Quartiles and the Interquartile Range for Ungrouped Data

Quartiles

The quartiles divide the data (in ascending order) into four quarters. Every quarter has the same number of items. There are three quartiles:
● The lower quartile (Q1)
● The middle quartile or median (Q2)
● The upper quartile (Q3)

lnterquartile range

The interquartile range is the difference between the third quartile and the first quartile. It measures the range of the middle 50% of the data (in ascending order). A large interquartile range indicates the data Values are widely dispersed.
● lnterquartile range = Third quartile – First quartile
● IQR=Q3Q1

Semi-interquartile range

The semi-interquartile range is half of the range of the middle half of the data. It is seldom influenced by extreme values in a set of data, and so is a good measure of dispersion.

Question 1. Given a data set in ascending order. Find the median, quartiles and the interquartile range (IQR).
2 5 5 6 9 11 11 11 14 15 19

Solution:
Square the median. (Because I didn’t know way to circle the median in the software ^_^)

The median represents the middle of the data. The data has been divided into two parts. Consider the set of data on each side of the median and circle the median again:

The circled numbers represent the quartiles — Quartile 1 (Q1), Quartile 2 (Q2), and Quartile 3 (Q3). These values split the set of data into quarters. Thus Quartile 2 can also be called the median).

The interquartile range (IQR) can be calculated by subtracting quartile 1 from quartile 3 (Q3Q1),. The IQR of this set of data is 14-5=9.

Question 2. For the data set: 7, 3, 1, 7, 6, 9, 3, 8, 5, 8, 6, 3, 7, 1, 9 find the:
a) median b) lower quartile c) upper quartile d) interquartile range
The ordered data set is:


solution:
a) As n=15, (n+1)/2=8 ∴ the median=8th data value=6
b/c) As the median is a data Value we now ignore it and split the remaining data into two:

Q1= median of lower half =3
Q3= median of upper half =8
d) IQR =Q3Q1=8-3=5

The data set in Question 2 can be summarised as the followong picture:


Question 3. For the data set: 6, 4, 9, 15, 5, 13, 7, 12, 8, 10, 4, 1, 13, 1, 6, 4, 5, 2, 8, 2 find:
a) the median b) Q1 c) Q3 d) the interquartile range
The ordered data set is:

solution:
a) As n=20, ½(n+1)=10.5

b/c) As we have an even number of data values, we split the data into two:

d) IQR =Q3Q1=9.5-4=5.5

Question 4. The marks obtained by 19 students of a class are given below:

27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.

Find: (i) median (ii) lower quartile
(iii) upper quartile (iv) interquartile range
solution:
Arranging in ascending order:
22, 24.25. 26. 26.27. 27.28. 28. 29.21. 32. 32, 33. 35. 35. 36. 36. 37

(i) Middle term is 10th term i.e. 29
Median=29
(ii) Lower quartile =

(iii) Upper quartile =

(iv) lnterquartile range =Q3Q1=35-26=9

Question 5. From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range

25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83

solution:
Arrange in ascending order:
0.7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median is the mean of 8th and 9th term

(ii) Upper quartile=

(iii) Interquartile range=

Interquartile range =Q3Q1=65-18=47

Question 6. The data below shows the number of laptops sold by 15 sales agents during the last financial year.

43 48 62 52 46 90 58 37 48 73 84 68 54 34 78

(a). Determine the median of the number of laptops sold.
(b). Calculate the range of the data.
(c). Calculate the interquartile range (IQR).
solution:
Arranging the data in ascending order, we obtain
34 37 43 46 48 48 52 54 58 62 68 73 78 84 90

Here, the numbers of observations are 15, which is odd.
(a). Median,Me=½(15+1)th observation=8th ob.
Me=54
(b). Range =90-34=56
(c). The median cuts the data into two sections. In section I we can get lower quartile and in section II we can get upper one.
We obtain the section I.
34 37 43 46 48 48 52

Q1=½(7+1)th observation=4th observation.
Q1=46
Then the section II.
58 62 68 73 78 84 90

Q3=73
IQR=Q3Q1=73-46=27

Question 7. The heights of 20 children were measured (in centimetres) and the results were recorded. The data collected is given in the table below.

127 128 129 130 131 133 134 134 135 136
137 138 139 140 141 142 142 143 144 145

(a). Write down the median height measured.
(b). Determine:
(b-i). The mean height
(b-ii). The range
(b-iiii). The interquartile range
solution:
The set of values is already in ascending order.
(a) Because n=20 is even, median

(b-i). Mean =2728/20=136.4
(b-ii). Range =145-127=18 cm
(b-iii). The median cuts the data into two sections. In section I we can get lower quartile and in section II we can get upper one.
We obtain the section I.
127 128 129 130 131 133 134 134 135 136

Lower quartile

Then the section II.
137 138 139 140 141 142 142 143 144 145

upper quartile

Interquartile range is equal to the difference between upper quartile and lower quartile.
=141.5-132=9.5 cm

Question 8. For each of the following data sets, compute the mean and all the quartiles. Round your answers to one decimal place.
a) -3.4;-3.1;-6.1;-1.5;-7.8;-3.4;-2.7;-6.2
b) -6;-99;90;81;13;-85;-60;65;-49
c) 7;45;11;3;9;35;31;7;16;40;12;6
Solution:
a) Mean:


To compute the quartiles, we order the data:
-7.8;-6.2;-6.1;-3.4;-3.4-;-3.1;-2.7;-1.5

We use the diagram below to find at or between which values the quartiles lie.

For the first quartile the position is between the second and third values. The second value is -6.2 and the third value is -6.1, which means that the first quartile is (-6.2-6.1)/2=-6,15.
For the median (second quartile) the position is halfway between the fourth and fifth values. Since both these values are -3.4, the median is -3.4.

For the third quartile the position is between the sixth and seventh values. Therefore the third quartile is -2.9.


To compute the quartiles, we order the data: -99;-85;-60;-49;-6;13;65;81;90
We use the diagram below to find at or between which values the quartiles lie.

We see that the quartiles are at -60; -6; and 65.

c) The mean is =18.5. To compute the quartiles, we order the data:

3;6;7;7;9;11;12;16;31;35;40;45

We use the diagram below to find at or between which values the quartiles lie.

For the first quartile the position is between the third and fourth values. Since both these values are equal to 7, the first quartile is 7.
For the median (second quartile) the position is halfway between the sixth and seventh values. The sixth value is 11 and the seventh value is 12, which means that the median is (11+12)/2=11.5.
For the third quartile the position is between the ninth and tenth values. Therefore the third quartile is (31+35)/2=33.

Question 9. The weight of 60 boys are given in the following distribution table:


Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
solution:

Number of terms=60
(i) median=the mean of the 30th and the 31st terms

ii) lower quartile (Q1)=60/4 th term =15th term =38
(iii) upper quartile (Q3) =(3×60)/4 th term =45th term =40
(iv) Interquartile range =Q3Q1=40-38=2.
Let’s read post The Quartile Common Formulae for Continuous or Discrete Distribution (Grouped Data).

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