is suitable for these data.

Section 5

Comparing two sets of data

To compare two data sets, make two comments: one about an average (mean or median) and one about the spread (range). Use the context, and compare the numbers. Example: Swift times have mean 2020 and range 44. Rapid times have mean 1818 and range 2424. Rapid is quicker on average, but Swift is more consistent because its range is smaller. In real-life (criterion D) problems, link the comparison to the situation (for example, exam papers must arrive within 2525 minutes) and say how reliable the data are. Small samples may not be typical.

Key termscompareconsistentreliable
Exam tip

Always write: average comment, spread comment, then a conclusion in context.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Averages and spread

  1. The ages, in years, of seven members of a chess club are 12, 15, 13, 12, 40, 14 and 12.
    Explain which average best represents the typical age of the members.2 marks
  2. A football team's goals scored per match over 20 matches: 0 goals in 4 matches, 1 goal in 6 matches, 2 goals in 5 matches, 3 goals in 3 matches and 4 goals in 2 matches.
    Find the mean number of goals per match.2 marks
  3. The heights of 40 plants are recorded in a grouped frequency table. In centimetres, 4 plants are in the class 0 ≤ h < 10, 10 plants in 10 ≤ h < 20, 15 plants in 20 ≤ h < 30, 8 plants in 30 ≤ h < 40 and 3 plants in 40 ≤ h < 50.
    Write down the modal class. Show that the median height lies in the class 20≤h<3020\leq h<30.3 marks
See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).