Connecting the Dots... · Lesson 5 of 13
Of Ends and the Essence
“Compare the centre, spread, and context of data rather than relying on one average.”
• Describe centre and spread together. • Compare a class with its subgroups using the recorded heights. • Explain why equal centres can hide different distributions. • Check whether a calculated mean or median is possible. • Distinguish closeness to a target from consistency.
Of Ends and the Essence
Two groups can have similar representative values but very different spreads. If a group’s average estimated minute is close to sixty seconds, that does not necessarily mean each person estimated accurately. To understand a collection, study its centre together with its minimum, maximum, clusters, and context.
How Tall is Your Class?
The question “How tall is the class?” cannot be answered by naming only its tallest student. We can describe the entire distribution and use mean or median as a central value. Separate dot plots also let us compare the boys and girls without assuming that every individual follows the group comparison.
| Group | Recorded heights (cm) |
|---|---|
| Boys | 147, 135, 130, 154, 128, 135, 134, 158, 155, 146, 146, 142, 140, 141, 144, 145, 150 |
| Girls | 143, 136, 150, 144, 154, 140, 145, 148, 156, 150, 150 |
| Group | Count | Mean (cm) | Median (cm) | Minimum–maximum (cm) | Range (cm) |
|---|---|---|---|---|---|
| Whole class | 28 | 144.5 | 145 | 128–158 | 30 |
| Boys | 17 | 142.94 | 144 | 128–158 | 30 |
| Girls | 11 | 146.91 | 148 | 136–156 | 20 |
Problem
How can the tallest student be a boy while the girls have the higher mean?
- 1.The maximum among boys is 158 cm, higher than the girls’ maximum of 156 cm.
- 2.Mean uses all observations. Boys’ mean is 2430 ÷ 17 ≈ 142.94 cm; girls’ mean is 1616 ÷ 11 ≈ 146.91 cm.
- 3.The girls’ group has the higher mean even though the single tallest observation is in the boys’ group. Individual heights overlap.
Problem
Find the whole class mean and count students taller than it.
- 1.Combine totals: 2430 + 1616 = 4046 cm. Combine counts: 17 + 11 = 28.
- 2.Whole class mean = 4046 ÷ 28 = 144.5 cm. Use the original table for an accurate value.
- 3.Nine boys and seven girls have heights greater than 144.5 cm, so 16 students are taller than the mean.
- 4.Do not simply average the two subgroup means: the groups contain different numbers of students.
A mean describes a group, not every person in it. Also, the combined mean must reflect the actual group sizes; an unweighted average of subgroup means can be wrong.
Comparing another class
The source gives another class’s heights as three dot plots. Its whole-class mean is 141.21 cm and median is 142.5 cm, both lower than the first class’s 144.5 cm and 145 cm. Its boys have higher centres than its girls, reversing the subgroup ordering in the first class. The overlapping dots show why group comparisons must not be turned into claims about every student.
How long is a minute?
In the source activity, children close their eyes and open them when they think sixty seconds have passed, without counting. Their observations are elapsed times in seconds. Here there is a target, 60 seconds, as well as a distribution to describe.
| Group in the source | Mean estimate (seconds) | Median estimate (seconds) |
|---|---|---|
| A | 58.21 | 60 |
| B | 59.28 | 59.5 |
The source dot plots show a wider spread for Group B even though its mean is closer to 60. Some members stop early and others late, so opposite errors can cancel in the mean. Group A is more clustered near 60. Compare the centre, spread, and deviations from the target rather than declaring the group with the nearest mean automatically best.
Problem
Compare estimates 58, 59, 60, 61, 62 with 40, 50, 60, 70, 80.
- 1.Both totals are 300 across five observations, so both means are 60 seconds. Both medians are also 60.
- 2.The first range is 62 − 58 = 4 seconds; the second is 80 − 40 = 40 seconds.
- 3.Distances from the target average 6 ÷ 5 = 1.2 seconds for the first group and 60 ÷ 5 = 12 seconds for the second.
- 4.Thus equal centres can conceal a substantial difference in estimation accuracy. These example datasets illustrate the idea; they are not the original activity’s observations.
What values can a centre take?
Mean and median cannot lie outside the minimum and maximum of the included numerical observations. Equal sharing cannot produce an amount greater than the largest original amount or less than the smallest. A median is either a recorded central value or an average between two central values.
Problem
Daily tap usage is 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4 litres. Could its mean or median lie between 25 and 30 litres?
- 1.The minimum is 3.09 litres and maximum is 20.5 litres.
- 2.Any mean lies between these extremes: if every observation is at most 20.5, their equal share cannot exceed 20.5.
- 3.Any median also lies between them. Therefore neither centre can lie between 25 and 30 litres.
This reasoning gives a useful calculation check. If your mean is larger than every recorded value, recheck the total, count, and units. If you are comparing unlike units, convert them before calculating a meaningful centre.
Pockets and distributions
The source uses clothing pockets as another distribution comparison. The two groups have overlapping pocket counts, but their centres and ranges differ. Inspect all dots rather than judging the groups only by their greatest value.
Problem
Which group has the larger range, median, and mean?
- 1.Boys: minimum 3, maximum 6, range 3. Girls: minimum 0, maximum 6, range 6. Girls have the larger range.
- 2.Boys total 56 across 12 observations, mean ≈ 4.67; median 5.0. Girls total 45 across 13, mean ≈ 3.46; median 4.
- 3.Thus the boys’ median is higher; the girls’ mean is not higher; and the maxima are equal. Of the four source statements, only the median comparison is true.
Quiz
Two groups have equal means. What must be true?
Can the mean of values from 3.09 to 20.5 be 26?
A group has the higher mean height. Which statement follows?
Why can a mean estimated minute of 60 seconds conceal poor estimates?
How should unequal-sized groups be combined?
Practice Problems
- Make a dot plot for the palm-tree heights listed here: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Find the centres and count values shorter than the mean.
- Describe a quicker mean calculation for these tree heights.
- Compare newborn weights (kg): boys 3.5, 4.1, 2.6, 3.2, 3.4, 3.8; girls 4.0, 3.1, 3.4, 3.7, 2.5, 3.4.
- The other class’s supplied centres are: whole class mean 141.21, median 142.5; boys mean 142.05, median 143; girls mean 140.14, median 140 cm. Compare with the first class.
- Approximately how many times heavier are the source’s sumo wrestlers than its ballet dancers? Wrestlers: 295.2, 250.7, 234.1, 221.0, 200.9 kg; dancers: 40.3, 37.6, 38.8, 45.5, 44.1, 48.2 kg.
- Seventeen heights are 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101 cm. Can one strict cutoff divide everyone into equal smaller-height and greater-height groups?
The total is 1621 feet across 29 trees. Mean = 1621 ÷ 29 ≈ 55.9 feet; median = 56 feet. 13 trees are shorter than the mean. Stack all repeated values; the heights span 43–67 feet.
Key Takeaways
• Read centres and spread together. • Equal means or medians do not imply identical distributions. • A whole-group mean uses the combined total and count. • Mean and median lie between minimum and maximum. • A group comparison does not determine every individual comparison. • A mean near a target can hide large opposite errors.