CLASS 7 · GANITA PRAKASH PART II · CHAPTER 5
Accessible steps for asking statistical questions, finding mean and median, reading graphs and explaining what data can show.
63 worked answers · short steps first, explanations when you need them
Textbook pages 98–134
Question 5.1-1
A statistical question expects different answers across a group or set of observations.
Shop prices can differ by seller, place, discounts and time.
Yes, if “in India” means prices from different shops or places. Prices can vary, so we need several observations.
Question 5.1-2
There is a group of dogs, not just one dog.
Their ages may vary.
Yes. The dogs can have different ages, so we collect several values.
Question 5.1-3
Survey each student.
Count “yes” answers and divide by the number of students surveyed.
Yes. Students may answer differently; count the students who like it and divide by the class total.
Question 5.1-4
The original question is about one person and has one response.
A question about a class would collect varying responses.
No, as phrased to one person: it asks for one answer. “How many students in our class like reading?” is statistical.
Question 5.1-5
Count the bricks in this particular wall.
The result is one value, not a distribution of data.
No. It asks for one fixed count for one wall; it does not anticipate a collection of varying answers.
Question 5.1-6
Collect rainfall measurements over the year.
Compare the measurements over time.
Yes. Rainfall varies from day to day or month to month, so many measurements are needed to describe its pattern.
Question 5.1-7
The question compares a group of bowlers, not one fixed value.
Collect each bowler’s relevant match figures and decide what “best” means (for example, most wickets or fewest runs per over).
Because the answer depends on varied data and a comparison, it is statistical.
Yes. To decide, we compare several bowlers’ match data, such as runs conceded, wickets and overs bowled; their results vary.
Question 5.2-1
Shubman total = 0+17+21+90 = 128; 128÷4=32; range=90−0=90.
Yashasvi total = 67+55+18+35 = 175; 175÷4=43.75; range=67−18=49.
Shubman: mean 32, range 90. Yashasvi: mean 43.75, range 49. Yashasvi had the higher mean.
Question 5.2-2
Add the scores: Shubman 110 across 5 matches; Yashasvi 96 across 4 matches. Do not count the missed match as zero.
Find fair per-match averages: 110÷5=22; 96÷4=24. Since 24>22, Yashasvi scored more per match on average.
A mean can represent a group with one number for comparison, but it hides variation between individual scores.
Shubman played 5 matches and averaged 110÷5=22 runs. Yashasvi played 4 (the dash is not a zero) and averaged 96÷4=24 runs. Vaishnavi’s total-only comparison is unfair; Yashasvi’s mean is higher. A mean gives one useful summary number, though it does not show every score or its spread.
Question 5.2-3
Add: 3+8+10+5+4=30.
Share 30 equally among 5 baskets: 30÷5=6.
30÷5=6 guavas per basket.
Question 5.2-4
Record a count for each of the seven days during the flowering season.
Add the seven counts to find the week’s total flowers.
Divide by 7. Label the answer in flowers per day and explain that it is an average, so it need not be a whole number.
Use your own seven daily counts. Add them and divide by 7; that is the average number of flowers per day. There is no single answer because the plant and week differ. For example, if your counts total 28, the mean is 28÷7=4 flowers per day.
Question 5.2-5
Add the eight counts: 6+2+9+5+4+6+3+5=40.
Divide by 8 trials: 40÷8=5.
40÷8=5 bounces.
Question 5.2-6
Add each runner’s seven times; each total is 122.
Divide each total by 7.
Both totals are 122 seconds, so both means are 122÷7 = 17 3/7 seconds (about 17.43 seconds). They tie by mean time.
Question 5.2-7
Add the six yearly counts: 11,154.
Divide by 6 years: 11,154÷6=1,859.
11,154÷6 = 1,859 students.
Question 5.2-8
Sort each town’s 12 prices.
For 12 values, average the 6th and 7th values.
Yahapur: (35+39)÷2; Wahapur: (38+39)÷2.
Yahapur median = (35+39)÷2 = ₹37. Wahapur median = (38+39)÷2 = ₹38.50.
Question 5.2-onion-costlier
Add each town’s 12 prices: Yahapur ₹458; Wahapur ₹450. Dividing by 12 gives approximate averages ₹38.17 and ₹37.50 per kg.
Compare months one by one: Yahapur is higher in six, Wahapur in five, and both are ₹35 in June.
Compare extremes: Wahapur ranges from ₹17 to ₹60; Yahapur ranges from ₹24 to ₹59. The overall yearly average favours Yahapur slightly.
Yahapur has the larger yearly total (₹458 vs ₹450) and higher prices in 6 months; Wahapur is higher in 5 months, and June is equal. So onions are slightly costlier in Yahapur overall. Wahapur has both the highest monthly price (₹60) and the lowest (₹17), so its prices vary more.
Question 5.2-onion-dotplot
Count the dots: 12 per town, matching the 12 monthly prices. The values are all represented.
A dot plot shows value frequency, not the month paired with each value; January cannot be identified from it alone.
Add and divide by 12: Yahapur 458÷12≈38.17; Wahapur 450÷12=37.50 rupees per kilogram.
It shows all 12 price values for each town, but sorting dots by price removes the month labels, so it cannot identify January’s value. Yahapur’s mean is ₹458÷12≈₹38.17/kg; Wahapur’s is ₹450÷12=₹37.50/kg.
Question 5.2-9
Ignore the two absent entries; they are missing data, not zero.
The total is 72 across 20 students, so mean=3.6.
In order, the 10th and 11th values are both 2, so the median is 2.
There are 20 present students. Mean = 72÷20=3.6; median = 2. Most counts are small, with 10 and 25 high values pulling the mean upward.
Question 5.2-10
Add the 29 heights to get 1,621.
Mean=1,621÷29≈55.90.
After sorting 29 values, the 15th value is 56.
Total 1,621 feet for 29 trees; mean ≈55.90 feet. Median=56 feet. The mean is close to the median, so the values are fairly balanced around the centre.
Question 5.2-11(a)
The minimum is 3.09 L and maximum is 20.5 L.
Neither the mean nor median can exceed the maximum.
No. Every value is at most 20.5 L. The mean lies between the smallest and largest values, and the median is one of the middle values.
Question 5.2-11(b)
The mean is an average of values, so it cannot fall outside their range.
The median is a middle value (or the average of two middle values), also within the range.
No. Both the mean and median lie between the minimum and maximum data values.
Question 5.2-12
Sort boys: 2.6,3.2,3.4,3.5,3.8,4.1; median=(3.4+3.5)÷2=3.45.
Sort girls: 2.5,3.1,3.4,3.4,3.7,4.0; median=(3.4+3.4)÷2=3.4.
Divide totals by six for the means.
Boys total 20.6 kg, mean ≈3.43 kg and median 3.45 kg. Girls total 20.1 kg, mean=3.35 kg and median 3.4 kg.
Question 5.2-13
Find each group’s mean.
Divide the sumo mean by the ballet mean.
Sumo mean = 1,201.9÷5=240.38 kg. Ballet mean = 254.5÷6≈42.42 kg. Ratio ≈240.38÷42.42≈5.7, or about 6 times.
Question 5.2-height-dotplots
Read the provided means and medians: boys are higher than girls in this section, but only by about 1.9 cm in mean.
Compare the plot ranges: girls 158−126=32 cm; boys 148−130=18 cm, so the girls’ values are more spread out.
Compare section summaries carefully: first section mean/median 144.4/145 cm; this section 141.21/142.5 cm. These are different groups, so do not claim individual growth or generalise to all students.
In this section boys are slightly taller on average (mean 142.05, median 143) than girls (mean 140.14, median 140). Girls’ heights have a wider spread (126–158 cm) than boys’ (130–148 cm). Compared with the first section, this section’s whole-class mean/median are lower (141.21/142.5 vs 144.4/145 cm). The two groups’ boys have similar means; the girls’ means differ more. This describes only these two sections.
Question 5.2-minute-estimates
Find each range: A 68−42=26 seconds; B 71−47=24 seconds.
Compare centres with the target 60 s: A mean 58.21 and median 60; B mean 59.28 and median 59.5.
Both are close to one minute. B is slightly closer by mean and has a slightly smaller range, but the difference is modest.
Both groups estimated a minute reasonably well: their means and medians are close to 60 seconds. Group B’s mean is closer to 60 (off by 0.72 s vs 1.79 s), and its range is slightly smaller (24 s vs 26 s), so B was a little more accurate and less spread out overall.
Question 5.2-story-outlier
Add the 15 counts: 122; mean=122÷15≈8.13. Sort them: 0,0,1,2,3,5,5,6,7,8,8,10,12,15,40; the 8th value is 6.
The value 40 sits far above the other counts, so it is an outlier.
Remove 40: total 82 across 14 values gives mean≈5.86. The middle pair is 5 and 6, so median=5.5. The mean falls by about 2.27; the median changes by only 0.5.
With all 15 values, mean=122÷15≈8.13 and median=6; the mean is greater, mainly because 40 is an outlier. Without 40, mean=82÷14≈5.86 and median=(5+6)÷2=5.5. The mean changes more than the median.
Question 5.2-newspaper-pages
Add the seven daily counts: 16+18+20+22+26+16+10=128; divide by 7 for mean≈18.29.
Sort: 10,16,16,18,20,22,26. With 7 values, the 4th is the median: 18.
Stack two dots above 16 and one above each of 10,18,20,22,26. The mean marker is just to the right of 18; the median marker is exactly at 18.
The mean is 128÷7=18 2/7≈18.29 pages. Ordered data are 10,16,16,18,20,22,26, so the median is 18 pages. On the dot plot, place two dots at 16 and one at each other listed count; mark the mean near 18.29 and median at 18.
Question 5.2-outliers-both-sides
With 2,20,22,24,26,28,46, the total is 168; mean=168÷7=24 and median=24. The low and high extremes balance.
Change 46 to 100: total=222, so mean≈31.71. The middle value remains 24.
Thus, the mean moves toward the more extreme outlier; the median depends on the middle position and changes little.
The low outlier pulls the mean down and the high outlier pulls it up; their effects can partly cancel. For 2,20,22,24,26,28,46, mean=median=24. If 46 is replaced by 100, mean=222÷7≈31.71 while the median stays 24. The median is much less affected.
Question 5.2-grade5-height-counts
Combine totals: 2,430+1,616=4,046 cm for 17+11=28 students. Mean=4,046÷28=144.5 cm.
Count each listed height greater than 144.5 cm (strictly taller); there are 15.
Among the boys, the heights above 144.5 are 147,154,158,155,146,146,145 and 150, so 8 boys.
There are 28 students and their total height is 4,046 cm, so class mean=4,046÷28=144.5 cm. Counting heights above 144.5 cm gives 15 students in total, including 8 boys.
Question 5.3-1
Compare each month’s pair of bars with the two prices in the table; every pair represents the same values.
Read the vertical scale: ₹10 between labelled major marks, with one intermediate division worth ₹5.
For each month compare the two bar heights. January favours Yahapur, October favours Wahapur and June is equal.
The bars match the table. The vertical axis measures rupees per kilogram: each major division is ₹10 and each smaller division is ₹5. Yes, side-by-side bars make it easier to compare towns: Yahapur is costlier in 6 months, Wahapur in 5 months, and June is equal.
Question 5.3-2
Check the bar for each organisation across all three years. A downward or unchanged bar disproves the claim that every one rose each year.
For totals and means, read the approximate counts: France 24; CASC 128÷3≈42.7; ISRO 14 and Galactic Energy 10; Russia 55.
Add the 2023 bars, including “Other”: 96+45+18+3+9+3+7+7+6+26≈220. This lies between 200 and 400.
(a) No: several organisations declined. (b) Yes: SpaceX is the only one shown above 50 in a year. (c) Yes: Arianespace’s values total about 15+6+3=24. (d) Yes: CASC averaged (48+35+45)÷3≈42.7. (e) Yes: ISRO 14 versus Galactic Energy 10. (f) No: Roscosmos totals about 16+21+18=55. SpaceX, ISRO and Galactic Energy increased every year. The 2023 total is about 220, so choose 200–400.
Question 5.3-3
Convert monthly totals to average daily hours by dividing by the days in the month.
Compare the month-by-month pattern.
The two cities are in opposite hemispheres, so seasons are reversed.
The cities are Helsinki in the Northern Hemisphere and Wellington in the Southern Hemisphere. Opposite seasons make daylight increase toward June in Helsinki but decrease toward June in Wellington.
Question 5.3-4
The legend identifies which bars belong to Blue and Red, but the graph gives no innings order or match result.
At over 12, the blue bar reaches 15 on the runs scale.
The shortest red bar is at over 4, about 2 runs. A target total is not shown directly by the over-by-over bars.
The graph alone does not tell which team batted first or which team won; it only compares over-by-over runs. Blue scored 15 runs in over 12. Red scored the least in over 4 (2 runs). A target total is not directly marked; the first innings’ over scores would need to be added.
Question 5.3-5
Count each code: w=water, a=air, n=none, s=space.
Use the same category order for both grades in the double-bar graph.
Grade 5: water 6, air 13, none 4, space 2. Grade 9: water 6, air 8, none 2, space 9.
Question 5.3-6
Label the eight times along the horizontal axis.
Use a vertical scale of 4°C per unit and two bars per time.
Compare the daily temperature patterns; treat month names as estimates.
Draw paired bars for each time using a 4°C scale. Day 1 is much cooler; it could be a winter day, while Day 2 could be a hot summer day. Exact months cannot be determined from these data alone.
Question 5.3-7
For the missing Gujarat bars, mark 69k, 89k and 78k in 2022–24; for Delhi mark 62k, 74k and 81k.
Compare each state’s first and last bars: all shown totals increase from 2022 to 2024, though the year-to-year pattern differs.
Assam’s change is about 60,000−40,000=20,000. West Bengal’s factor is about 43,000÷11,000≈3.9, or about four times. For Uttarakhand distinguish a modest increase from a small total number of registrations.
Plot the six given bars using the graph’s 25,000-registration major intervals (small lines are about 5,000). Every displayed state has more registrations in 2024 than in 2022. Gujarat rises then falls; Delhi rises each year. Assam adds about 20,000 from 2022 to 2023. West Bengal grows to about 4 times its 2022 count. The claim that Uttarakhand had “very few” new registrations because the bars grew little is misleading: its increase is small compared with the graph scale, but the 2023 and 2024 totals are still about 16,000 and 19,000.
Question 5.3-8
The axis goes 0,16,32,…, so one large division represents 16 km/h.
Compare ratios: 322÷170≈1.9 and 109÷56≈1.95; both are close to 2.
For sailfish and humpback whale, 109÷26≈4.2. A selected list cannot prove a worldwide superlative.
Yes. Each bar length represents speed, and the horizontal scale increases by 16 km/h per major division. The peregrine falcon (322 km/h) is about twice the spine-tailed swift (170 km/h); the sailfish (109 km/h) is about twice the flying fish (56 km/h) and about four times the humpback whale (26 km/h). The graph cannot show that the sailfish is the fastest aquatic animal in the world because it includes only selected animals.
Question 5.3-school-height-plots
Read the labelled means: School A boys/girls are 134.8/137.78, 141.8/141.83 and 149.35/147.81 cm; School B values are 149.84/150.2, 156.14/155.41 and 156.14/156.83 cm.
Compare grade by grade: both School B means exceed the matching School A means, and means generally increase across Grades 6–8.
Avoid claiming that the school caused the difference or that each individual grew by these amounts; the plots compare different student groups.
School B’s means are higher than School A’s for boys and girls in every grade. Mean heights generally rise from Grade 6 to Grade 8 at both schools. Within one grade, the boys–girls difference is small compared with the gap between schools. The dot plots show association, not why the school averages differ; each grade is a different group, not the same children followed over time.
Question 5.3-tall-tales
Compare the same age in 1989 and 2019: every 2019 mean is higher. For example, 5-year-old girls rise 100→107.2 cm (+7.2), while 19-year-old boys rise 163.5→166.5 (+3.0). In 2019, girls average taller at ages 10–12; boys average taller from 13 onward.
Check each claim against exactly what is measured: (1) true; (2) false because 143.2 cm<148 cm; (3) true because 159 cm>158.9 cm; (4) averages cannot prove every individual girl is taller; (5) false since girls’ mean is higher at ages 10–12; (6) the table ends at 19, so it says nothing about later growth.
Compare consecutive 2019 rows: boys’ largest increase is 148.4−142.2=6.2 cm from 12 to 13; girls’ is 138.6−132.8=5.8 cm from 10 to 11.
Both averages increase from 1989 to 2019 at every listed age; gains are larger at younger ages, and girls gain slightly more over the full period. In 2019, girls’ mean exceeds boys’ at ages 10–12; boys are higher from age 13 onward. Claims: (1) supported; (2) false, 143.2<148; (3) supported, 159>158.9; (4) cannot be inferred from averages; (5) false; (6) cannot be inferred beyond age 19. Largest 2019 annual gain: boys 12→13, 6.2 cm; girls 10→11, 5.8 cm.
Question 5.3-tall-tales-estimates
For ages 1–4, spread the approximate 57 cm gain from birth (50 cm) to age 5 (107 cm) evenly over five years: 57÷5=11.4 cm/year. This gives estimates 61.4, 72.8, 84.2 and 95.6 cm.
For each sex and age, calculate the last decade’s change: 2019 value−2009 value. Add that change once more to the 2019 value to estimate 2029.
This continuation estimates ages 5–19 only. Clearly label it a “what if the last decade repeats” estimate; actual future growth may differ.
One simple birth-to-age-5 estimate uses 50 cm at birth and 107 cm at age 5: the 57 cm increase averages 11.4 cm/year, giving about 61, 73, 84 and 96 cm at ages 1–4. These are rough estimates. For 2029, add each 2009–2019 change to the corresponding 2019 value. Example: 5-year-old boys: 107.1+(107.1−105.1)=109.1 cm; 5-year-old girls: 107.2+(107.2−104)=110.4 cm. Repeat this method for each age; it assumes the last decade’s change continues and is not a prediction guaranteed to occur.
Question 5.3-tall-tales-2029-table
For each age and sex, subtract the 2009 mean from the 2019 mean to get the decade change.
Add that change to the 2019 mean. For example, boys age 12: 142.2−136.2=6.0; 142.2+6.0=148.2 cm. (Use the table’s full row data for every age.)
Repeat row by row. The results describe only the assumption that the last decade’s change repeats; real growth can differ.
Using 2029 = 2019 + (2019−2009), estimated boys’/girls’ heights in cm for ages 5–19 are: 5 109.1/110.4; 6 115.2/116.1; 7 121.0/121.2; 8 126.1/125.8; 9 131.0/130.7; 10 135.8/135.7; 11 140.3/141.5; 12 145.5/146.5; 13 151.6/150.3; 14 157.3/152.8; 15 161.7/154.7; 16 164.7/156.0; 17 166.6/156.8; 18 167.7/157.4; 19 167.9/157.4. These are arithmetic extrapolations, not observed heights or reliable forecasts.
Question 5.4-1
Range = greatest − least: boys 6−3=3; girls 6−0=6.
Boys have 12 values, so median=(6th+7th)/2=(5+5)/2=5. Girls have 13 values, so the 7th value is 4. The boys’ median is greater.
Boys’ mean=56÷12≈4.67; girls’ mean=45÷13≈3.46. Both groups have maximum 6.
(a) False: ranges are 3 for boys and 6 for girls. (b) True: medians are 5 and 4. (c) False: boys’ mean is 56÷12≈4.67 and girls’ mean is 45÷13≈3.46. (d) False: both maxima are 6. Only (b) is true.
Question 5.4-2(a)
Add: 14+16+10+10=50.
Divide by 4 games.
50÷4=12.5 points per game.
Question 5.4-2(b)
C’s total is 8+11+13=32.
There are three played games, so mean=32÷3≈10.67.
Divide by 3, because C played three games. A game not played is missing, not a zero score.
Question 5.4-2(c)
Calculate each player’s mean over games actually played.
Compare 12.5, 4.5 and 10.67.
By average points per game: A=12.5, B=4.5, C≈10.67. Player A had the highest average.
Question 5.4-3
Group 1 total 730÷10=73; its middle pair is 76 and 81.
Group 2 total 681÷9≈75.67; its middle value is 79.
Group 1 mean=73; median=78.5. Group 2 mean≈75.67; median=79. Group 2 is slightly higher by both measures; group 1 has a low outlier (39) that lowers its mean.
Question 5.4-4
Add each row for the totals.
Use a common scale and paired bars for the two activities.
Compare categories using the supplied counts.
Watching total=2,900; participating total=1,615. Cricket leads in both (1,240 watching; 620 participating); athletics is lowest in both (250; 105). Draw paired bars for each sport.
Question 5.4-5
Put the heights in order: 101,102,106,109,110,110,112,115,115,115,115,115,117,120,120,123,125.
The 9th value, the median, is 115 cm. There are 7 values below it, 5 equal to it and 5 above it.
Because 17 is odd, it cannot be split into two same-sized groups while including everyone. Tied heights also prevent a strict below/above split at this cutoff.
No. The list has an odd total of 17 students, so two groups cannot have the same size if everyone must be included. At the median height 115 cm, 7 are shorter, 5 are exactly 115 cm and 5 are taller. The equal-group instruction needs adjustment (for example, leave one student out or allow ties to be split by another rule).
Question 5.4-6
Make a table with every student’s height in centimetres; include each student once.
Mean = total of all heights ÷ number of students. Sort the heights to find the middle value (or average the two middle values).
On a number line, place one dot for every student above that height. Explain whether values cluster near the centre or spread out.
This depends on your class data. Record one height per student, order the heights, and calculate both measures. For example, with 130,132,135,135,138,140,142 cm, the mean is 136 cm and the median is 135 cm. Draw one dot above each student’s height; stack dots when heights repeat.
Question 5.4-7
A mean describes only the measured group.
No heights or total height for the other section are given.
Therefore none of “equal”, “lower” or “higher” must be true.
It cannot be determined from the first section’s mean. The other section could have a lower, equal or higher mean.
Question 5.4-8
Estimate each bar by comparing it with labelled values: New York is between 251 and 367 (about 300); Tokyo is near 165; London is below Moscow’s 46 (about 30–40).
Count bars higher than Mumbai’s 86: there are 12. Count bars lower: there are 7.
The vertical measure is a count of skyscrapers, not the height of the tallest building. A high count cannot prove that city has the tallest single building.
Approximate readings are New York 300, Tokyo 165 and London about 30–40 skyscrapers. Both count statements are true for the cities shown: 12 bars are above Mumbai’s 86 and 7 are below it. The tallest-building claim does not follow: the graph counts buildings taller than 150 m in each city; it does not compare the heights of individual buildings.
Question 5.4-9
Before using a ruler, write your estimate for each length; then measure to the nearest centimetre or millimetre.
Subtract measured length from estimate and take the positive value so the error cannot be negative.
Add all five errors and divide by 5. A smaller average error means your estimates were closer overall.
This depends on your own measurements. For each object, positive error = |estimate − measured length|. The average error is the sum of the five positive errors divided by 5. Example: if a pen estimate is 15 cm and its measured length is 14 cm, its error is |15−14|=1 cm.
Question 5.4-10
Week 1 total=3,240; divide by 9 to get 360 s. Its ordered middle (5th) value is 360 s.
Week 2 total=2,230; divide by 8 to get 278.75 s. Its 4th and 5th ordered values are 270 and 280, so median=275 s.
Lower times mean faster solving. Compare both mean and median; the high 380 s value pulls up Week 2’s mean.
Week 1: mean 3,240÷9=360 s and median 360 s. Week 2: mean 2,230÷8=278.75 s and median (270+280)÷2=275 s. Both measures are lower in Week 2, so Aditi was generally faster. Week 2 also has a wide range, 380−220=160 s, including one unusually slow 380 s attempt.
Question 5.4-11
Choose the two text pages or class list, state your sample, and record every observation in a table.
Make dot plots for numerical counts. For names, tally starting letters and group names by whether they begin/end with vowels or consonants.
Calculate mean and median for numerical data; use the most frequent category for the most popular letter/pattern. Describe spread and any limits in your sample.
These are personal investigations, so answers depend on the books and classmates you choose. For sentences, count words in several consecutive sentences from two subjects. For names, record each classmate’s name length and starting letter, then count vowel/consonant patterns. Use your collected values to calculate means, medians and frequencies, then describe the results.
Question 5.4-12
Keep a daily record and do not skip days; write 0 on a day you do not leave home.
Mean = total trips ÷ 30; sort the 30 daily counts and average the 15th and 16th values for the median.
Range = largest count − smallest count. Place one dot for each day at its trip count and explain any pattern you notice.
Record one daily count for all 30 days. Then find the mean and median, calculate the range, make a dot plot and describe any pattern such as more trips on school days or weekends. Use your own observations rather than a made-up universal answer.
Question 5.4-13
Agree on the measurement method and record units. For the time activity, do not count while estimating; repeat for both one and three minutes.
Calculate each family’s mean height or each person’s mean estimate when repeat trials are used; use medians when helpful.
Make the requested plots, compare ranges and centres, and state that conclusions apply to your group’s data only.
Choose one project and gather each group member’s observations consistently. For heights, compare each student with their family’s mean; for time, record each person’s estimate of one and three minutes. Use dot plots and a double-bar graph where appropriate, then discuss centre and spread without treating a small group as representative of everyone.
Textbook pages 94–134
Question 5.1-height-guess
A group-level pattern may say that one height is more common for one group.
People’s heights overlap, so an individual can differ from the pattern.
No. Height alone cannot reliably tell us either person’s gender; the guess may be wrong.
Question 5.2-second-guava-group
First group: 3+8+10+5+4=30; 30÷5=6 guavas each.
Second group: 5+4+6+3+4+8=30; 30÷6=5 guavas each.
The first group gets more: 6 guavas per person compared with 5 in the second group.
Question 5.2-hibiscus-example
Add the five counts: 2+7+9+4+3=25 flowers.
Divide by the five days: 25÷5=5.
The mean was 5 flowers per day. It is the equal daily count that would give the same total of 25 flowers.
Question 5.2-bounce-your-own-data
Write down the bounce count for every attempt, including low and high results.
Add all counts and divide by the number of attempts.
For example, counts 4, 6, 3, 5, 7, 5, 6 total 36; 36÷7=5 1/7.
Answers depend on your own trials. In the example, the mean is 5 1/7 bounces per attempt.
Question 5.2-family-heights-outlier
Yaangba total=986 cm, so mean=986÷6≈164.33 cm. Sorted values are 155,160,164,165,169,173; median=(164+165)÷2=164.5 cm.
Poovizhi total=801 cm, so mean=801÷5=160.2 cm. Sorted values are 118,165,170,173,175; median=170 cm.
Without 118 cm, Poovizhi’s mean=683÷4=170.75 cm and median=(170+173)÷2=171.5 cm.
With all values, Poovizhi’s mean is lower than Yaangba’s even though most of her family heights are greater. The unusually low 118 cm value pulls the mean down. Without it, her mean rises to 170.75 cm; her median changes only from 170 to 171.5 cm.
Question 5.2-compare-three-outlier-examples
With the family heights 169,173,155,165,160,164, the mean is about 164.33 and the median is 164.5: they are close.
For the story counts 6,8,5,15,40,0,8,3,0,2,7,12,10,5,1, the high value 40 pulls the mean (about 8.13) above the median (6).
For newspaper pages 16,18,20,22,26,16,10, the mean is about 18.29 and median is 18; the low value 10 shifts the mean slightly down.
When values are fairly balanced, the mean and median can be close. A high or low outlier pulls the mean toward itself more than it moves the median.
Question 5.4-magic-trick
Call the chosen number n. After multiplying by 2 and adding 10, the result is 2n+10.
Divide by 2: (2n+10)÷2=n+5.
Subtract n: n+5−n=5. Then add 3.
The trick always gives 8, whatever number you start with.