Quartiles from frequency tables
Answers are for parents and teachers.
1
The ages of 80 members of a sports club are in the classes 20–, 30–, 40–, 50–, 60–70 with frequencies 9, 21, 28, 16, 6. Build the cumulative column and find Q₁, Q₂ and Q₃.
Cumulative: 0, 9, 30, 58, 74, 80. Orders 20, 40, 60. Q₁ = 30 + (20 − 9) ÷ 21 × 10 = 35.24; Q₂ = 40 + (40 − 30) ÷ 28 × 10 = 43.57; Q₃ = 50 + (60 − 58) ÷ 16 × 10 = 51.25.
2
Heights of 40 seedlings in cm: classes 10–, 20–, 30–, 40–, 50–60 with frequencies 5, 9, 13, 8, 5. Find the three quartiles with the formula.
Cumulative 0, 5, 14, 27, 35, 40; orders 10, 20, 30. Q₁ = 20 + 5 ÷ 9 × 10 = 25.56; Q₂ = 30 + 6 ÷ 13 × 10 = 34.62; Q₃ = 40 + 3 ÷ 8 × 10 = 43.75 cm.
3
The curve shows the daily steps, in hundreds, of 60 walkers. Read Q₁, Q₂ and Q₃ (orders 15, 30 and 45). The table behind it has the classes 0–, 20–, 40–, 60–, 80–100 with frequencies 6, 14, 20, 12, 8: check one of your readings with the formula.
Readings about 33, 50 and 68. Formula: Q₁ = 20 + (15 − 6) ÷ 14 × 20 = 32.86; Q₂ = 40 + (30 − 20) ÷ 20 × 20 = 50; Q₃ = 60 + (45 − 40) ÷ 12 × 20 = 68.33.
4
A pupil has 50 values in a frequency table and finds the orders (50 + 1) ÷ 4 = 12.75, 25.5 and 38.25. What should she use instead, and what are the orders?
For a frequency table the orders are n ÷ 4, n ÷ 2 and 3n ÷ 4: 12.5, 25 and 37.5.
5
A table has the classes 0–, 10–, 20–30 with frequencies 10, 20 and 10. Find the three quartiles. Notice what happens with the order 10.
n = 40, cumulative 0, 10, 30, 40; orders 10, 20, 30. Q₁ = 0 + 10 ÷ 10 × 10 = 10 (exactly the boundary, since 10 values lie below it); Q₂ = 10 + 10 ÷ 20 × 10 = 15; Q₃ = 10 + 20 ÷ 20 × 10 = 20.