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VCE 12 General 2023

1.03 Measures of centre and spread

Lesson

Measures of centre

The mean is often referred to as the average. To calculate the mean, add all the scores in a data set, then divide this by number of scores.

The mean of a data set, denoted \overline{x}, is given by \overline{x}=\dfrac{\Sigma x}{n}.

The median is the middle score in a data set, when the data is sorted to be in order. If there are two middle scores, we take the average of them to find the median.

To find the median of a data set:

  • Sort the data points into order

  • If there are an odd number of scores, the median is the middle score (the \dfrac{n+1}{2} score

  • If there are an even number of scores, the median is calculated by finding the mean of the two middle scores (the \dfrac{n}{2}th and \left(\dfrac{n}{2} + 1\right) th scores.)

The mode describes the most frequently occurring score.

Suppose that 10 people were asked how many pets they had. 2 people said they didn't own any pets, 6 people had one pet and 2 people said they had two pets.

In this data set, the most common number of pets that people have is one pet, and so the mode of this data set is 1.

A data set can have more than one mode, if two or more scores are equally tied as the most frequently occurring.

The mode of a data set is the most frequently occurring score.

Examples

Example 1

Find the mean of the following scores: -14,\,0,\,-2,\,-18,\,-8,\,0,\,-15,\,-1

Worked Solution
Create a strategy

We need to add all the score and divide it by the number of scores.

Apply the idea
\displaystyle \text{Mean}\displaystyle =\displaystyle \dfrac{-14+0-2-18-8+0-15-1}{8}Add all the scores and divide by 8
\displaystyle =\displaystyle \dfrac{-58}{8}Evaluate the addition
\displaystyle =\displaystyle -7.25Evaluate the division

Example 2

Find the median of the nine numbers below:1,\,1,\,3,\,5,\,7,\,9,\,9,\,10,\,15

Worked Solution
Create a strategy

The median in an odd set of scores is the \left( \dfrac{n+1}{2} \right)th score, where n is the total number of scores.

Apply the idea

There are 9 scores.

\displaystyle \text{Median position}\displaystyle =\displaystyle \dfrac{9+1}{2} Substitute the values
\displaystyle =\displaystyle \dfrac{10}{2} Evaluate the numerator
\displaystyle =\displaystyle 5\text{th score} Perform the division

Therefore the median will be the 5th score. \text{Median}=7

Reflect and check

This data set had an odd number of scores, so the median was just the middle score.

Example 3

Find the median from the frequency distribution table:

\text{Score }(x)\text{Frequency }(f)
232
2426
2537
2624
2725
Worked Solution
Create a strategy

To find the median we need to add a cumulative frequency column.

Apply the idea

To find the median, we create a cumulative frequency column.

ScoreFrequencyCumulative frequency
2322
24262+26=28
253728+37=65
262465+24=89
272589+25=114

Since there are 114 scores, the median will be the average of the 57th and 58th score. Looking at the cumulative frequency table, there are 28 scores less than or equal to 24 and 65 scores less than or equal to 25. This means that the 57th and 58th scores are both 25, so the median is 25.

\text{Median}=25

Example 4

Find the mode of the following scores:8,\,18,\,5,\,2,\,2,\,10,\,8,\,5,\,14,\,14,\,8,\,8,\,10,\,18,\,14,\,5

Worked Solution
Create a strategy

Choose which number in the set appeared the most.

Apply the idea

8 appeared in the set 4 times.

18 appeared in the set 2 times.

5 appeared in the set 3 times.

2 appeared in the set 2 times.

10 appeared in the set 2 times.

14 appeared in the set 3 times.

\text{Mode}= 8

Idea summary

The mean of a data set, denoted \overline{x}, is given by \overline{x}=\dfrac{\Sigma x}{n}

To find the median of a data set:

  • Sort the data points into order

  • If there are an odd number of scores, the median is the middle score (the \dfrac{n+1}{2} score

  • If there are an even number of scores, the median is calculated by finding the mean of the two middle scores (the \dfrac{n}{2}th and \left(\dfrac{n}{2} + 1\right) th scores.)

The mode of a data set is the most frequently occurring score.

Measure of spread

The range is the simplest measure of spread in a quantitative (numerical) data set. It is the difference between the maximum and minimum scores in a data set.

For example, at one school the ages of students in Year 7 vary between 11 and 14. So the range for this set is 14-11=3.

As a different example, if we looked at the ages of people waiting at a bus stop, the youngest person might be a 7 year old and the oldest person might be a 90 year old. The range of this set of data is 90-7=83, which is a much larger range of ages.

The Range of a numerical data set is given by: \text{Range }=\text{maximum score}-\text{minimum score}.

When we are trying to the understand what our data is telling us, we might want to find measures of centre (the median, mean, and modes) as well as measures of spread, such as the range. However, the range is easily affected by outliers. To get a better picture of the spread in a data set, we can instead find the set's quartiles.

Each quartile represents 25\% of the data set. The Lower quartile marks 25\% of the way through data. The median, which we already know, marks 50\% of the way through the data, and the upper quartile marks 75\% of the way through the data. We can use these to measure the spread of the data. For example, 50\% of the scores in a data set lie between the lower and upper quartiles.

We already know how to calculate the median of a data set. Calculating quartiles is exactly the same, but we use only the lower half of the scores (for the lower quartile), or the upper half of the scores (for the upper quartiles).

Here is a data set with 8 scores:

A data set with 8 scores. The scores are 1, 3, 4, 7, 11, 12, 14, 19.

First locate the median, between the 4\text{th} and 5\text{th} scores:

A data set with 8 scores. The scores are 1, 3, 4, 7, 11, 12, 14, 19, the median is located between 7 and 11.

Now there are four scores in each half of the data set, so split each of the four scores in half to find the quartiles. We can see the first quartile, Q_{1} is between the 2\text{nd} and 3\text{rd} scores, so there are two scores on either side of Q_{1}. Similarly, the third quartile, Q_{3} is between the 6\text{th} and 7\text{th} scores:

Scores 1, 3, 4, 7, 11, 12, 14, 19. Quartile 1 is between 3 and 4, quartile 3 is between 12 and 14.

Now let's look at a situation with 9 scores:

Scores 8, 8, 10, 11, 13, 14, 18, 22, 25. Quartile 1 is between 8 and 8, the median is 13, quartile 3 is between 18 and 22.

This time, the 5\text{th} term is the median. There are four terms on either side of the median, like for the set with eight scores. So Q_{1} is still between the 2\text{nd} and 3\text{rd} scores and Q_{3} is between the 6\text{th} and 7\text{th} scores.

Finally, let's look at a set with 10 scores:

A data set scores 12, 13, 14, 19, 19, 21, 22, 22, 28, 30. Quartile 1 is 14, the median is between 19 and 21, quartile 3 is 22

For this set, the median is between the 5\text{th} and 6\text{th} scores. This time, however, there are 5 scores on either side of the median. So Q_{1} is the 3\text{rd} term and Q_{3} is the 8\text{th} term.

  • The first quartile is also called the lower quartile. It is the middle score between the lowest score and the median and it represents the 25th percentile. The first quartile score is the \dfrac{n+1}{4}th score, where n is the total number of scores.

  • The second quartile is the median, which we have already learnt about and it represents the 50th percentile.

  • The third quartile is also called the upper quartile. It is the middle score between the median and the highest score. It represents the 75th percentile. The third quartile is the \dfrac{3(n+1)}{4}th score, where n is the total number of scores.

The interquartile range (IQR) is the difference between the upper quartile and the lower quartile. 50\% of scores lie within the IQR because two full quartiles lie in this range.

Subtract the first quartile from the third quartile. That is, IQR = Q_{3}-Q_{1}

Examples

Example 5

Find the range of the following set of scores: 10,\, 7,\, 2,\, 14,\, 13,\, 15,\, 11,\, 4

Worked Solution
Create a strategy

Find the difference between the highest and lowest score.

Apply the idea

From the set of scores, the highest score is 15 and the lowest score is 2.

\displaystyle \text{Range}\displaystyle =\displaystyle 15-2Subtract 2 from 15
\displaystyle =\displaystyle 13Evaluate

Example 6

Consider the following set of scores:33,\,38,\,50,\,12,\,33,\,48,\,41

a

Sort the scores in ascending order.

Worked Solution
Create a strategy

Arrange the scores from smallest to largest.

Apply the idea

12,\,33,\,33,\,38,\,41,\,48,\,50

b

Find the number of scores.

Worked Solution
Create a strategy

Count the total number of scores.

Apply the idea

\text{Number of scores} = 7

c

Find the median.

Worked Solution
Create a strategy

The median in an odd set of scores is the \left( \dfrac{n+1}{2} \right)th score, where nn is the total number of scores.

Apply the idea

There are 7 scores.

\displaystyle \text{Median position}\displaystyle =\displaystyle \dfrac{7+1}{2} Substitute the values
\displaystyle =\displaystyle \dfrac{8}{2} Evaluate the numerator
\displaystyle =\displaystyle 4\text{th score} Perform the division

Therefore the median will be the 4th score. \text{Median}=7

d

Find the first quartile of the set of scores.

Worked Solution
Create a strategy

Use the first half of the scores excluding the median.

Apply the idea

The first half of the scores are: 12,\,33,\,33

The median of this set is 33.

So, the first quartile of the original set of scores is 33.

e

Find the third quartile of the set of scores.

Worked Solution
Create a strategy

Use the second half of the scores excluding the median.

Apply the idea

The second half of the scores are: 41,\,48,\,50

The median of this set is 48.

So, the third quartile of the original set of scores is 48.

f

Find the interquartile range.

Worked Solution
Create a strategy

We can use the interquartile range formula: \text{IQR} = Q_{3} - Q_{1}

Apply the idea
\displaystyle \text{IQR}\displaystyle =\displaystyle 48-33Substitute the quartiles
\displaystyle =\displaystyle 15Evaluate
Idea summary
\displaystyle \text{{Range = Maximum score - Minimum score}}
\bm{\text{Range}}
is the difference between the smallest and largest scores in the set set.
\displaystyle \text{IQR} = Q_{3} - Q_{1}
\bm{\text{IQR}}
is the interquartile range which is difference between third quartile and first quartile

Outcomes

U3.AoS1.5

mean and sample standard deviation

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