2.4 Box Plots

Reading the three plots. Every value below is read off the shared number line in Figure 2.4.7.

• China — the row is a bare line from 0 to 5 with no box at all. A box of zero width means \(Q_1\), the median and \(Q_3\) are all the same value: 0. Five-number summary \(0,\ 0,\ 0,\ 0,\ 5\).
• Germany — box from 4 to 8 with the median drawn on the box's right edge, whiskers out to 0 and 11. Five-number summary \(0,\ 4,\ 8,\ 8,\ 11\).
• United States — box from 0 to 5 with the median at 2 and the right whisker out to 11. The box starts flush against the minimum, so there is no left whisker. Five-number summary \(0,\ 0,\ 2,\ 5,\ 11\).

a — What each shape implies.

The Chinese plot is collapsed at zero: at least 75 percent of the 20-year-olds surveyed had visited no foreign country at all, and the whole upper quarter stretches from 0 out to only 5. The distribution is piled hard against the left end of the scale.

The German plot sits highest of the three. Its box runs from 4 to 8 and the median lands on the box's right edge, which tells us at least a quarter of German respondents had visited exactly eight countries. The middle half is up in the 4-to-8 range — travel is common and fairly evenly spread.

The U.S. plot is spread wide but low. Its box runs from 0 to 5 with the median at 2, so half the American respondents had visited two or fewer countries; the long right whisker to 11 says a small group had travelled a great deal. This is a right-skewed distribution.

b — Over eight countries. For Germany, \(Q_3 = 8\), so about 25 percent of Germans surveyed had been to more than eight countries. For the United States, \(Q_3 = 5\) and the maximum is 11, so the entire top quarter spans 5 to 11 and only part of it lies beyond 8 — fewer than 25 percent.

Answer (b): More Germans surveyed had been to over eight foreign countries.

c — The three compared. German 20-year-olds had travelled the most: their box sits farthest right and three quarters of them had visited at least four countries. Americans reach just as high at the top end (both maxima are 11) but their middle half is far lower, so the American figures are dragged up by a well-travelled minority rather than being typical. Chinese respondents had travelled the least by a wide margin — most had not left the country at all, and the most-travelled respondent had visited five countries, fewer than the German or American median-to-max reach.

Reading the plot. In Figure 2.4.8 the box runs from 0 to about 95, the median line sits at 20, and the right whisker reaches 150. There is no left whisker: the box starts flush against the minimum, so the smallest value and \(Q_1\) are both 0. The five-number summary is roughly \(0,\ 0,\ 20,\ 95,\ 150\).

a — An example that would fit. Suppose we ask 200 people how many minutes they spent commuting by bus yesterday. A large share did not take a bus at all, so their answer is 0 — that is why at least a quarter of the data sits at zero and there is no left whisker. Among the people who did ride, most had short trips of ten to thirty minutes, which pulls the median down to 20. A smaller group commutes across the county and reports an hour, ninety minutes, or in one case two and a half hours, which is what stretches the box out to 95 and the whisker all the way to 150. Any variable with a big pile-up at zero and a long thin upper tail — rainfall by day, dollars donated per person, minutes of overtime worked — would produce this same shape.

b — Quartiles close together, then far apart. The gap between two of the five numbers tells you how much of the number line one quarter of the data is spread across, not how many values it holds. Every quarter holds about 25 percent no matter how wide it looks.

So \(Q_1 = 0\) and the median \(= 20\) sitting close together means that a full quarter of the values are squeezed into the narrow band from 0 to 20 — those observations are tightly bunched and nearly identical to one another. The median at 20 and \(Q_3\) at 95 being far apart means the next quarter of the values — the same number of observations — is smeared across 75 units of the scale.

Answer (b): The data below the median is densely packed near zero, while the data just above the median is thinly spread over a wide range. Equal counts, wildly unequal spread — the signature of a right-skewed distribution.

Reading the two plots. From Figures 2.4.8 and 2.4.9:

• Data 1 — box from 2 to about 4.7 with the median at 4, whiskers out to 0 and 7. Five-number summary about \(0,\ 2,\ 4,\ 4.7,\ 7\).
• Data 2 — a narrow box from about 1.5 to about 2.4 with the median at 2, whiskers out to 0 and 7. Five-number summary about \(0,\ 1.5,\ 2,\ 2.4,\ 7\).

a i — "Data 1 has more data values above two than Data 2 has above two."

False, because a box plot reports percentages, not counts. It tells us that 75 percent of Data 1 lies above 2 (since \(Q_1 = 2\)) and that 50 percent of Data 2 lies above 2 (since the median is 2) — but neither plot says how many observations are in its data set. If Data 1 holds 20 values and Data 2 holds 200, then 75 percent of Data 1 is 15 values while 50 percent of Data 2 is 100. The graph gives us no way to rule that out, so the comparison of counts cannot be made at all.

a ii — "The data sets cannot have the same mode."

False, because a box plot does not show the mode. The mode is the most frequently occurring value, and none of the five numbers a box plot is built from carries frequency information. Both data sets could easily have a mode of 2 — Data 2's median is 2, and Data 1's box covers 2 — and the two pictures would look exactly as they do.

a iii — "For Data 1, there are more data values below four than there are above four."

False, because 4 is the median of Data 1. The median is the value that splits the ordered data into halves: about 50 percent below and about 50 percent above. They are equal, not unbalanced.

b — For which group is 7 more likely to be an outlier?

Use the \(1.5 \times IQR\) rule from the previous section on each set.

$$ IQR_{\text{Data 1}} \approx 4.7 - 2 = 2.7 \qquad Q_3 + 1.5 \times IQR \approx 4.7 + 4.05 = 8.75 $$ $$ IQR_{\text{Data 2}} \approx 2.4 - 1.5 = 0.9 \qquad Q_3 + 1.5 \times IQR \approx 2.4 + 1.35 = 3.75 $$

For Data 1 the upper fence lands at about 8.75, so a value of 7 is comfortably inside it and would not be flagged. For Data 2 the fence lands at about 3.75, and 7 is nearly twice that.

Answer (b): Data 2. Its middle half is packed into a band less than one unit wide, so the fences sit close in and a value of 7 is far outside them. In Data 1 the same value of 7 is just the top of a genuinely wide distribution. Note that this is exactly why the plots are drawn with whiskers reaching 7 in both cases — the same number can be ordinary in one data set and extraordinary in another, and it is the \(IQR\) that decides which.

Reading the three plots. All values below are estimates read off the shared number line in Figure 2.4.11.

• BMW 3 series — about \(23,\ 29,\ 33,\ 41,\ 66\)
• BMW 5 series — about \(31,\ 41,\ 42,\ 55,\ 64\)
• BMW 7 series — about \(36,\ 41,\ 46,\ 59,\ 67\)

a — What each shape implies. The 3 series distribution is the youngest and the most right-skewed: its box sits between 29 and 41, its median is 33, and yet the right whisker runs all the way to 66. Most 3 series buyers were in their late twenties and thirties, with a thin tail of much older buyers. The 5 series is shifted up the scale, and its median sits almost on top of \(Q_1\) — a dense cluster of buyers right around 41 or 42 with a wide, thinly populated upper half. The 7 series is the oldest and the most symmetric of the three: the median sits near the middle of a box running from 41 to 59, with whiskers of similar length on both sides.

b — Most likely to have an outlier. Apply the \(1.5 \times IQR\) fence to each.

$$ \text{3 series: } IQR \approx 41 - 29 = 12, \quad Q_3 + 1.5 \times IQR \approx 41 + 18 = 59 $$ $$ \text{5 series: } IQR \approx 55 - 41 = 14, \quad Q_3 + 1.5 \times IQR \approx 55 + 21 = 76 $$ $$ \text{7 series: } IQR \approx 59 - 41 = 18, \quad Q_3 + 1.5 \times IQR \approx 59 + 27 = 86 $$

Only the 3 series has a maximum (about 66) beyond its own upper fence (about 59).

Answer (b): The BMW 3 series. Its middle half is narrow, which pulls the fences in close, while its maximum reaches 66 — the combination is what makes a value extreme, not the raw size of the number.

c — The three compared. Buyer age climbs steadily across the series: median about 33 for the 3 series, about 42 for the 5 series, about 46 for the 7 series. The spread of the middle half climbs too, from an \(IQR\) of about 12 years up to about 18. Read together, the plots say the cheaper 3 series draws a younger and more tightly clustered group of buyers, while the more expensive 7 series draws older buyers across a broader age range — and all three overlap heavily in the forties, so knowing a buyer's age would not let you guess their series with any confidence.

d — 5 series, smallest spread. The four quarters of the 5 series span about \(41 - 31 = 10\), \(42 - 41 = 1\), \(55 - 42 = 13\), and \(64 - 55 = 9\) years.

Answer (d): The second quarter, spanning roughly 1 year. A quarter of all 5 series buyers were essentially the same age, right around 41 or 42.

Answer (e): The third quarter, spanning roughly 13 years (about 42 to 55).

f — Interquartile range.

$$ IQR \approx Q_3 - Q_1 = 55 - 41 = 14 \text{ years} $$

g — 31 to 38 or 45 to 55? Both intervals sit inside a single quarter, and each quarter holds about 25 percent of the buyers, so compare how much of its quarter each interval covers. The interval 45 to 55 covers 10 of the 13 years of the third quarter — most of it. The interval 31 to 38 covers 7 of the 10 years of the first quarter, but it stops 3 years short of \(Q_1\) at both the busy end and is spread over a slightly wider share of empty scale.

Answer (g): More data lies in the interval 45 to 55, because it takes in nearly the whole third quarter, while 31 to 38 leaves out the part of the first quarter nearest \(Q_1\) where those buyers are most densely packed.

h — Fewest data. Option iii, 41 to 64, runs from \(Q_1\) all the way to essentially the maximum, so it holds about 75 percent of the buyers — the most, not the fewest. Options i and ii both sit inside the first quarter, which holds about 25 percent in total, spread across the 10 years from 31 to 41. Option i, 31 to 35, takes in 4 of those 10 years; option ii, 38 to 41, takes in 3.

Answer (h): ii, the interval 38 to 41 holds the fewest — it is the narrowest slice of the single quarter that both candidate intervals fall inside. All of these are estimates read off a graph; with the underlying ages we could count exactly.

Step 1 — Turn the frequency table back into an ordered list. There are \(5 + 9 + 6 + 4 + 1 = 25\) students. Listing every response in order gives five 0s, then nine 1s, then six 2s, then four 3s, then one 4 — so positions 1–5 hold 0, positions 6–14 hold 1, positions 15–20 hold 2, positions 21–24 hold 3, and position 25 holds 4.

Step 2 — Extremes. Minimum \(= 0\), maximum \(= 4\).

Step 3 — Median. With 25 values, an odd count, the median is the 13th value. Position 13 falls inside the run of 1s, so the median is 1.

Step 4 — First quartile. \(Q_1\) is the median of the 12 values below the median (positions 1 through 12), so it is the average of the 6th and 7th values. Both are 1.

$$ Q_1 = \frac{1 + 1}{2} = 1 $$

Step 5 — Third quartile. \(Q_3\) is the median of the 12 values above the median (positions 14 through 25), so it is the average of the 19th and 20th values. Both are 2.

$$ Q_3 = \frac{2 + 2}{2} = 2 $$

Step 6 — Draw it. Scale a number line from 0 to 4. The box runs from 1 to 2 — just one unit wide. The median is 1, which is also \(Q_1\), so the median line lands exactly on the left edge of the box and no dashed line appears inside it. The left whisker runs from 1 down to 0 and the right whisker from 2 out to 4.

Answer: The five-number summary is \(0,\ 1,\ 1,\ 2,\ 4\) with \(IQR = 1\). This is the coinciding-values case from earlier in the section: because \(Q_1\) and the median are both 1, at least a quarter of the students watched exactly one movie, and the missing median line inside the box is what tells you so.