Chapter 5: Connecting the Dots… (Data Handling)

A statistical question needs data that varies to answer it.

• Statistical: (a) prices differ by shop, brand and time; (b) the dogs have different ages; (c) we must ask each student; (g) rainfall differs day to day; (f) can be answered by comparing the bowlers' figures (wickets, runs given).
• Not statistical: (d) asks one person for one answer; (e) is a single count or estimate of one wall.

Statistical: (a), (b), (c), (f), (g). Not statistical: (d), (e).

Vaishnavi's comparison is not fair: Shubman played 5 matches and Yashasvi only 4. A better single number is the average (runs per match): Shubman $110÷5=22$, Yashasvi $96÷4=24$. So Yashasvi did slightly better on average.

(The book's line "110 ÷ 5 = 21 runs" should read 22.)

Totals are unfair because the numbers of matches differ; averages: Shubman 22, Yashasvi 24.

$(6+2+9+5+4+6+3+5)÷8=40÷8=$ 5 bounces

5 bounces

Sample data for 8 attempts: 4, 7, 5, 9, 6, 3, 8, 6. Total $=48$, so the average $=48÷8=6$ bounces. (Use your own counts in the same way: add them and divide by the number of attempts.)

Add your counts and divide by the number of attempts (e.g. 48 ÷ 8 = 6).

Sample record for a jasmine plant: 5, 8, 6, 9, 7, 4, 10 flowers. Total $=49$; average $=49÷7=7$ flowers per day.

E.g. 49 flowers in 7 days, an average of 7 a day.

Nikhil: $122÷7\approx 17.43$ s. Sunil: $122÷7\approx 17.43$ s. Their averages are equal: neither was quicker on average. (Nikhil was more consistent, between 16 and 19 s; Sunil varied from 16 to 20 s.)

Both average 122 ÷ 7 ≈ 17.4 s: equal.

$(1555+1670+1750+2013+2040+2126)÷6=11154÷6=$ 1859

1859

Average (mean) price: Yahapur $458÷12\approx 38.2$, i.e. ₹38.2 per kg; Wahapur $450÷12=37.5$, i.e. ₹37.5 per kg. On average onions are slightly costlier in Yahapur, though the two are very close; Wahapur's prices vary more (₹17 to ₹60).

Other ways to compare: the median price (Yahapur ₹37, Wahapur ₹38.5), the range (35 and 43), the number of months above some price (e.g. above ₹40), or comparing the season-wise (summer, monsoon, winter) averages.

Means ₹38.2 (Yahapur) and ₹37.5 (Wahapur): Yahapur is very slightly costlier; Wahapur's prices are more spread out.

It shows all 24 prices (every value appears as a dot), but it loses the months: from the dot plot alone we cannot tell which price belongs to January. (From the table, January in Yahapur is ₹25.)

It shows every value but not the month, so January's price cannot be read from it.

Without 118: 165, 170, 173, 175. Mean $=683÷4=170.75$ cm; median $=(170+173)÷2=171.5$ cm.

The mean jumps from 160.2 cm to 170.75 cm, while the median changes only a little (170 → 171.5). The outlier affects the mean much more than the median.

Mean 170.75 cm, median 171.5 cm: the mean rises a lot, the median hardly changes.

Sorted: 0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40 (15 values).

Mean $=122÷15\approx$ 8.13; median $=$ 6 (8th value). The mean is greater than the median because of the large value 40, the outlier.

Without 40: mean $=82÷14\approx 5.86$, median $=(5+6)÷2=5.5$. The mean drops a lot (8.13 → 5.86) while the median changes slightly (6 → 5.5).

Mean ≈ 8.13, median 6 (mean > median because of the outlier 40); without 40: mean ≈ 5.86, median 5.5.

Mean $=128÷7\approx 18.3$; sorted 10, 16, 16, 18, 20, 22, 26, so median $=18$.

Mean ≈ 18.3 pages, median 18 pages.

If there is one very low and one very high outlier, their effects on the mean partly cancel, so the mean may still be close to the median. E.g. 2, 20, 21, 22, 23, 24, 50: mean = 162 ÷ 7 ≈ 23.1, median = 22. But if one outlier is much more extreme than the other, the mean shifts towards it. The median stays near the middle either way.

Low and high outliers pull the mean in opposite directions and may balance out; the median is hardly affected.

Boys above 144.4 cm: 147, 154, 158, 155, 146, 146, 145, 150 → 8 boys. Girls above: 150, 154, 145, 148, 156, 150, 150 → 7 girls.

So 15 students in all are taller than the average.

15 students; 8 of them are boys.

Both groups did well on the whole: their centres are close to 60 seconds. In Group A the median is exactly 60, but the mean (58.21) is lower, so a few children opened their eyes much too early (low values pull the mean down). Group B's mean and median are both close to 60 and to each other, so its estimates are more balanced, though spread on both sides. Overall Group B's estimates were slightly closer to a minute on average.

Both centred near 60 s; A has some early openers (mean < median), B is more balanced with mean and median both ≈ 59.5 s.

Yes. With 11 players, the median is the 6th score in order. If at least 6 players scored 0 (ducks or did not need to bat much), the median is 0, while the remaining 5 players and extras can still make 407.

Mangoes (only the fruiting months, March to July): 8, 24, 41, 16, 5. Mean $=94÷5=$ 18.8 mangoes a month; median $=$ 16.

Yes, if at least 6 players scored 0; mangoes: mean 18.8, median 16 (over the 5 fruiting months).

Yahapur sorted: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59 → median $=(35+39)÷2=$ ₹37

Wahapur sorted: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60 → median $=(38+39)÷2=$ ₹38.5

Yahapur ₹37, Wahapur ₹38.5

Leave out the 2 absent students: 20 values, total 72.

Mean $=72÷20=$ 3.6. Sorted: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25 → median $=$ 2.

Most students have 0 to 4 animals (6 have none); a few have many (10 and 25 are outliers), which pulls the mean above the median. The median (2) describes a typical student better.

Mean 3.6, median 2; most have 0–4 animals, the outliers 10 and 25 raise the mean.

29 trees; total 1621 ft. Mean $=1621÷29\approx$ 55.9 ft; median (15th value) $=$ 56 ft.

The heights range from 43 ft to 67 ft, spread fairly evenly, with most trees between 50 and 62 ft and no outliers, so the mean and median are almost equal.

Quicker mean: subtract a base, e.g. 50, from each height, add the differences ($0-5-7+2+\cdots$, total 171), divide by 29 (≈ 5.9) and add back 50: ≈ 55.9.

Shorter than the average: 13 trees (all those of 55 ft or less).

Mean ≈ 55.9 ft, median 56 ft; heights 43–67 ft, evenly spread; 13 trees are shorter than the mean.

(a) No. Every value is at most 20.5. The mean is the "fair share", so it cannot exceed the largest value; the median is one of the middle values. Both lie between 3.09 and 20.5. (Here the mean ≈ 9.71 L, median = 8 L.)

(b) No. Both always lie between the minimum and the maximum (they can equal them only if all values are equal).

(a) No; both must lie between 3.09 and 20.5 (mean ≈ 9.71, median 8) (b) No, never.

Boys: mean $=20.6÷6\approx 3.43$ kg, median $=(3.4+3.5)÷2=3.45$ kg. Girls: mean $=20.1÷6=3.35$ kg, median $=3.4$ kg.

Both groups range from about 2.5 to 4.1 kg with similar spreads. The boys are very slightly heavier on average (by under 0.1 kg), a difference too small to matter with only 6 babies each.

Boys: mean ≈ 3.43, median 3.45; girls: mean 3.35, median 3.4. Very similar, boys slightly heavier.

• In this section the boys are slightly taller on average than the girls (142.05 vs 140.14), the opposite of the first section, where girls were taller (146.9 vs 142.94).
• For the boys and the whole class the mean is a little below the median, so a few shorter students pull the mean down; for the girls the mean and median are nearly equal.
• The first section is taller overall (mean 144.4 vs 141.21; median 145 vs 142.5).

So whether boys or girls are taller depends on the group; we cannot generalise from one class.

Here boys are slightly taller (opposite of section 1); section 1 is taller overall (mean 144.4 vs 141.2).

Mean sumo $=1201.9÷5\approx 240.4$ kg; mean dancer $=254.5÷6\approx 42.4$ kg. Ratio $\approx 240.4÷42.4\approx 5.7$: about 6 times heavier.

About 6 times (≈ 240 kg vs ≈ 42 kg on average).

The vertical scale is 1 unit = ₹10 (gridlines at 0, 10, 20, …, 60). Yes: with the two bars side by side for each month, we can see at a glance which town is costlier in each month.

1 unit = ₹10; yes, side-by-side bars make monthly comparison easy.

Approximate readings (2021, 2022, 2023): SpaceX 31, 61, 96; CASC (China) 48, 35, 45; Roscosmos (Russia) 16, 21, 18; Arianespace (France) 15, 6, 3; Rocket Lab (USA) 6, 6, 9; ULA (USA) 5, 8, 3; ISRO (India) 2, 5, 7; Galactic Energy (China) 1, 2, 7; Expace (China) 4, 4, 6; Others 18, 17, 26.

(a) Not justified: e.g. Arianespace and CASC launched fewer in some years.

(b) Justified: only SpaceX (USA) crossed 50 (61 and 96); CASC's highest is about 48.

(c) Justified: about $15+6+3=24$.

(d) Justified (roughly): $(48+35+45)÷3\approx 43$.

(e) Justified: about 14 for ISRO against about 10 for Galactic Energy.

(f) Not justified: about $16+21+18=55$.

Justified: (b), (c), (d), (e). Not justified: (a), (f).

Increasing every year: SpaceX, ISRO and Galactic Energy (Rocket Lab and Expace stayed level from 2021 to 2022 before rising).

2023 total ≈ 96 + 45 + 18 + 3 + 9 + 3 + 7 + 7 + 6 + 26 ≈ 220, so (b) 200 to 400.

SpaceX, ISRO, Galactic Energy; 2023 total ≈ 220, option (b).

Yes. City 1 has long days in June and short days in December, so it is in the Northern Hemisphere, far from the Equator (it is Helsinki). City 2 has the opposite pattern, so it is in the Southern Hemisphere (Wellington). Near the Equator, daylight stays close to 12 hours all year.

Things to explore: daylight in your own town month by month; where the "midnight sun" can be seen; why the Equator has nearly equal days and nights.

City 1 is in the Northern and City 2 in the Southern Hemisphere, both far from the Equator.

(1) The blue team batted all 20 overs; the red team's bars stop after 18 overs, with no wickets marked. An innings that stops early without being all out usually means the chasing side reached its target, which suggests blue batted first and red chased and won. (Note: adding the bars as drawn gives blue about 163 and red about 152, so the bars in the figure are only approximate.)

(2) 15 runs in over 12.

(3) Over 4 (only about 2 runs).

(4) Not easy: we would have to read and add all 20 blue bars (about 163, so a target of about 164). A cumulative (running-total) graph would show it directly.

(1) Blue batted first; red's innings ends at 18 overs, suggesting red chased and won (2) 15 (3) over 4 (4) No, the 20 bars must be added.

Yes, it is a bar graph (each animal's speed is shown by the length of a bar, with colours for air, land and water).

(a) The speed axis is marked every 16 km/h (16, 32, 48, …, 322): 1 unit = 16 km/h.

(b) For example: the peregrine falcon (322 km/h) is about 3 times as fast as the cheetah (103 km/h); the tiny tiger beetle runs only 8 km/h but covers 120 body lengths a second.

(c) Pairs at about twice: sailfish (109) and flying fish (56); peregrine falcon (322) and spine-tailed swift (170); pronghorn antelope (88) and dolphin (40); flying fish (56) and humpback whale (26).

(d) $109÷26\approx 4.2$, so yes, about 4 times faster. But we cannot say from this graph that the sailfish is the fastest aquatic animal in the world: only five water animals are shown.

Yes, a bar graph; (a) 1 unit = 16 km/h (c) e.g. sailfish 109 and flying fish 56 (d) yes, about 4 times, but the graph does not prove it is the fastest in the world.

Counts: Grade 5: aquatic 6, aerial 13, spaceborne 2, none 4 (25 students). Grade 9: aquatic 6, aerial 8, spaceborne 9, none 2 (25 students).

Aerial is the most popular in Grade 5, while spaceborne is the most popular in Grade 9; aquatic is equally liked in both.

Grade 5: 6, 13, 2, 4; Grade 9: 6, 8, 9, 2 (aquatic, aerial, spaceborne, none).

Day 1 ranges from 16°C to 34°C: a cool night and warm afternoon, typical of winter (around December–January). Day 2 ranges from 30°C to 43°C: very hot, typical of summer (around May–June).

Day 1 is likely a winter day (Dec–Jan) and Day 2 a summer day (May–June).

(a)

(b) States are along the horizontal line; for each state three bars show 2022, 2023 and 2024. The vertical scale is 1 unit = 25,000 registrations, with fainter guidelines every 5,000.

(c) Every state registered more vehicles in 2024 than in 2022. West Bengal and Odisha grew the most sharply; Assam and Andhra Pradesh grew steadily; Gujarat rose in 2023 and then fell a little in 2024; Uttarakhand grew only slightly.

(d) About $60,000-40,000=$ 20,000 more.

(e) From about 10,000 to about 43,000: roughly 4 times.

(f) Not correct. Each bar shows the registrations in that year (not a running total). Uttarakhand had about 16,000 new registrations in 2023 and 19,000 in 2024; the small change in bar length only means the number per year grew slowly.

(c) All grew; WB and Odisha most (d) ≈ 20,000 (e) ≈ 4 times (f) No: each bar is that year's new registrations (≈ 16,000 and 19,000).

(1) Justified: for every age, the 2019 averages are higher than the 1989 averages (e.g. girls aged 5: 100 → 107.2).

(2) Not justified: 143.2 cm (1989, 13) is less than 148 cm (2009, 14).

(3) Justified, just barely: 159 cm > 158.9 cm.

(4) Not justified: the table gives averages only; some 11-year-old girls are taller than some 13-year-old girls.

(5) Not justified: at ages 10–12 (and age 5 in 2019) the girls' average is more than the boys'.

(6) Not justified: the table stops at 19; growth is slowing (166 → 166.5 cm from 18 to 19 in 2019).

Justified: (1) and (3). Not justified: (2), (4), (5), (6).

In 2019, boys grew most between 12 and 13 (+6.2 cm); girls between 10 and 11 (+5.8 cm). Girls have their growth spurt earlier; afterwards boys keep growing longer.

Ages 1–4 (from about 50 cm at birth to about 107 cm at 5, growing fastest in the first year): about 75 cm (1), 86 cm (2), 95 cm (3), 101 cm (4).

2029: from 2009 to 2019 the averages rose by about 2–3 cm; so for 2029 add about 2–3 cm to each 2019 value, e.g. boys aged 10 ≈ 135 cm, girls aged 10 ≈ 135 cm, boys aged 19 ≈ 168.5 cm, girls aged 19 ≈ 157 cm.

Boys grew most from 12 to 13 (6.2 cm), girls from 10 to 11 (5.8 cm); ages 1–4 ≈ 75, 86, 95, 101 cm; 2029 ≈ 2019 values + 2 to 3 cm.

Boys (12): range 3 to 6; median $=(5+5)÷2=5$; mean $=56÷12\approx 4.7$. Girls (13): range 0 to 6; median $=4$; mean $=45÷13\approx 3.5$.

(a) False: the girls' data is more spread out (0 to 6). (b) True: 5 > 4. (c) False: 3.5 < 4.7. (d) False: both maximums are 6.

Only (b) is true.

(a) $50÷4=$ 12.5

(b) C did not play game 3, so divide by 3: $32÷3\approx 10.67$. B played all 4 games (a score of 0 is still a game played), so divide by 4: $18÷4=4.5$.

(c) A has the highest average (12.5).

(a) 12.5 (b) C: by 3 (≈ 10.67); B: by 4 (4.5) (c) A

Group 1: mean $=730÷10=73$; median $=(76+81)÷2=78.5$. Group 2: mean $=681÷9\approx 75.7$; median $=79$.

Group 2 did slightly better on both measures. In group 1 the low scores 39 and 48 pull the mean well below the median; group 2's scores are more even.

Group 1: mean 73, median 78.5; group 2: mean ≈ 75.7, median 79; group 2 did slightly better.

Observations: cricket is by far the most watched (1240) and played (620). For every sport, fewer people participate than watch, roughly half or less. Athletics is the least watched and played. Basketball and swimming have equal participation (320).

Cricket leads in both; for each sport, participants are about half (or fewer) of the watchers; athletics is lowest.

Sort: 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125. The median is 115 cm (9th value).

With 17 students, keep the median student aside: the 8 students before the 9th position form the shorter group and the 8 after it the taller group. Since five students are exactly 115 cm, the teacher must split them by position in the sorted line (or measure them more precisely).

Heights around 101–125 cm, with a median of 115 cm, match the average heights of children aged about 6 to 7 years in the table.

Use the median (115 cm): 8 students below it, 8 above (the five 115 cm students split by order). They are probably about 6–7 years old.

Measure every student, draw a number line (say 130 to 170 cm) and put one dot per student. For example, for 10 students: 142, 145, 147, 148, 150, 150, 151, 153, 156, 162: mean $=1504÷10=150.4$ cm, median $=(150+150)÷2=150$ cm. Close values of mean and median show the heights are evenly spread with no outlier.

E.g. mean 150.4 cm, median 150 cm for a sample class; mark them on your dot plot.

(d) It cannot be determined: the other section's heights could be anything.

(d)

(a) New York ≈ 300 (its bar is between Shenzhen's 367 and Dubai's 251); Tokyo ≈ 170 (between 183 and 154); London ≈ 30 (shorter than Moscow's 46).

(b) (i) Valid for the cities shown: 12 cities are above Mumbai (86). (ii) Valid for the cities shown: 7 cities are below Mumbai (Seoul to London). (iii) Not valid: the graph shows the number of tall buildings, not their heights. (In fact the tallest building, the Burj Khalifa, is in Dubai.)

(a) New York ≈ 300, Tokyo ≈ 170, London ≈ 30 (b) (i) and (ii) valid for the cities listed; (iii) not valid.

Sample record:

Average difference $=(0.8+0.5+3.5+1.2+2.5)÷5=8.5÷5=1.7$ cm. Draw estimate and measure as two bars for each object (as in the graphs above).

Find each positive difference and average them (sample: 8.5 ÷ 5 = 1.7 cm).

Week 1: mean $=3240÷9=360$ s, median $=360$ s. Week 2: mean $=2230÷8\approx 278.8$ s, median $=(270+280)÷2=275$ s.

Aditi became much faster in week 2 (about 80 seconds quicker on average). Her times also kept falling within each week; the 380 s in week 2 stands out as an unusually slow puzzle.

Week 1: mean = median = 360 s; week 2: mean ≈ 278.8 s, median 275 s; she improved by about 80 s.

These are data-collection projects; the method is the same each time:

1. Collect the data carefully and write it in a table.
2. Draw a dot plot (or a double-bar graph when comparing two groups).
3. Find the minimum, maximum, mean and median.
4. Look for outliers and describe how the data is spread or clustered.

For example, for "What is in a name?": if the 30 names in a class have 3 to 11 letters with mean 6.2 and median 6, most names have 5 to 7 letters; the median starting letter tells you whether more names start with A–M or with N–Z.

Collect, tabulate, plot (dot plot or double-bar graph), then compare using min, max, mean and median.

• From 036: 0, 3, 6 are not in the code. So in 064, the correct digit must be 4, and it is not in the 3rd place.
• In 542, two digits are correct but misplaced: 4 is one of them (not in the 2nd place). With 265 having one digit in the right place and 6 ruled out, the other is 2 or 5.
• If 2 were in the code: from 265 it would have to be in the 1st place, but from 271 the correct digit must be misplaced, so 2 cannot be first. Hence 5 is in the code, and from 265 it is in the 3rd place.
• Now 4 is not in the 2nd or 3rd place, so 4 is 1st. From 271, the correct misplaced digit is 1 (2 and 7 are out), and it is not 3rd, so it is 2nd.

The code is 415. (Check: 265 → 5 right place; 271 → 1 wrong place; 542 → 5 and 4 wrong places; 036 → none; 064 → 4 wrong place.)

415