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# Summarising data

## Summarising Data

Once the data has been collected, it must be organised in a manner that allows for the information to be extracted most efficiently. For this reason it is useful to be able to summarise the data set by calculating a few quantities that give information about how the data values are spread and about the central values in the data set. Other methods of summarising and representing data will be covered in grade 11.

### Measures of Central Tendency

#### Mean or Average

The mean, (also known as arithmetic mean), is simply the arithmetic average of a group of numbers (or data set) and is shown using the bar symbol


¯¯. So the mean of the variable xx is x¯x¯ pronounced "x-bar". The mean of a set of values is calculated by adding up all the values in the set and dividing by the number of items in that set. The mean is calculated from the raw, ungrouped data.

Definition 1: Mean

The mean of a data set, xx, denoted by x¯x¯, is the average of the data values, and is calculated as:

x ¯ = sum of all values number of all values = x 1 + x 2 + x 3 + ... + x n n x ¯ = sum of all values number of all values = x 1 + x 2 + x 3 + ... + x n n
(1)

Method: Calculating the mean

1. Find the total of the data values in the data set.
2. Count how many data values there are in the data set.
3. Divide the total by the number of data values.
##### Exercise 1: Mean

What is the mean of x={10,20,30,40,50}x={10,20,30,40,50}?

###### Solution
1. Step 1. Find the total of the data values :
10 + 20 + 30 + 40 + 50 = 150 10 + 20 + 30 + 40 + 50 = 150
(2)
2. Step 2. Count the number of data values in the data set :

There are 5 values in the data set.

3. Step 3. Divide the total by the number of data values. :
150 ÷ 5 = 30 150 ÷ 5 = 30
(3)
4. Step 4. Answer :

the mean of the data set x={10,20,30,40,50}x={10,20,30,40,50} is 30.

#### Median

Definition 2: Median

The median of a set of data is the data value in the central position, when the data set has been arranged from highest to lowest or from lowest to highest. There are an equal number of data values on either side of the median value.

The median is calculated from the raw, ungrouped data, as follows.

Method: Calculating the median

1. Order the data from smallest to largest or from largest to smallest.
2. Count how many data values there are in the data set.
3. Find the data value in the central position of the set.
##### Exercise 2: Median

What is the median of {10,14,86,2,68,99,1}{10,14,86,2,68,99,1}?

###### Solution
1. Step 1. Order the data set from lowest to highest :

1,2,10,14,68,86,99

2. Step 2. Count the number of data values in the data set :

There are 7 points in the data set.

3. Step 3. Find the central position of the data set :

The central position of the data set is 4.

4. Step 4. Find the data value in the central position of the ordered data set. :

14 is in the central position of the data set.

5. Step 5. Answer :

14 is the median of the data set {1,2,10,14,68,86,99}{1,2,10,14,68,86,99}.

This example has highlighted a potential problem with determining the median. It is very easy to determine the median of a data set with an odd number of data values, but what happens when there is an even number of data values in the data set?

When there is an even number of data values, the median is the mean of the two middle points.

##### Tip: Finding the Central Position of a Data Set:
An easy way to determine the central position or positions for any ordered data set is to take the total number of data values, add 1, and then divide by 2. If the number you get is a whole number, then that is the central position. If the number you get is a fraction, take the two whole numbers on either side of the fraction, as the positions of the data values that must be averaged to obtain the median.
##### Exercise 3: Median

What is the median of {11,10,14,86,2,68,99,1}{11,10,14,86,2,68,99,1}?

###### Solution
1. Step 1. Order the data set from lowest to highest :

1,2,10,11,14,68,85,99

2. Step 2. Count the number of data values in the data set :

There are 8 points in the data set.

3. Step 3. Find the central position of the data set :

The central position of the data set is between positions 4 and 5.

4. Step 4. Find the data values around the central position of the ordered data set. :

11 is in position 4 and 14 is in position 5.

5. Step 5. Answer :

the median of the data set {1,2,10,11,14,68,85,99}{1,2,10,11,14,68,85,99} is

( 11 + 14 ) ÷ 2 = 12 , 5 ( 11 + 14 ) ÷ 2 = 12 , 5
(4)

#### Mode

Definition 3: Mode

The mode is the data value that occurs most often, i.e. it is the most frequent value or most common value in a set.

Method: Calculating the mode Count how many times each data value occurs. The mode is the data value that occurs the most.

The mode is calculated from grouped data, or single data items.

##### Exercise 4: Mode

Find the mode of the data set x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}

###### Solution
1. Step 1. Count how many times each data value occurs. :
 data value frequency data value frequency 1 1 6 1 2 1 7 1 3 1 8 2 4 3 9 1 5 1 10 2
2. Step 2. Find the data value that occurs most often. :

4 occurs most often.

3. Step 3. Answer :

The mode of the data set x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}x={1,2,3,4,4,4,5,6,7,8,8,9,10,10} is 4. Since the number 4 appears the most frequently.

A data set can have more than one mode. For example, both 2 and 3 are modes in the set 1, 2, 2, 3, 3. If all points in a data set occur with equal frequency, it is equally accurate to describe the data set as having many modes or no mode.

Figure 1
Khan academy video on statistics

### Measures of Dispersion

The mean, median and mode are measures of central tendency, i.e. they provide information on the central data values in a set. When describing data it is sometimes useful (and in some cases necessary) to determine the spread of a distribution. Measures of dispersion provide information on how the data values in a set are distributed around the mean value. Some measures of dispersion are range, percentiles and quartiles.

#### Range

Definition 4: Range

The range of a data set is the difference between the lowest value and the highest value in the set.

Method: Calculating the range

1. Find the highest value in the data set.
2. Find the lowest value in the data set.
3. Subtract the lowest value from the highest value. The difference is the range.
##### Exercise 5: Range

Find the range of the data set x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}

###### Solution
1. Step 1. Find the highest and lowest values. :

10 is the highest value and 1 is the lowest value.

2. Step 2. Subtract the lowest value from the highest value to calculate the range. :
10 - 1 = 9 10 - 1 = 9
(5)
3. Step 3. Answer :

For the data set x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}x={1,2,3,4,4,4,5,6,7,8,8,9,10,10}, the range is 9.

#### Quartiles

Definition 5: Quartiles

Quartiles are the three data values that divide an ordered data set into four groups containing equal numbers of data values. The median is the second quartile.

The quartiles of a data set are formed by the two boundaries on either side of the median, which divide the set into four equal sections. The lowest 25% of the data being found below the first quartile value, also called the lower quartile. The median, or second quartile divides the set into two equal sections. The lowest 75% of the data set should be found below the third quartile, also called the upper quartile. For example:

 22 24 48 51 60 72 73 75 80 88 90 ↓ ↓ ↓ ↓ ↓ ↓ Lower quartile Median Upper quartile (Q1Q1) (Q2Q2) (Q3Q3)

Method: Calculating the quartiles

1. Order the data from smallest to largest or from largest to smallest.
2. Count how many data values there are in the data set.
3. Divide the number of data values by 4. The result is the number of data values per group.
4. Determine the data values corresponding to the first, second and third quartiles using the number of data values per quartile.
##### Exercise 6: Quartiles

What are the quartiles of {3,5,1,8,9,12,25,28,24,30,41,50}{3,5,1,8,9,12,25,28,24,30,41,50}?

###### Solution
1. Step 1. Order the data set from lowest to highest :

{ 1 , 3 , 5 , 8 , 9 , 12 , 24 , 25 , 28 , 30 , 41 , 50 } { 1 , 3 , 5 , 8 , 9 , 12 , 24 , 25 , 28 , 30 , 41 , 50 }

2. Step 2. Count the number of data values in the data set :

There are 12 values in the data set.

3. Step 3. Divide the number of data values by 4 to find the number of data values per quartile. :
12 ÷ 4 = 3 12 ÷ 4 = 3
(6)
4. Step 4. Find the data values corresponding to the quartiles. :
 1 3 5 ∥ ∥ 8 9 12 ∥ ∥ 24 25 28 ∥ ∥ 30 41 50 Q 1 Q 1 Q 2 Q 2 Q 3 Q 3

The first quartile occurs between data position 3 and 4 and is the average of data values 5 and 8. The second quartile occurs between positions 6 and 7 and is the average of data values 12 and 24. The third quartile occurs between positions 9 and 10 and is the average of data values 28 and 30.

5. Step 5. Answer :

The first quartile = 6,5. (Q1Q1)

The second quartile = 18. (Q2Q2)

The third quartile = 29. (Q3Q3)

#### Inter-quartile Range

Definition 6: Inter-quartile Range

The inter quartile range is a measure which provides information about the spread of a data set, and is calculated by subtracting the first quartile from the third quartile, giving the range of the middle half of the data set, trimming off the lowest and highest quarters, i.e. Q3-Q1Q3-Q1.

The semi-interquartile range is half the interquartile range, i.e. Q3-Q12Q3-Q12

##### Exercise 7: Medians, Quartiles and the Interquartile Range

A class of 12 students writes a test and the results are as follows: 20, 39, 40, 43, 43, 46, 53, 58, 63, 70, 75, 91. Find the range, quartiles and the Interquartile Range.

###### Solution
1. Step 1. :
 20 39 40 ∥ ∥ 43 43 46 ∥ ∥ 53 58 63 ∥ ∥ 70 75 91 Q 1 Q 1 M M Q 3 Q 3
2. Step 2. The Range :

The range = 91 - 20 = 71. This tells us that the marks are quite widely spread. (Remember, however, that 'wide' and 'large' are relative terms. If you are considering one hundred people, a range of 71 would be 'large', but if you are considering one million people, a range of 71 would likely be 'small', depending, of course, on what you were analyzing).

3. Step 3. The median lies between the 6th and 7th mark :

i.e. M=46+532=992=49,5M=46+532=992=49,5

4. Step 4. The lower quartile lies between the 3rd and 4th mark :

i.e. Q1=40+432=832=41,5Q1=40+432=832=41,5

5. Step 5. The upper quartile lies between the 9th and 10th mark :

i.e. Q3=63+702=1332=66,5Q3=63+702=1332=66,5

6. Step 6. Analysing the quartiles :

The quartiles are 41,5, 49,5 and 66,5. These quartiles tell us that 25%% of the marks are less than 41,5; 50%% of the marks are less than 49,5 and 75%% of the marks are less than 66,5. They also tell us that 50%% of the marks lie between 41,5 and 66,5.

7. Step 7. The Interquartile Range :

The Interquartile Range = 66,5 - 41,5 = 25. This tells us that the width of the middle 50%% of the data values is 25.

8. Step 8. The Semi-interquatile Range :

The Semi-interquartile Range = 252252 = 12,5

#### Percentiles

Definition 7: Percentiles

Percentiles are the 99 data values that divide a data set into 100 groups.

The calculation of percentiles is identical to the calculation of quartiles, except the aim is to divide the data values into 100 groups instead of the 4 groups required by quartiles.

Method: Calculating the percentiles

1. Order the data from smallest to largest or from largest to smallest.
2. Count how many data values there are in the data set.
3. Divide the number of data values by 100. The result is the number of data values per group.
4. Determine the data values corresponding to the first, second and third quartiles using the number of data values per quartile.

### Five number summary

We can summarise a data set by using the five number summary. The five number summary gives the lowest data value, the highest data value, the median, the first (lower) quartile and the third (higher) quartile. Consider the following set of data: 5, 3, 4, 6, 2, 8, 5, 4, 6, 7, 3, 6, 9, 4, 5. We first order the data as follows: 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 8, 9. The lowest data value is 2 and the highest data value is 9. The median is 5. The first quartile is 4 and the third quartile is 6. So the five number summary is: 2, 4, 5, 6, 9.

### Box and whisker plots

The five number summary can be shown graphically in a box and whisker plot. The main features of the box and whisker diagram are shown in Figure 2. The box can lie horizontally (as shown) or vertically. For a horizonatal diagram, the left edge of the box is placed at the first quartile and the right edge of the box is placed at the third quartile. The height of the box is arbitrary, as there is no y-axis. Inside the box there is some representation of central tendency, with the median shown with a vertical line dividing the box into two. Additionally, a star or asterix is placed at the mean value, centered in the box in the vertical direction. The whiskers which extend to the sides reach the minimum and maximum values. This is shown for the data set: 5, 3, 4, 6, 2, 8, 5, 4, 6, 7, 3, 6, 9, 4, 5.

#### Exercise 8

Draw a box and whisker diagram for the data set: x={1,25;1,5;2,5;2,5;3,1; 3,2;4,1;4,25;4,75;4,8;4,95;5,1}x={1,25;1,5;2,5;2,5;3,1;3,2;4,1;4,25;4,75;4,8;4,95;5,1}.

##### Solution
1. Step 1. Determine the five number summary:
Minimum=1,25Minimum=1,25
Maximum=5,10Maximum=5,10
The position of first quartile is between 3 and 4.
The position of second quartile is between 6 and 7.
The position of third quartile is between 9 and 10.
The data value between 3 and 4 is: 12(2,5+2,5)=2,512(2,5+2,5)=2,5
The data value between 6 and 7 is: 12(3,2+4,1)=3,6512(3,2+4,1)=3,65
The data value between 9 and 10 is: 12(4,75+4,8)=4,77512(4,75+4,8)=4,775
2. Step 2. Draw a box and whisker diagram and mark the positions of the minimum, maximum and quartiles:

### Exercises - Summarising Data

1. Three sets of data are given:
1. Data set 1: 9 12 12 14 16 22 24
2. Data set 2: 7 7 8 11 13 15 16 16
3. Data set 3: 11 15 16 17 19 19 22 24 27
For each one find:
1. the range
2. the lower quartile
3. the interquartile range
4. the semi-interquartile range
5. the median
6. the upper quartile
Click here for the solution
2. There is 1 sweet in one jar, and 3 in the second jar. The mean number of sweets in the first two jars is 2.
1. If the mean number in the first three jars is 3, how many are there in the third jar?
2. If the mean number in the first four jars is 4, how many are there in the fourth jar?
Click here for the solution
3. Find a set of five ages for which the mean age is 5, the modal age is 2 and the median age is 3 years.
Click here for the solution
4. Four friends each have some marbles. They work out that the mean number of marbles they have is 10. One of them leaves. She has 4 marbles. How many marbles do the remaining friends have together?
Click here for the solution
5. Jason is working in a computer store. He sells the following number of computers each month: 27; 39; 3; 15; 43; 27; 19; 54; 65; 23; 45; 16 Give a five number summary and a box and whisker plot of his sales.
Click here for the solution
6. Lisa works as a telesales person. She keeps a record of the number of sales she makes each month. The data below show how much she sells each month. 49; 12; 22; 35; 2; 45; 60; 48; 19; 1; 43; 12 Give a five number summary and a box and whisker plot of her sales.
Click here for the solution
7. Rose has worked in a florists shop for nine months. She sold the following number of wedding bouquets: 16; 14; 8; 12; 6; 5; 3; 5; 7
1. What is the five-number summary of the data?
2. Since there is an odd number of data points what do you observe when calculating the five-numbers?
Click here for the solution

We can apply the concepts of mean, median and mode to data that has been grouped. Grouped data does not have individual data points, but rather has the data organized into groups or bins. To calculate the mean we need to add up all the frequencies and divide by the total. We do not know what the actual data values are, so we approximate by choosing the midpoint of each group. We then multiply those midpoint numbers by the frequency. Then we add these numbers together to find the approximate total of the masses. The modal group is the group with the highest frequency. The median group is the group that contains the middle terms.

Measures of dispersion can also be found for grouped data. The range is found by subtracting the smallest number in the lowest bin from the largest number in the highest bin. The quartiles are found in a similar way to the median.

#### Exercise 9: Mean, Median and Mode for Grouped Data

Consider the following grouped data and calculate the mean, the modal group and the median group.

 Mass (kg) Frequency 41 - 45 7 46 - 50 10 51 - 55 15 56 - 60 12 61 - 65 6 Total = 50
##### Solution
1. Step 1. Calculating the mean :

To calculate the mean we need to add up all the masses and divide by 50. We do not know actual masses, so we approximate by choosing the midpoint of each group. We then multiply those midpoint numbers by the frequency. Then we add these numbers together to find the approximate total of the masses. This is show in the table below.

 Mass (kg) Midpoint Frequency Midpt ×× Freq 41 - 45 (41+45)/2 = 43 7 43 ×× 7 = 301 46 - 50 48 10 480 51 - 55 53 15 795 56 - 60 58 12 696 61 - 65 63 6 378 Total = 50 Total = 2650
2. Step 2. Answer :

The mean = 265050=53265050=53.

The modal group is the group 51 - 53 because it has the highest frequency.

The median group is the group 51 - 53, since the 25th and 26th terms are contained within this group.

#### More mean, modal and median group exercises.

In each data set given, find the mean, the modal group and the median group.

1. Times recorded when learners played a game.
 Time in seconds Frequency 36 - 45 5 46 - 55 11 56 - 65 15 66 - 75 26 76 - 85 19 86 - 95 13 96 - 105 6
Click here for the solution
2. The following data were collected from a group of learners.
 Mass in kilograms Frequency 41 - 45 3 46 - 50 5 51 - 55 8 56 - 60 12 61 - 65 14 66 - 70 9 71 - 75 7 76 - 80 2
Click here for the solution

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