12.6 Beyond Lines: Infographics and Strips

Infographics

An infographic crams information into a single, eye-catching picture so the insight leaps out fast. A map of India shaded by the difference between rice and wheat consumption (running from -100 for “mostly wheat” to +100 for “mostly rice”) exposes a striking north–south divide, with a clean dividing line. A score of +100 doesn’t mean zero wheat, it means rice wins by the widest margin. Reading an infographic begins with decoding its colour scale, after which the pattern (here, geography) tells its own story.

Activity strips

Suppose a student records an entire day on a paper strip of 48 boxes, each box a 30-minute slice from midnight to midnight. Different colours flag different activities, sleeping, eating, classes and study, friends/hobbies/media, getting ready or exercise, and travelling. Reading the colours, we can tell when they slept, when school paused for lunch (a brief eating band mid-day), and which day held a long film (an unusually wide “friends/media” band). One compact strip captures a whole day’s tale.

Try This Build an activity strip of 48 boxes for one of your ordinary days. Colour each 30-minute block by activity. How does a school day strip differ from a holiday strip? Make one for an adult at home and compare, what surprises you?

A data story: sleep across ages

A line graph of typical sleep duration for people from age 6 to 75 looks like a smooth curve, because it joins about 80 tightly packed data points. (A column graph would call for 70 columns and look cluttered; the line graph stays light and catches the pattern cleanly.) The curve shows daily sleep near 9.5 hours at age 6, falling through the teens to roughly 8 hours between ages 30 and 50, then easing back up to about 8.5 hours after 50.

When to Use a Line Graph A line graph shines when you have many data points along a continuous scale (usually time) and want to show a trend. The closer together the points, the more the line looks like a smooth curve, ideal for seeing the overall shape without clutter.

Figure it Out: Reasoning with Data

Practice

  1. Mean grid. Fill a 3\times 3 grid with 9 distinct numbers so the average along every row, column and both diagonals is 10. (Hint: each line of 3 must total 30.)

  2. Give two examples each: (i) 3 numbers with mean 8; (ii) 4 numbers with median 15.5; (iii) 5 numbers with mean 13.6; (iv) 6 numbers with mean = median; (v) 6 numbers with mean > median.

  3. Fill the blanks so the median of 5, 21, 14, \_\,, \_\,, \_ is 13. How many ways if only counting numbers are allowed?

  4. Check each statement, justifying with algebra if needed: (i) the average of two even numbers is even; (ii) the average of any two multiples of 5 is a multiple of 5; (iii) the average of any five multiples of 5 is a multiple of 5.

  5. A long-jump squad recorded how many attempts each athlete took to clear a marker:

    Attempts 1 2 3 4 5 6 7 8 9 10
    Athletes 1 0 1 2 5 8 11 14 9 9

    Describe this data using its minimum, maximum, mean and median.

  1. Centre the grid on 10 and use balanced offsets. One solution is a magic-square pattern with centre 10: \begin{array}{ccc} 9 & 14 & 7 \\ 8 & 10 & 12 \\ 13 & 6 & 11 \end{array} Every row, column and diagonal sums to 30, so every average is 10. (Many other answers exist, e.g. take any magic square of sum 15 and add 5 to each cell.)
  2. Sample answers: (i) 6, 8, 10 and 1, 8, 15 (each mean 8). (ii) 10, 15, 16, 20 and 14, 15, 16, 21 (median =\frac{15+16}{2}=15.5). (iii) 13, 13, 13, 14, 15 (sum 68, mean 13.6) and 10, 12, 14, 15, 17. (iv) any symmetric set, e.g. 2, 4, 6, 8, 10, 12 (mean = median =7). (v) skew right, e.g. 1, 2, 3, 4, 5, 21 (median 3.5, mean 6).
  3. Sorted, the three known values are 5, 14, 21. With three blanks there are 6 values, so the median is the average of the 3rd and 4th. To land on 13, the middle pair must average 13. Many fills work, e.g. blanks 12, 13, 13 (sorted \dots 12,13,13,14\dots, median \frac{13+13}{2}=13), or 1, 12, 14. Working through counting-number options systematically gives several possibilities; the key constraint is simply that the two central values average 13.
    1. Sometimes false, e.g. \frac{2+4}{2}=3 is odd, so “always even” is false. (ii) Always true: \frac{5a+5b}{2}=\frac{5(a+b)}{2}=5\cdot\frac{a+b}{2}, which is 5\times(\text{something}), hence a multiple of 5. (iii) The average of five multiples of 5 is \frac{5(a+b+c+d+e)}{5}=a+b+c+d+e, an integer that is not always a multiple of 5 (e.g. \frac{5+5+5+5+10}{5}=6).
  5. Minimum =1, maximum =10. Total athletes =1+0+1+2+5+8+11+14+9+9 = 60. Mean =\dfrac{(1\cdot1)+(3\cdot1)+(4\cdot2)+(5\cdot5)+(6\cdot8)+(7\cdot11)+(8\cdot14)+(9\cdot9)+(10\cdot9)}{60}=\dfrac{445}{60}\approx \mathbf{7.42}. Median is the average of the 30th and 31st values; cumulative counts reach 27 by 7 attempts and 41 by 8 attempts, so both land on 8, median =8.