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THE QUARTILES



Let x1, x2, x3, ..., xn  be the data under consideration and they are ordered so that x1 ≤ x2 ≤ x3 ≤ ... ≤ xn  (i.e. they have been sorted from the smallest to the largest). The first quartile (Q1) and the third quartile (Q3) of the data are defined as follows.
Q1 = xL where $L = \frac{1}{4} (n+1)$
Q3 = xU where $U = \frac{3}{4} (n+1)$

Example 1
The following are the grades of a Social Statistics assignment achieved by 15 students.
47  56  71  65  29 68  78  73 80  75 29  38  65  90  95
Find the first and third quartiles of the data.

Answer
Firstly, sort the data from the lowest to the highest. So we have:
29  29  38  47  56  65  65  68  71  73 75  78  80  90  95
Now, let x1 = 29, x2 = 29, x3 = 38, ..., x15 = 95. To determine the first quartile, we calculate:
$L = \frac{1}{4} (n+1) = \frac{1}{4}(15+1) = 4$
So, Q1 = xL = x4 = 47
To find the third quartile, calculate $U = \frac{3}{4}(n+1) = \frac{3}{4}(15+1) = 12$. From this, we get Q3 = x12 = 78.
Conclusion: the first quartile is 47 and the third quartile is 78.

How to calculate the quartiles if U and L are not a whole numbers? Suppose that L is not a whole number. Find the whole number k such that k < L < (k+1). Then $Q_{1} = \frac{j}{4}  \cdot (x_{k}) + \frac{4-j}{4} \cdot (x_{k+1})$. Substitute j = 1 if L is closer to (k+1) than to k, j = 2 if L is in the middle, j = 3 if L is closer to k than to (k+1). The similar procedure is applied for U: Suppose that U is not a whole number. Find the whole number k such that k < U < (k+1). Then $Q_{3} = \frac{j}{4}  \cdot (x_{k}) + \frac{4-j}{4} \cdot (x_{k+1})$. The substitute for j is determined with the same procedure as before.

Example 2
The following are the grades of a Social Statistics assignment achieved by 14 students.
47  56  71  65  29 68  78  73 80  75 29  38  65  90
Find the first and third quartiles of the data.

Answer
Firstly, sort the data from the lowest to the highest. So we have:
29  29  38  47  56  65  65  68  71  73 75  78  80  90
Now, let x1 = 29, x2 = 29, x3 = 38, ..., x14 = 90. To determine the first quartile, we calculate:
$L = \frac{1}{4} (n+1) = \frac{1}{4}(14+1) = 3 \frac{3}{4}$
Here, k = 3 because $3 < 3 \frac{3}{4} < (3+1)$, i.e. $3 < 3 \frac{3}{4} < 4$.
L is closer to (k+1) = 4, so we substitute j = 1. Then we have:
$ Q_{1} = \frac{1}{4} (x_{3}) + \frac{3}{4} (x_{4}) = \frac{1}{4} \cdot 38 + \frac{3}{4} \cdot 47 = 44.75$
To find the third quartile, we calculate:
$U = \frac{3}{4} (n+1) = \frac{3}{4}(14+1) = 11 \frac{1}{4}$
Here, k = 11 because $11 < 11 \frac{1}{4} < (11+1) \$, i.e. \$ 11 < 11 \frac{1}{4} < 12$.
U is closer to k = 11, so we substitute j = 3. Then we have:
$Q_{3} = \frac{3}{4} (x_{11}) + \frac{1}{4} (x_{12}) = \frac{3}{4} \cdot 75 + \frac{1}{4} \cdot 78 = 75.75$
Conclusion: the first quartile is 44.75 and the third quartile is 75.75.



PROBLEMS PART I

  1. The following are scores of a Social Statistics test: 28, 40, 87, 47, 59, 72, 84, 82, 19, 55, 64, 87. Calculate the first and third quartiles.
  2. The following are the lengths of 14 conversations using mobile phones, in seconds: 16, 62, 45, 90, 55, 49, 100, 89, 72, 115, 45, 59, 128, 219. Find the first and third quartiles.
  3. The following are 11 data on the number of students attending Social Statistics class, from the first lecture to the 11th : 37, 40, 35, 36, 40, 40, 42, 37, 39, 40, 41. What are the quartiles of the data?

PROBLEMS PART II
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In this part, you may need to learn the concept of median. Click here to learn it.


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