Introduction

The range of a data set is the difference between the greatest and smallest values in the data set.

To demonstrate, let's find the range of the following data set: 2, \: 3, \: 3, \: 6, \: 7, \: 9

Notice that our data set is already arranged from smallest to greatest.

The smallest value is \color{red}2 and the greatest is {\color{blue}9}\mathbin{:}

{\color{red}\underline{2}}, \: 3, \: 3, \: 6, \: 7, \: {\color{blue}\underline{9}}

Therefore, the range of the data set is:

{\color{blue}\text{greatest}} - {\color{red}\text{smallest}} = {\color{blue}9} - {\color{red}2} = 7

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Example: Finding the Range

Suppose that the temperatures registered yesterday in the southwest of the United States were as follows: 19^\circ\text{C}, \quad 29^\circ\text{C}, \quad 15^\circ\text{C}, \quad 12^\circ\text{C}, \quad 16^\circ\text{C}, \quad 26^\circ\text{C}, \quad 29^\circ\text{C} What is the range of temperatures?

EXPLANATION

Our data set is not arranged from smallest to greatest. So, let's first order the numbers:

12, \: 15, \: 16, \: 19, \: 26, \: 29, \: 29

Now, we see that the smallest value is \color{red}12 and the greatest is {\color{blue}29} \mathbin{:}

{\color{red}\underline{12}}, \: 15, \: 16, \: 19, \: 26, \: 29, \: {\color{blue}\underline{29}}

Therefore, the range of the data set is:

{\color{blue}\text{greatest}} - {\color{red}\text{smallest}} = {\color{blue}29} - {\color{red}12} = 17

So, the range of temperatures is 17^{\circ}\text{C}.

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Practice: Finding the Range

Find the range of the following data set: \[ 1, \: 2, \: 3, \: 3, \: 7 \]

a
 $4$
b
 $1$
c
 $8$
d
 $3$
e
 $6$
Practice: Finding the Range

Find the range of the following data set: \[ 2.2, \: 2.8, \: 3.5, \: 4.4, \: 5.7, \: 6.9 \]

a
 $4.4$
b
 $0.9$
c
 $3.5$
d
 $1.2$
e
 $4.7$
Lower and Upper Quartiles

To compute the lower quartile and upper quartile of a data set, we first need to order the data set from smallest to greatest and then split the data set into two halves: the half below the median and the half above the median. Then,

  • the lower quartile is the median of the lower half, and

  • the upper quartile is the median of the upper half.

For example, consider the following data set:

7, \: 13, \: 15, \: 21, \: 24, \: 29

Notice that our data set is already arranged from smallest to greatest.

The lower half is what's below the median, and the upper half is what's above the median:

\underbrace{7, \: 13, \: 15}_{\text{lower half}}, \: \underbrace{21, \: 24, \: 29}_{\text{upper half}}

To find the lower quartile, we find the median of the lower half: 7, \: {\color{blue}\underline{13}}, \: 15

So, the lower quartile is {\color{blue}{13}}.

Now, to find the upper quartile, we find the median of the upper half:

21, \: {\color{red}\underline{24}}, \: 29

So, the upper quartile is {\color{red}{24}}.

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Example: Finding the Lower or Upper Quartile: Even Number of Elements

Arthur recorded how many hours he worked per day each day over a period of 6 days and got the following data: 7, \: 5, \: 6, \: 7, \: 6, \: 8 Find the lower quartile of this data set.

EXPLANATION

Our data set is not arranged from smallest to greatest. So, let's first order the numbers: 5, \: 6, \: 6, \: 7, \: 7, \: 8

Now, we consider the lower half of the data: \underbrace{5, \: 6, \: 6,}_{\text{lower half}} \: 7, \: 7, \: 8

Next, we find the median of the lower half. In this case, it's the middle number: 5, \: {\color{blue}\underline{6}}, \:6

So, the lower quartile is \bbox[3pt,Gainsboro]{\color{blue}6}.

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Practice: Finding the Lower or Upper Quartile: Even Number of Elements

What is the lower quartile of the data set below?

\[ 19, \: 20, \: 22, \: 24, \: 28, \: 29, \: 30, \: 33 \]

a
 $22$
b
 $21.5$
c
 $20.5$
d
 $20$
e
 $21$
Practice: Finding the Lower or Upper Quartile: Even Number of Elements

What is the upper quartile of the given data set?

\[ 46, \: 60, \: 41, \: 34, \: 55, \: 48, \: 51, \: 37 \]

a
 $52$
b
 $53$
c
 $51$
d
 $48$
e
 $54$
Example: Finding the Lower or Upper Quartile: Odd Number of Elements

Marlon asked his friends about their weights and got the following data (in kilograms):

63.2, \: 50, \: 48.4, \: 45.6, \: 45, \: 55, \: 58.9, \: 92.1, \: 88

What is the lower quartile of the given data set?

EXPLANATION

Our data set is not arranged from smallest to greatest. So, let's first order the numbers:

45, \: 45.6, \: 48.4, \: 50, \: 55, \: 58.9, \: 63.2, \: 88, \: 92.1

The median of the data set is the middle number ( \color{blue}55 ).

Now, we consider the lower half of the data:

\underbrace{45, \: 45.6, \: 48.4, \: 50,}_{\text{lower half}} \: {\color{blue}55}, \: 58.9, \: 63.2, \: 88, \: 92.1

We want to find the lower quartile, so we need to find the median of the lower half. In this case, it's the mean of the two middle numbers:

45, \: {\color{red}\underline{45.6}}, \: {\color{red}\underline{48.4}}, \: 50

So, the lower quartile is: \dfrac{{\color{red}45.6}+{\color{red}48.4}}{2} = \dfrac{94}{2} = 47

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Practice: Finding the Lower or Upper Quartile: Odd Number of Elements

What is the upper quartile of the given data set?

\[ 10, \: 14, \: 16, \: 20, \: 24, \: 26, \: 27, \: 28, \: 30 \]

a
 $28$
b
 $27.5$
c
 $28.5$
d
 $26$
e
 $27$
Practice: Finding the Lower or Upper Quartile: Odd Number of Elements

Karen traveled to Buenos Aires and it was colder than usual. So, she decided to record the temperature each day and she got the following data:

\[ 9, \: 11, \: 11, \: 12, \: 13, \: 14, \: 15 \] What is the lower quartile of the given data set?

a
 $10$
b
 $11$
c
 $9$
d
 $12$
e
 $11.5$
Interquartile Range

The interquartile range (IQR) of a data set is the difference between its upper and lower quartiles.

For example, suppose that the lower and upper quartiles of a data set are 3.8 and 6 , respectively.

Then to find the interquartile range, we take the difference:

\text{IQR} = \text{upper quartile} - \text{lower quartile} = 6 - 3.8 = 2.2

Therefore, the interquartile range is 2.2.

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Example: Finding the Interquartile Range

What is the interquartile range of the data set below?

19, \: 21, \: 21, \: 25, \: 28, \: 33, \: 36

EXPLANATION

Notice that our data set is already arranged from smallest to greatest.

The median of the data set is the middle number ( \color{blue}25 ), so the lower and upper halves of the data set are as follows:

\underbrace{19, \: 21, \: 21,}_{\text{lower half}} \: {\color{blue}{25}}, \: \underbrace{28, \: 33, \: 36}_{\text{upper half}}

The lower quartile is the median of the lower half: 19, \: {\color{red}\underline{21}}, \: 21 So, the lower quartile is 21.

Likewise, the upper quartile is the median of the upper half: 28, \: {\color{red}\underline{33}}, \: 36 So, the upper quartile is 33.

Finally, the interquartile range of the data set is: \text{IQR} = \text{upper quartile} - \text{lower quartile} = 33 - 21 =12

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Practice: Finding the Interquartile Range

Find the interquartile range of a data set if its lower and upper quartiles are $5.2$ and $6.1$, respectively.

a
 $1$
b
 $5.2$
c
 $0.9$
d
 $6.1$
e
 $1.1$
Practice: Finding the Interquartile Range

What is the interquartile range of the data set below?

\[ 21, \: 33, \: 18, \: 16, \: 52, \: 26, \: 21, \: 17 \]

a
 $3$
b
 $21$
c
 $16$
d
 $12$
e
 $33$
Practice: Finding the Interquartile Range

What is the interquartile range of the data set below?

\[ 12, \: 13, \: 22, \: 24, \: 27, \: 34, \: 45 \]

a
 $21$
b
 $13$
c
 $24$
d
 $34$
e
 $18$
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