Sum \(= 6+2+9+5+4+6+3+5 = 40\)

\[\text{Average} = \dfrac{40}{8} = 5\]

These are hands-on data-collection activities. Once you record your own data (e.g. 7 bounce counts or 7 days of flower counts), use the same method: Average = (Sum of all values) ÷ (Number of values). Add up your recorded numbers and divide by how many readings you took.

Nikhil: Sum \(=17+18+17+16+19+17+18 = 122\); Average \(=\dfrac{122}{7}\approx 17.43\)s

Sunil: Sum \(=20+18+18+17+16+16+17 = 122\); Average \(=\dfrac{122}{7}\approx 17.43\)s

Sum \(=1555+1670+1750+2013+2040+2126 = 11154\)

\[\text{Mean} = \dfrac{11154}{6} = 1859\]

Yahapur sorted: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59 (12 values)

\[\text{Median} = \dfrac{35+39}{2} = 37\]

Wahapur sorted: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60 (12 values)

\[\text{Median} = \dfrac{38+39}{2} = 38.5\]

Ignore the two dashes (students absent, not zero). Valid data (20 values): 0,1,0,4,8,0,0,2,1,1,5,3,4,0,0,10,25,2,2,4

Sum \(= 72\); Mean \(=\dfrac{72}{20} = 3.6\)

Sorted: 0,0,0,0,0,0,1,1,1,2,2,2,3,4,4,4,5,8,10,25 (20 values)

Median \(=\dfrac{2+2}{2} = 2\) (average of 10th and 11th values)

Sum of all 29 heights \(= 1621\)

\[\text{Mean} = \dfrac{1621}{29} \approx 55.9 \text{ ft}\]

Sorted (29 values): 43,44,45,46,49,50,51,52,52,54,54,55,55,56,56,57,58,59,60,60,60,60,61,61,62,63,65,66,67

Median = 15th value (middle of 29) = 56 ft

Trees shorter than average (55.9 ft): 43,44,45,46,49,50,51,52,52,54,54,55,55 = 13 trees

Quicker way to find the mean: Group repeated values into a frequency table (value × frequency), sum those products, then divide by total count — this avoids adding 29 numbers one by one.

(a) No. The maximum value in the data is only 20.5 L. Both mean and median are always computed from within the data range, so neither can exceed the maximum value (20.5). Hence they cannot lie between 25–30.

(b) No, in general the mean and median can never be less than the minimum value or greater than the maximum value of a dataset — they always lie somewhere between (or equal to) the smallest and largest values, since they represent a “balance point” or “middle point” of the data.

Boys: Sum \(=3.5+4.1+2.6+3.2+3.4+3.8=20.6\); Mean \(=\dfrac{20.6}{6}\approx 3.43\) kg

Sorted: 2.6,3.2,3.4,3.5,3.8,4.1 → Median \(=\dfrac{3.4+3.5}{2}=3.45\) kg

Girls: Sum \(=4.0+3.1+3.4+3.7+2.5+3.4=20.1\); Mean \(=\dfrac{20.1}{6}=3.35\) kg

Sorted: 2.5,3.1,3.4,3.4,3.7,4.0 → Median \(=\dfrac{3.4+3.4}{2}=3.4\) kg

In the first section (from the text), girls were taller on average (146.9 cm) than boys (142.94 cm). In this new section, the trend is reversed: boys are taller on average (142.05 cm) than girls (140.14 cm).

Sumo mean: Sum \(=295.2+250.7+234.1+221.0+200.9=1201.9\); Mean \(=\dfrac{1201.9}{5}\approx 240.4\) kg

Ballet mean: Sum \(=40.3+37.6+38.8+45.5+44.1+48.2=254.5\); Mean \(=\dfrac{254.5}{6}\approx 42.4\) kg

\[\text{Ratio} = \dfrac{240.4}{42.4} \approx 5.67\]

(a) This is technically a bar chart, but drawn horizontally with icons; the scale used along the axis is approximately 1 unit = 16 kph.

(b) The peregrine falcon (322 kph, diving) is the fastest animal shown overall — far ahead of every land and water animal.

(c) Peregrine falcon (322 kph) is roughly twice as fast as the spine-tailed swift (170 kph): \(322 \div 170 \approx 1.9\), close to 2.

(d) Sailfish = 109 kph, Humpback whale = 26 kph. \(\dfrac{109}{26} \approx 4.2\) — yes, about 4 times faster.

Day 1: 20,18,16,20,26,34,30,24 °C (moderate range, cooler nights) — likely a winter month, e.g. December/January.

Day 2: 37,34,30,33,37,43,42,39 °C (consistently very hot, peaking at 43°C) — likely peak summer, e.g. May/June in the Rajasthan desert.

(a) Gujarat: 2022 = 69000, 2023 = 89000, 2024 = 78000. Delhi: 2022 = 62000, 2023 = 74000, 2024 = 81000. Place these bar-tops between the appropriate gridlines on the chart.

(d) Approximate reading: Assam ≈ 40,000 (2022) → ≈ 60,000 (2023), an increase of about 20,000 registrations.

(e) West Bengal ≈ 10,000 (2022) → ≈ 43,000 (2024), roughly a 4× increase.

(f) Not quite correct. Even though Uttarakhand’s bars look visually short and similar in length, because the graph’s scale is large (each gridline represents 25,000 vehicles), a “small” visual increase can still represent a few thousand actual vehicle registrations — so describing it as “very few new registrations” understates the real numbers without checking the scale.

Reading the dot plots: Boys’ pocket counts cluster in a narrower range (mostly 3–6), while Girls’ pocket counts spread more widely (from 0 up to 6).

Note: exact counts should be verified by counting dots directly on your copy of the dot plot.

(a) A’s average \(=\dfrac{14+16+10+10}{4}=\dfrac{50}{4}=12.5\) points/game

(b) C should be divided by 3, because C only played 3 games (the “did not play” game must be excluded, just like the batting series example earlier). B should be divided by 4, because B played all 4 games — scoring 0 in one game still counts as a played match.

C’s average \(=\dfrac{8+11+13}{3}=\dfrac{32}{3}\approx 10.67\); B’s average \(=\dfrac{0+8+6+4}{4}=\dfrac{18}{4}=4.5\)

Group 1: Sum \(=730\), \(n=10\); Mean \(=73\). Sorted: 39,48,56,75,76,81,85,85,90,95 → Median \(=\dfrac{76+81}{2}=78.5\)

Group 2: Sum \(=681\), \(n=9\); Mean \(=\dfrac{681}{9}\approx75.67\). Sorted: 47,59,68,73,79,86,86,90,93 → Median = 5th value = 79

Use scale 1 unit = 100 (or 200) people on the vertical axis, with two bars (“Watching” and “Participating”) for each of the 5 sports.

Sorted: 101,102,106,109,110,110,112,115,115,115,115,115,117,120,120,123,125 (17 values)

The value 115 cm repeats 5 times — right around the middle of the data. Using 115 cm as the splitting height: 7 students are below 115 cm, and 5 students are above 115 cm, with the 5 students at exactly 115 cm excluded as “the particular height” itself.

Comparing these heights (roughly 101–125 cm) to the 1989–2019 height table, this range best matches children around ages 6 to 8.

This requires your own class’s actual height data. Method: record every student’s height, then compute Mean = (sum of all heights) ÷ (number of students), and find the Median by sorting all heights and picking the middle value (or averaging the two middle values if the class size is even). Plot the values as a dot plot to see how spread out or clustered your class’s heights are, and note whether mean ≈ median (balanced data) or they differ a lot (possible outliers).

(a) Approximate readings from the bar chart: New York ≈ 300, Tokyo ≈ 170, London ≈ 15 (the shortest bar on the chart).

Fill in your own estimated and actual measured lengths using a ruler. The Positive Difference for each row = |Estimate − Measure| (ignore the sign). Then:

\[\text{Average difference} = \dfrac{\text{Sum of all Positive Differences}}{5}\]

This tells you, on average, how accurate your estimates were — a smaller average difference means better estimation skills!

Week 1: Sum \(=410+400+370+340+360+400+320+330+310=3240\); Mean \(=\dfrac{3240}{9}=360\)s. Sorted: 310,320,330,340,360,370,400,400,410 → Median = 360

Week 2: Sum \(=320+290+380+280+270+230+220+240=2230\); Mean \(=\dfrac{2230}{8}=278.75\)s. Sorted: 220,230,240,270,280,290,320,380 → Median \(=\dfrac{270+280}{2}=275\)

These are hands-on data-collection projects — there is no single fixed numeric answer since the data depends on your own textbooks, classmates’ names, or personal daily habits. For each project, follow the same core method used throughout this chapter:

1. Collect your raw data carefully (word counts, name lengths, step-out counts, heights, or time estimates).
2. Make a dot plot to visualise the spread (minimum, maximum, clustering).
3. Compute Mean = (sum of values) ÷ (number of values).
4. Sort the data and find the Median (middle value, or average of the two middle values if the count is even).
5. Compare mean and median — if they’re close, the data is balanced; if they differ a lot, look for outliers.
6. For comparisons across two groups (e.g., boys vs girls names, or Week 1 vs Week 2), draw a double-bar graph with a sensible scale.

Step 1 — Eliminate digits using “036 → nothing correct”: Digits 0, 3, 6 do not appear anywhere in the code.

Step 2 — Use “064 → one digit correct, wrong place”: Since 0 and 6 are eliminated, the correct digit must be 4, and it is not in position 3 (where it appears in “064”).

Step 3 — Use “265 → one digit correct & well-placed”: Since 6 is eliminated, the correct digit is either 2 (position 1) or 5 (position 3). If 2 were correct at position 1, digit 4 (from Step 2) would have nowhere left to go (it can’t be position 3). So it must be 5, correctly placed at position 3.

Step 4 — Use “542 → two digits correct, wrong place”: We already know 5 is in the code (at position 3) — in “542” it sits at position 1, which is indeed the wrong place, so that’s one correct-but-misplaced digit. We need one more correct digit from {5,4,2}; since 4 is confirmed in the code (Step 2), 4 is the second one. In “542”, 4 sits at position 2 — but we already know (Step 2) 4 is not at position 3, and now also not at position 2 (since it’s “wrongly placed” here too). So 4 must be at position 1. This also tells us 2 is NOT in the code (since only two of {5,4,2} were correct, and we’ve already found both: 5 and 4).

Step 5 — Use “271 → one digit correct, wrong place”: Since 2 is eliminated (Step 4), the correct digit must be 1 or 7. If 7 were correct, it would need a position other than position 2 (guessed position) — but positions 1 and 3 are already taken by 4 and 5, leaving no room for 7. So it must be 1, and since position 3 is taken by 5, 1 must go in the only remaining slot: position 2.

Verification against all 5 clues with code 4-1-5: