Statistics with Grouped Data
Senior 2 · Mathematics
Statistics with Grouped Data
Lesson 1 of 5
20 min
A head teacher in Huye wants to know how long her 40 Senior 2 students take to walk to school, so that she can decide whether morning classes should start later. She collects 40 numbers. Forty separate numbers are hard to read at a glance, and a school survey may have hundreds. Putting the values into a few groups (classes) and counting how many fall in each makes the pattern visible. The price is a little lost detail: once a value is inside a class, we no longer know it exactly.
Grouped data appears in census reports, in health records from RSSB, in examination results and in every agricultural survey. In this unit you will organise it, calculate its averages and draw it.
Walking times to school, in whole minutes, of 40 students:
26, 44, 12, 39, 21, 20, 28, 20, 33, 25, 10, 19, 31, 28, 50, 40, 27, 8, 30, 5, 25, 7, 23, 35, 45, 18, 16, 36, 34, 30, 38, 53, 24, 15, 32, 22, 24, 17, 42, 29.
The smallest value is 5 and the largest is 53. The range is largest − smallest = 53 − 5 = 48 minutes.
With a range of 48, classes 10 minutes wide are convenient: 48 ÷ 10 is about 5 classes, and 5 to 15 classes is a sensible number. Start at the smallest value:
5–14, 15–24, 25–34, 35–44, 45–54
The classes must not overlap and must not leave gaps, so that every value belongs to exactly one class. Go through the list once, making a tally mark for each value:
| Time (min) | Tally | Frequency |
|---|---|---|
| 5–14 | 5 | |
| 15–24 | 12 | |
| 25–34 | 13 | |
| 35–44 | 7 | |
| 45–54 | III | 3 |
| Total | 40 |
Always check that the frequencies add up to the number of values: 5 + 12 + 13 + 7 + 3 = 40.
Take the class 15–24:
Measurements such as heights or masses can take any value, so classes are often written with inequalities, for example 150 ≤ h < 155 for heights in centimetres. Here the limits ARE the boundaries: 150 and 155. The width is 5, the midpoint is 152.5, and a height of exactly 155 cm belongs to the next class, 155 ≤ h < 160.
Tip
Boundaries, not limits, for width. For whole-number classes, subtract 0.5 from the lower limit and add 0.5 to the upper limit to get the boundaries. Then width = difference of the boundaries, and midpoint = average of the limits (or of the boundaries: it is the same number).
For the class 60–69 (whole-number marks), give the limits, the boundaries, the width and the midpoint.
Solution: limits 60 and 69; boundaries 59.5 and 69.5; width 69.5 − 59.5 = 10; midpoint (60 + 69) ÷ 2 = 64.5.
For the class 150 ≤ h < 155, give the width and the midpoint.
Solution: width 155 − 150 = 5; midpoint (150 + 155) ÷ 2 = 152.5.
A set of test marks runs from 12 to 87. What is the range? How many classes of width 10 are needed, starting with 10–19?
Solution: range = 87 − 12 = 75. The classes 10–19, 20–29, …, 80–89 are 8 classes.
With classes 15 ≤ x < 25 and 25 ≤ x < 35, where do the values 24.5 and 25 go?
Solution: 24.5 goes in 15 ≤ x < 25, because 24.5 < 25; the value 25 goes in 25 ≤ x < 35.
Remember
Sources and credits
The Rwandan examples, practice items and exercises are written by Ganzaa.
