Year 7 Mathematics | Victorian Curriculum 2.0
Summary statistics
Topic 14 | Statistics & Probability | Practice

What you will learn

  • calculate the mean (average) of a data set,
  • find the median (middle value when sorted),
  • find the mode (most common value),
  • find the range (max −-− min) as a simple measure of spread,
  • choose the most appropriate measure for a situation.
Why three different averages?

One number can’t tell the whole story about a data set. The mean is the balance point — it uses every value but gets pulled by extremes. The median sits in the middle and ignores outliers. The mode shows the most common value. Each answers a different question, so knowing all three lets you choose the one that gives the fairest picture for the situation.

Where you'll see this
  • Sports: batting averages, goals per match, winning percentages.
  • School reports: class test means and medians.
  • News: “the average Australian worker earns $…”
  • Economics: median house price, average household income.
Worked example 0 Real-world example: which 'average' house price?

Five houses on a street sold for $420 000, $440 000, $450 000, $460 000, and $1 800 000 (a mansion).

400k600k800k1.0M1.8Moutliermean 714kmedian 450k
  1. Mean: (420+440+450+460+1800)÷5=3570÷5=714(420 + 440 + 450 + 460 + 1800) \div 5 = 3570 \div 5 = 714(420+440+450+460+1800)÷5=3570÷5=714 (thousands).
  2. Median (middle value when sorted): 450450450 (thousands) — that’s $450 000.
  3. The mansion pulls the mean up to $714 000 — far above what four of the five houses actually cost.
  4. A news headline saying “average house price $714 000” is technically correct but misleading. The median ($450 000) better represents a typical house on this street.

Key idea: one outlier can drag the mean far from the centre. That is why property reports use the median — it resists extreme values.

1. Mean

The mean adds all the values and divides by how many there are.

Mean
mean  =  sum of all valuesnumber of values.\text{mean} \;=\; \frac{\text{sum of all values}}{\text{number of values}}.mean=number of valuessum of all values​.
Worked example 1 Mean of five numbers

Find the mean of 7,9,11,10,87, 9, 11, 10, 87,9,11,10,8.

mean=7+9+11+10+85=455=9.\text{mean} = \frac{7 + 9 + 11 + 10 + 8}{5} = \frac{45}{5} = 9.mean=57+9+11+10+8​=545​=9.

2. Median

The median is the middle value after sorting the data from smallest to largest. If the count is even, the median is the mean of the two middle values.

Worked example 2 Median - odd count

Find the median of 5,8,3,9,75, 8, 3, 9, 75,8,3,9,7.

  1. Sort: 3,5,7,8,93, 5, 7, 8, 93,5,7,8,9.
  2. The middle (3rd of 5) is 777.
median=7.\text{median} = 7.median=7.
Worked example 3 Median - even count

Find the median of 2,4,6,82, 4, 6, 82,4,6,8.

Sort: 2,4,6,82, 4, 6, 82,4,6,8. The two middle values are 444 and 666.

median=4+62=5.\text{median} = \frac{4 + 6}{2} = 5.median=24+6​=5.

3. Mode

The mode is the value that appears most often. A data set can have one mode, two modes (bimodal), or no mode (all values unique).

1,2,2,3,4,4,4,5  →  mode=41, 2, 2, 3, 4, 4, 4, 5 \;\to\; mode = 41,2,2,3,4,4,4,5→mode=4.

1,2,2,3,3,4  →  modes=2and31, 2, 2, 3, 3, 4 \;\to\; modes = 2 and 31,2,2,3,3,4→modes=2and3 (bimodal).

4. Range

The range is a simple measure of spread.

Range
range=largest−smallest.\text{range} = \text{largest} - \text{smallest}.range=largest−smallest.

5. Which measure to use?

Choosing a measure

Mean

Uses every value. Sensitive to outliers (extreme values can pull it away from the middle).

Median

Only position matters. Not sensitive to outliers - useful when the data has extreme values.

Mode

The typical or most-common value. Useful for categorical data, where mean and median do not apply.

Worked example 4 Effect of an outlier

A small business has five salaries (in thousands): $40, $45, $50, $55, $300 (the owner).

Mean=40+45+50+55+3005=4905=98.\text{Mean} = \frac{40 + 45 + 50 + 55 + 300}{5} = \frac{490}{5} = 98.Mean=540+45+50+55+300​=5490​=98.

So the mean salary is $98 000. The median (middle after sorting) is $50 000.

The mean is pulled up sharply by the owner’s high salary; the median gives a better picture of “typical” pay.

Symmetricmean ≈ medianRight-skewed (e.g. salaries)medianmeantail pullsmean right
Symmetric vs skewed data. When data is symmetric, mean ≈ median. When skewed, the mean is pulled toward the tail — use the median instead.

Practice

Fluency

Tier 1: basic skills

    1. Find the mean of 4,6,8,10,124, 6, 8, 10, 124,6,8,10,12.
    2. Find the mean of 3,7,5,9,63, 7, 5, 9, 63,7,5,9,6.
    3. Find the mean of 12,14,18,2012, 14, 18, 2012,14,18,20.
    4. Find the median of 1,3,5,7,91, 3, 5, 7, 91,3,5,7,9.
    5. Find the median of 4,8,6,2,104, 8, 6, 2, 104,8,6,2,10.
    6. Find the median of 3,5,8,103, 5, 8, 103,5,8,10.
    7. Find the median of 11,13,14,17,18,2011, 13, 14, 17, 18, 2011,13,14,17,18,20.
    8. Find the mode of 2,3,3,5,7,3,92, 3, 3, 5, 7, 3, 92,3,3,5,7,3,9.
    9. Find the mode of 4,4,6,6,7,84, 4, 6, 6, 7, 84,4,6,6,7,8.
    10. Find the range of 2,7,5,9,32, 7, 5, 9, 32,7,5,9,3.
    11. Find the range of 45,60,72,38,5545, 60, 72, 38, 5545,60,72,38,55.
    12. A data set: 10,12,14,16,18,2010, 12, 14, 16, 18, 2010,12,14,16,18,20. Find the mean.
    13. For the set 5,5,6,8,115, 5, 6, 8, 115,5,6,8,11, find the mean, median, mode, range.
Reasoning

Tier 2: mixed practice

    Use this data for questions 1-5: 12,15,18,15,14,20,17,15,16,1812, 15, 18, 15, 14, 20, 17, 15, 16, 1812,15,18,15,14,20,17,15,16,18.

    1. Sort the data and find the median.

    2. Find the mean.

    3. Find the mode.

    4. Find the range.

    5. If you added the value 100100100 to this data set, which of mean, median, mode, range would change most? Explain briefly.

    6. Five students scored an average of 707070 on a test. Four of the scores are 65,72,68,7565, 72, 68, 7565,72,68,75. Find the fifth score.

    7. The mean of 666 numbers is 181818. Five of them are 15,20,12,18,2215, 20, 12, 18, 2215,20,12,18,22. Find the sixth.

    8. A data set has mean 151515 and 101010 values. If one value is wrongly recorded as 888 but should be 181818, what is the correct mean?

    9. Give an example of a data set with mean 101010, median 101010, mode 101010 and range 000.

    10. Give an example of a data set with 555 values where mean >>> median.

Reasoning

Tier 3: explain and spot the mistake

    1. A student writes: “the mode is 101010 because 101010 is the biggest number in the list”. Explain the confusion and give the correct definition.
    2. Explain why the median is usually a better measure than the mean when a data set has a single extreme value.
    3. Can a data set have a mode but no mean? Can it have a mean but no mode? Explain both.
    4. Is the median always in the data set? Give an example where it is not.
Problem solving

Tier 4: real-world problems

    1. A cricket batter’s last 777 scores are 23,45,12,38,62,51,2923, 45, 12, 38, 62, 51, 2923,45,12,38,62,51,29. Find the mean and median.
    2. A family has children aged 4,7,10,134, 7, 10, 134,7,10,13. Another child aged 161616 joins. Find the new mean age.
    3. The daily temperatures ( degC) for a week: 22,25,21,28,30,24,2322, 25, 21, 28, 30, 24, 2322,25,21,28,30,24,23. Find the mean temperature and the range.
    4. Seven students scored an average of 606060 marks. Adding a new student with a score of 767676 changes the class size to 888. What is the new mean?
    5. A data set of 101010 values has a mean of 505050. The smallest value is 202020 and the largest is 909090. What is the mean of the middle 888 values (the 101010 values with the min and max removed)?
Year 7 Mathematics study companion | Practice