Week 15 · Term 1 review: correlation and regression, quartiles and spread, probability, conditional probability and independence · 3 lessons
2
Review of Unit 2: stem-and-leaf plots, quartiles from three kinds of data, box plots and the semi-interquartile range
Statistics Sec 3 Term 1, Units 1–3, Review
GoalReview Unit 2 with new data: stem-and-leaf plots, quartiles from a list and a frequency table, box plots and the semi-interquartile range Q.
New words
quartile orders · for a list the orders are (n + 1) ÷ 4, (n + 1) ÷ 2 and 3(n + 1) ÷ 4; for a table n ÷ 4, n ÷ 2 and 3n ÷ 4five numbers · minimum, Q₁, Q₂, Q₃, maximum: they draw the box plotsemi-interquartile range · Q = (Q₃ − Q₁) ÷ 2, not moved by extreme values
Explanation
Unit 2 describes the middle and the spread of one set of data. A stem-and-leaf plot keeps every value: the stem is all the digits but the last, the leaf is the last digit, and a key says how to read them. For quartiles we use the rule of the book. For a list or a plot the orders are (n + 1) ÷ 4, (n + 1) ÷ 2 and 3(n + 1) ÷ 4; a whole order is that value, and otherwise we go part of the way between two values. For a frequency table the orders are n ÷ 4, n ÷ 2 and 3n ÷ 4 and the quartile is lower boundary + (order − cumulative frequency before) ÷ class frequency × class width. The five numbers draw a box plot, whose box holds the middle half of the data, and Q = (Q₃ − Q₁) ÷ 2 measures the spread without being moved by extreme values. The box plot below belongs to the list in the first example: fifteen pupils’ daily reading minutes.
Worked examples
key 2|4 = 24 minutes
Stem
Leaves
0
9
1
2 4 5 7 8
2
0 1 2 4 5 7 9
3
1 5
A list. Minutes of reading per day for 15 pupils: 12, 18, 25, 9, 31, 22, 15, 27, 20, 35, 14, 24, 29, 17, 21. In order: 9, 12, 14, 15, 17, 18, 20, 21, 22, 24, 25, 27, 29, 31, 35. Here n + 1 = 16, so the orders 4, 8 and 12 are whole: Q₁ = 15, Q₂ = 21, Q₃ = 27. The five numbers are 9, 15, 21, 27, 35, and Q = (27 − 15) ÷ 2 = 6 minutes. The stem-and-leaf plot keeps all 15 values.
Mass (kg)
Frequency
Cumulative
10–
4
4
15–
9
13
20–
14
27
25–
8
35
30–35
5
40
A frequency table. The masses of 40 parcels in kilograms, in classes of width 5. The orders are 10, 20 and 30. Q₁ is in the class 15– (4 before, frequency 9): 15 + (10 − 4) ÷ 9 × 5 ≈ 18.33. Q₂ is in 20– (13 before, frequency 14): 20 + (20 − 13) ÷ 14 × 5 = 22.5. Q₃ is in 25– (27 before, frequency 8): 25 + (30 − 27) ÷ 8 × 5 = 26.875. So Q = (26.875 − 18.33) ÷ 2 ≈ 4.27 kg.
Class
Q₁
Median
Q₃
Q
A
29
36
41
6
B
28.75
36
43.25
7.25
Two classes. Class A (11 marks): 23, 27, 29, 31, 34, 36, 38, 40, 41, 44, 46 gives, at the places 3, 6, 9, Q₁ = 29, Q₂ = 36, Q₃ = 41 and Q = 6. Class B (10 marks): 18, 25, 30, 33, 35, 37, 39, 42, 47, 48 gives, at the places 2.75, 5.5, 8.25, Q₁ = 25 + 0.75 × 5 = 28.75, Q₂ = 35 + 0.5 × 2 = 36 and Q₃ = 42 + 0.25 × 5 = 43.25, so Q = 7.25. The medians are equal, but B has the larger Q: B is more spread out and A is steadier.
Class activity
Time the journey to school (in minutes) of 11 classmates or neighbours. Order the times, find Q₁, Q₂, Q₃ with the orders 3, 6 and 9, draw the box plot to scale on squared paper, and compute Q. Then add one extreme value and see how much Q and the range change.
Practice
1
Seven pupils borrowed 6, 2, 9, 4, 11, 5, 8 books. Find the five numbers and Q.
Sixty pupils’ times in classes 0–, 10–, 20–, 30–, 40–50 minutes have frequencies 6, 15, 21, 12, 6. Find Q₁ and Q₃.
Cumulative: 6, 21, 42, 54, 60. Orders 15 and 45. Q₁: class 10– (6 before, frequency 15): 10 + (15 − 6) ÷ 15 × 10 = 16. Q₃: class 30– (42 before, frequency 12): 30 + (45 − 42) ÷ 12 × 10 = 32.5.
5
Box plot A has the five numbers 10, 20, 30, 45, 60 and box plot B has 5, 22, 30, 38, 70. Which data set has the larger range, and which has the larger Q?
Ranges: A 50, B 65, so B has the larger range. Q: A (45 − 20) ÷ 2 = 12.5, B (38 − 22) ÷ 2 = 8, so A has the larger Q. B’s middle half is tighter; its extreme values stretch only its range.
