Averages (Mean, Median, Mode and Range): Question 1

Worked solution

Setting up

It helps to put the marks in order (smallest to largest) before doing any of the calculations, because the median (and a clear view of the mode and range) depends on the ordered list.

$1, \quad 3, \quad 4, \quad 5, \quad 7, \quad 8, \quad 8, \quad 8, \quad 10$

There are $n = 9$ values.

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(a) Mode

The mode is the value that occurs most often.

Counting the ordered list, the value $8$ appears three times, while every other value appears only once.

$\text{Mode} = \boxed{8}$

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(b) Median

The median is the middle value of the ordered data.

With $n = 9$ values, the middle position is

$\frac{n+1}{2} = \frac{9+1}{2} = 5\text{th value}.$

Counting to the 5th value in the ordered list:

$1,\ 3,\ 4,\ \underbrace{5}_{4\text{th}},\ \underbrace{7}_{5\text{th}},\ 8,\ 8,\ 8,\ 10$

$\text{Median} = \boxed{7}$

Tip: with $4$ values below it and $4$ values above it, $7$ is exactly in the middle, a quick check that you have located the median correctly.

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(c) Range

The range measures the spread:

$\text{Range} = \text{largest} - \text{smallest} = 10 - 1 = \boxed{9}$

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(d) Mean

The mean is the total of all the marks divided by how many marks there are.

First find the total:

$1 + 3 + 4 + 5 + 7 + 8 + 8 + 8 + 10 = 54$

Then divide by the number of quizzes, $n = 9$:

$\text{Mean} = \frac{54}{9} = \boxed{6}$

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Summary

All four are different here, which is a useful reminder that mode, median and mean are three distinct “averages” and the range describes spread rather than a typical value.