Misleading Histograms and Stem-and-Leaf Diagrams

Solution

(a) The axis starts at 24. The bar for $10 \le t < 15$ rises $34 - 24 = 10$ units above it, and the bar for $0 \le t < 5$ only $26 - 24 = 2$ units. It looks $10 \div 2 = 5$ times as tall.

(b) $34 \div 26 = 1.30\ldots \approx 1.3$ (to 1 decimal place): only about 1.3 times as many.

(c) Cutting off the bottom of the axis removes the same 24 from every bar, so small differences look large. A fair histogram starts the axis at 0:

If the axis must start higher, a zigzag at the bottom warns the reader that part of it is left out. The bars are then still not in proportion, so the reader must compare the numbers, not the heights:

Solution

(a) $4$, $5$, $4$, $5$, $4$: the numbers are equally far apart on the page, but the steps are not equal.

(b) Equal distances on the axis stand for different numbers of cyclists, so a height cannot be read by counting steps, and two bars cannot be compared by their heights. Only the last bar ends on a number, 18; every other bar ends between two numbers whose step changes, so its value has to be worked out from how far along the step it ends. Reading each bar against its axis numbers: 5 (a fifth of the way from 4 to 9), 10 (a quarter of the way from 9 to 13), 20 (half way from 18 to 22), 15 (two fifths of the way from 13 to 18) and 18. Check: $5 + 10 + 20 + 15 + 18 = 68\text{,}$ the number of cyclists. A fair axis goes up in equal steps:

Solution

The last class is 20 minutes wide, twice the others, and its bar is as tall as its frequency. So the bar is the tallest and also twice as wide: it looks as if far more runners took 20 to 40 minutes than anything else.

Its 12 runners are spread over twice as many minutes. Grouped in classes of equal width, the times of the last 20 minutes are 8 and 4, both fewer than the 10 of $10 \le t < 20$:

The fair histogram shows the times rising to the class $10 \le t < 20$ and falling after it, which Histogram C hid.

Solution

The classes leave gaps: 2 kg is in no class, and nor is 1.95 kg (it is more than 1.9 and less than 2.1). Masses in the gaps are lost from the table, so the histogram shows fewer parcels than there are.

Written as $1 \le m < 2$, $2 \le m < 3$, $3 \le m < 4$, the classes meet with no gaps and no overlaps: 1.95 kg is in $1 \le m < 2$, and 2 kg in $2 \le m < 3$.

Solution

Histogram D has two bars, 18 and 22. It suggests the times rise steadily, with a few more pupils taking longer.

Histogram E has four bars, 12, 6, 4 and 18. It shows two groups: 12 pupils who finished in 20 to 30 minutes, and 18 who took 50 minutes or more, with only 10 in between.

Both are counted correctly from the same 40 times, but Histogram D's wide classes hide the two groups. Someone who wanted to show that the test's length suited everyone might choose D. Histogram E is fairer.

Solution

(a) The gaps suggest that the classes are separate categories, with times between them that belong to no class. The classes written under the middle of the bars, like names, suggest the same, and are a fault too: the boundaries are not numbered. Time is a number scale with no gaps, so the bars of a histogram touch, with the boundaries numbered at their edges (Example 1).

(b) Touching bars suggest a number scale, as if the subjects were values in order, with Mathematics "between" English and Science. Subjects are separate categories: a bar graph with gaps is right.

Solution

(a) Row 2 looks the longest, but it has 5 leaves; row 1 has 8, squeezed together. Every leaf must take the same width, or the lengths of the rows mean nothing. Fairly drawn:

(b) Without a key, 4 | 2 could be 42 kg or 4.2 kg; for a baby only 4.2 kg makes sense, but the reader should not have to guess. The leaves are also out of order. Fairly drawn:

Solution

Diagram K leaves 4, 6 and 7 off the line, so 5 and 8 sit next to 3 and the pupil who borrowed 8 books does not look far from the rest.

Diagram L draws the dots at 1 larger, so its column of 4 looks bigger than the column of 5 at 2.