Week 8 · Quartiles from a frequency table: ascending cumulative frequency, the quartile class and the formula, and the graphical method · 3 lessons
2
The quartile formula: lower boundary + (order − previous cumulative frequency) ÷ class frequency × class width, for all three quartiles
Statistics Sec 3 Term 1, Unit 2, Lesson 2
GoalUse the quartile formula, lower boundary + (order − previous cumulative frequency) ÷ class frequency × class width, for Q₁, Q₂ and Q₃.
New words
lower boundary · the first number of the class, for example 5 in the class 5–class width · the difference between two boundaries, for example 10 − 5 = 5
Explanation
Inside the quartile class we do not know where each value lies, so we assume they are spread evenly. Then the quartile is a proportional part of the class. Take Q₁ of the bakery data: the order is 15, and before the class 5– there are 6 values, so we need 15 − 6 = 9 more values. The class holds 14 values, so we go 9 ÷ 14 of the way across the class, whose width is 5. This gives the formula: quartile = lower boundary + (order − cumulative frequency before the class) ÷ class frequency × class width. It is the same idea as in lists: a fraction of the distance between two ends. The answer must lie inside the quartile class, which is a good check. Round at the end, to two decimal places.
Quartile
Order
Class
Cumulative before
Frequency
Result
Q₁
15
5–
6
14
5 + 9 ÷ 14 × 5 = 8.21
Q₂
30
10–
20
18
10 + 10 ÷ 18 × 5 = 12.78
Q₃
45
15–
38
12
15 + 7 ÷ 12 × 5 = 17.92
Worked examples
Q₁ of the bakery data. Order 15, class 5–, 6 values before it, frequency 14, width 5: Q₁ = 5 + (15 − 6) ÷ 14 × 5 = 5 + 3.21 = 8.21 minutes. It lies in the class 5–, as it should.
Q₂ and Q₃. Q₂: order 30, class 10– (20 before, frequency 18): 10 + (30 − 20) ÷ 18 × 5 = 12.78 minutes. Q₃: order 45, class 15– (38 before, frequency 12): 15 + (45 − 38) ÷ 12 × 5 = 17.92 minutes. The three quartiles are 8.21, 12.78 and 17.92.
Class
Frequency
Less than
Cumulative frequency
0
0
0–
5
10
5
10–
12
20
17
20–
18
30
35
30–
10
40
45
40–50
5
50
50
The cycling distances (n = 50). Q₁: order 12.5, class 10– (5 before, frequency 12, width 10): 10 + (12.5 − 5) ÷ 12 × 10 = 16.25 km. Q₂: order 25, class 20– (17 before, frequency 18): 20 + 8 ÷ 18 × 10 = 24.44 km. Q₃: order 37.5, class 30– (35 before, frequency 10): 30 + 2.5 ÷ 10 × 10 = 32.5 km.
Class activity
Use the table of 40 school-journey times you made in the last lesson. Find Q₁, Q₂ and Q₃ with the formula, and check that each lies inside its class.
Practice
1
Order 10, class 20–, 4 values before it, frequency 9, width 10. Find the quartile.
20 + (10 − 4) ÷ 9 × 10 = 26.67.
2
Order 20, class 30–, 13 values before it, frequency 13, width 10. Find the quartile. What is special about the answer?
30 + (20 − 13) ÷ 13 × 10 = 35.38. It lies between 30 and 40, inside the class.
3
Find Q₁, Q₂, Q₃ for the classes 10–, 15–, 20–, 25–, 30–, 35–40 with frequencies 3, 9, 14, 12, 7, 5.
