Data Handling — Questions and Answers
Welcome to HSLC Guru. This page gives the complete solutions to ASSEB Class 7 New Mathematics Chapter 13, Data Handling — statistical questions and statements, average (arithmetic mean), median, outliers, range, dot plots, double bar graphs and clustered column graphs — with every “Work it Out” box, all solved examples, the Project and the full Exercise 13 answered.
Summary
The systematic collection, organisation, analysis, interpretation and representation of data is the subject of Statistics. A question that can be answered only by collecting and analysing data, and whose answer varies, is a statistical question; a claim or summary based on collected data and expressed in numbers is a statistical statement.
The average (arithmetic mean) is the sum of all observations divided by the number of observations; it gives the idea of a fair or representative share. When the values are arranged in ascending or descending order, the middle value is the median. A value lying far away from the rest is an outlier; it affects the mean but not the median much.
A dot plot shows each value as a dot on a line, making the maximum, minimum, range, density and variability easy to read. Two related data sets shown side by side make a double bar graph or clustered column graph, which makes comparison quick. Exploring the further questions that data raises makes us a data detective.
Summary: This ASSEB Class 7 New Mathematics Chapter 13 (Data Handling) guide explains statistical questions and statements, arithmetic mean (average), median, outliers, range, dot plots, double bar graphs and clustered column graphs. It reproduces every “Work it Out” box (13.1–13.6), all solved examples, the Project and the full Exercise 13 with worked answers, plus inline SVG bar graphs, clustered column graphs and a dot plot drawn from the textbook data.
Textbook Questions and Answers
Work it Out 13.1
1. Which of the following is a statistical question? (a) What are the ages of the students in your class? (b) How many pieces of bread did Neshim eat this morning? (c) Which are the most and least populated states of India? (d) What is the height of Naveen?
Answer: A statistical question needs data to be collected and analysed, and has varying answers. (a) needs many students’ ages, answer varies — statistical question. (b) has one fixed answer — not statistical. (c) needs population data to compare — statistical question. (d) one fixed height — not statistical. So (a) and (c) are statistical questions.
2. Form five statistical statements about your school campus.
Answer: Only data-based statements expressed in numbers are statistical. For example —
- There are more girls than boys in our school.
- Most students in our class come to school by cycle.
- There are more than 10 species of trees in the school campus.
- Our library has more books in 2025 than in 2024.
- The favourite colour of most students in our class is blue.
Average or Arithmetic Mean — Solved Examples
The runs scored by two players in six matches are given below.
| Runs Scored | 1st | 2nd | 3rd | 4th | 5th | 6th | Total |
|---|---|---|---|---|---|---|---|
| Pratika Rawal | 37 | 31 | 37 | 75 | 6 | 122 | 308 |
| Smriti Mandhana | 8 | 23 | 23 | 80 | 88 | 109 | 331 |
Both played the same number of matches, so comparing totals shows Smriti Mandhana did better (331 > 308). But when the number of matches differs, totals are not a fair comparison — we use the average.
| Runs Scored | 1st | 2nd | 3rd | 4th | 5th | Total | Average |
|---|---|---|---|---|---|---|---|
| Harleen Deol | 48 | 46 | 13 | 38 | — | 145 | 36.25 |
| Harmanpreet Kaur | 21 | 19 | 9 | 22 | 89 | 160 | 32 |
Harleen Deol’s average = $\frac{48+46+13+38}{4} = \frac{145}{4} = 36.25$ runs; Harmanpreet Kaur’s = $\frac{21+19+9+22+89}{5} = \frac{160}{5} = 32$ runs. Though her total is smaller, Harleen Deol’s average is higher, so by average her performance is better.
Average = Sum of all observations ÷ Total number of observations. Also, Sum of all observations = Average × Total number of observations.
Example 1: The weights (kg) of five Class 7 students are 38, 37, 40, 42, 43. Find their average weight.
Answer: Total weight = 38 + 37 + 40 + 42 + 43 = 200 kg. Average = $\frac{200}{5}$ = 40 kg. This 40 kg is the centre of the data where the total weight balances on both sides.
Example 2: The average age of 10 students is 13.9 years. After a new student joins, the average becomes 14 years. Find the new student’s age.
Answer: Total age of 10 students = 13.9 × 10 = 139 years. Total age of 11 students = 14 × 11 = 154 years. New student’s age = 154 − 139 = 15 years.
Example 3: A player’s average in four matches is 50. He scored 45, 60 and 55 in the first three. How many runs in the fourth match?
Answer: Total in four matches = 50 × 4 = 200. Total in three matches = 45 + 60 + 55 = 160. Runs in the fourth match = 200 − 160 = 40 runs.
Average as fair share: Tripti and her five friends collected 4, 2, 5, 3, 6, 4 chocolates; Mrinmay and his four friends collected 3, 11, 9, 6, 1. Each group shares equally — which group gets more per person?
Find the share per member of each group.
Answer: Tripti’s group total = 4 + 2 + 5 + 3 + 6 + 4 = 24; members 6; per member = $\frac{24}{6}$ = 4. Mrinmay’s group total = 3 + 11 + 9 + 6 + 1 = 30; members 5; per member = $\frac{30}{5}$ = 6. So each member of Mrinmay’s group gets more (6). Thus the average expresses the idea of a fair share.
Example 4: Rubul read 8, 10, 15, 5, 12, 18, 9 pages on the days of a week. If he read the same number every day, how many per day?
Answer: Total pages = 8 + 10 + 15 + 5 + 12 + 18 + 9 = 77; days = 7; per day = $\frac{77}{7}$ = 11 pages.
Everyday use of the mean: the average is a representative value used in statistics, science, economics, sports and more — for example, the average weight of Class 7 students is about 40 kg, Ramen walks about 6000 steps a day, and Anup’s scooter gives about 46 km per litre on average.
Ancient Indian mathematicians used beautiful words for the idea of average —
| Term | Meaning | Used by |
|---|---|---|
| Samarajju | Average size of a line segment | Brahmagupta (628 CE) |
| Samikarana | Equalisation, levelling | Mahaviracharya (850 CE) |
| Samya | Equality, impartiality | Sripati (1039 CE) |
| Samamitti | Mean measure | Bhaskaracharya (1150) & Ganesha (1545) |
Work it Out 13.2
1. Eliza recorded her skips in seven attempts — 25, 30, 18, 22, 27, 35, 25. Find the average number of jumps per attempt.
Answer: Total jumps = 25 + 30 + 18 + 22 + 27 + 35 + 25 = 182; attempts = 7. Average = $\frac{182}{7}$ = 26 jumps.
2. Play the same game yourself for 7 or more attempts and find the average.
Answer: This is an activity. Record your jumps each attempt, add them all and divide by the number of attempts (Average = total jumps ÷ number of attempts).
3. The monthly maximum and minimum temperatures of Tezpur in 2023 are given. Find the average maximum and average minimum temperature.
| Temp (°C) | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Maximum | 25 | 26 | 29 | 31 | 32 | 33 | 33 | 34 | 35 | 32 | 30 | 27 |
| Minimum | 12 | 16 | 18 | 21 | 23 | 25 | 26 | 25 | 26 | 22 | 17 | 15 |
Average maximum and minimum temperature:
Answer: Total maximum = 367; average = $\frac{367}{12}$ ≈ 30.6 °C. Total minimum = 246; average = $\frac{246}{12}$ = 20.5 °C.
Comparison of Data using Dot Plot
The glasses of sugarcane juice sold by two shops at Barua Tiniali in the first ten days of July 2025 are given in Table 3.
| Date | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| First shop | 157 | 140 | 146 | 92 | 95 | 120 | 134 | 122 | 122 | 112 |
| Second shop | 135 | 114 | 131 | 120 | 145 | 110 | 115 | 121 | 135 | 104 |
First shop total = 1240 glasses, second shop = 1230 glasses. The difference between the maximum and minimum value is the range — first shop 157 − 92 = 65, second shop 145 − 104 = 41. Plotting each value as a dot on a horizontal line gives a dot plot. Larger gaps between dots mean higher variability and lower density; smaller gaps mean higher density and more consistent data. Thus the second shop’s sales are more consistent than the first’s.
Work it Out 13.3
1. The heights (cm) of 9 Class 7 students are 120, 150, 128, 125, 132, 122, 135, 132, 128. Represent them by a dot plot and find the arithmetic mean.
Answer: Total height = 120 + 150 + 128 + 125 + 132 + 122 + 135 + 132 + 128 = 1172 cm; students 9. Mean = $\frac{1172}{9}$ ≈ 130.2 cm. On a line from 120 to 150 put one dot above each height; 128 and 132 get two dots each (they occur twice).
Median of Data — Solved Examples
Book prices at the Assam Book Fair — Geet: 280, 300, 230, 200, 70 (rupees); Kriti: 230, 250, 150, 270, 200, 220. Geet’s average = $\frac{1080}{5}$ = 216, Kriti’s = $\frac{1320}{6}$ = 220.
Although most of Geet’s books are costlier, one book costing only Rs 70 pulls his average down, so the average does not represent all values well. Arranging the values in ascending order and taking the middle value gives the median.
Find the median price for Geet and Kriti.
Answer: Geet ascending: 70, 200, 230, 280, 300 — 5 values (odd), so the middle (3rd) value is the median = Rs 230. Kriti ascending: 150, 200, 220, 230, 250, 270 — 6 values (even), so the median is the average of the middle two = $\frac{220+230}{2}$ = Rs 225.
For Geet the gap between mean (216) and median (230) is 14, but for Kriti it is only 5. Geet’s Rs 70 book lies far from the rest — such a value is an outlier. Outliers affect the mean but not the median, so the median is more suitable for data with large deviations.
Example 4: Each student wrote a whole number below 50 — 19, 10, 0, 3, 7, 2, 0, 19, 48, 9, 6, 12, 8. Find the mean and median and mark them on a dot plot.
Answer: Sum = 143; count 13. Mean = $\frac{143}{13}$ = 11. Descending: 48, 19, 19, 12, 10, 9, 8, 7, 6, 3, 2, 0, 0 — the middle (7th) value is 8, so median = 8. Since the outlier 48 is much larger, the mean (11) is greater than the median (8).
Try out: find the mean and median excluding the outlier 48.
Answer: Removing 48, the remaining 12 numbers sum to 95; mean = $\frac{95}{12}$ ≈ 7.92. The middle two (6th and 7th) values are 8 and 7, so median = $\frac{8+7}{2}$ = 7.5. Removing the outlier drops the mean sharply (11 → 7.92) but the median barely changes (8 → 7.5).
Example 5: Ages of 40 picnic-goers — Males: 37, 30, 42, 8, 30, 9, 3, 45, 30, 42, 4, 13, 12, 23, 29, 35, 39, 38, 22, 4, 35, 42 and Females: 34, 38, 33, 26, 29, 22, 44, 4, 19, 21, 16, 6, 31, 34, 41, 24, 11, 35. Find the mean and median.
Answer: Males (22): total = 572; mean = $\frac{572}{22}$ = 26. The middle two (11th and 12th) values are 30 and 30, so median = 30. Females (18): total = 468; mean = $\frac{468}{18}$ = 26. The middle two (9th and 10th) values are 26 and 29, so median = $\frac{26+29}{2}$ = 27.5.
Observation (role of zero): Runs 22, 0, 17, 118, 0, 32, 0, 9, 0, 0, 0 with 5 extras (total 203). Players’ mean = $\frac{198}{11}$ = 18 (extras not counted), but median = 0. Again, runs 49, 20, 16, (did not play), 0, 80 — dropping the unplayed match leaves 5 matches, total 165; mean = $\frac{165}{5}$ = 33, and the median of (0, 16, 20, 49, 80) = 20. For Bivan’s jackfruit tree Jan–Dec — 0, 0, 0, 30, 44, 26, 20, 5, 0, 0, 0, 0 — ignoring the zero months leaves 5 months; mean = $\frac{125}{5}$ = 25, and the median of (5, 20, 26, 30, 44) = 26.
Work it Out 13.4
1. Goals in 10 football matches — 2, 3, 4, 5, 0, 1, 3, 3, 4, 3. Find the median.
Answer: Ascending: 0, 1, 2, 3, 3, 3, 3, 4, 4, 5 — 10 values (even); the middle two (5th and 6th) are 3 and 3, so median = $\frac{3+3}{2}$ = 3.
2. The incomes (rupees) of 5 members of Maria’s family — 10,000, 7,000, 0 (does not earn), 4,000, 50,000. Find the mean and median. Which is the outlier?
Answer: Total income = 71,000; members 5. Mean = $\frac{71000}{5}$ = Rs 14,200. Ascending: 0, 4,000, 7,000, 10,000, 50,000 — the middle (3rd) value is the median = Rs 7,000. The value far above the rest, 50,000, is the outlier.
3. Spelling mistakes — 2, 7, 3, 6, 2, 6, 2, 3, 3, 4, 1, 5, 2, 6, 7, 2, 1, 2. Find the mean and median.
Answer: Total mistakes = 64; count 18. Mean = $\frac{64}{18}$ ≈ 3.56. Ascending: 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 5, 6, 6, 6, 7, 7 — the middle two (9th and 10th) values are 3 and 3, so median = 3. On a dot plot from 1 to 7 put dots by repetition and mark the mean (3.56) and median (3).
Visual Representation of Data — Double Bar Graph
The sugarcane-juice sales of Table 3 can be drawn as bars. Showing both shops’ daily sales together makes it easy to see which shop sold more on each day — such a graph is a double bar graph (scale: 1 unit = 10 glasses).
Similarly, the average maximum temperature of City A and City B for each month (Table 5) is shown as a double bar graph (scale: 1 unit = 5 °C). City A rises from January to a maximum of 41.2 °C in June and falls to a minimum of 22.7 °C in December; City B is the opposite. This suggests City A is in the Northern Hemisphere and City B in the Southern Hemisphere, where the seasons are reversed.
Work it Out 13.5
The clustered column graph below shows the number of foreign tourists at some popular destinations of Assam in 2021-22, 2022-23 and 2023-24 (scale: 1 unit = 2000 tourists).
1. Verify the following statements from the graph.
Answer: (a) Kaziranga National Park received the most foreign tourists in all three years — True (654, 9183, 13904, the highest). (b) Tourists to Majuli decreased over three years — False (0 → 305 → 600, increased). (c) Historical places have fewer foreign tourists than natural places — True (Kaziranga and Manas are far higher). (d) Foreign tourists to the historical places of Sivsagar have increased — True (634 → 1197 → 1540). (e) Kamakhya Temple in 2021-22 was very low compared with other years — True (36 vs 315, 701). (f) Srimanta Sankardev Kalakshetra had consistently more tourists than other destinations — False (11, 216, 231 — among the lowest).
2. Arrange the destinations in descending order by 2023-24 tourists. Which cultural site drew the most foreign tourists?
Answer: Descending: Kaziranga (13904) > Historical places of Sivsagar (1540) > Manas (1453) > Kamakhya Temple (701) > Majuli (600) > Srimanta Sankardev Kalakshetra (231). Among cultural sites, the Historical places of Sivsagar (1540) attracted the most foreign tourists.
3. Which was the least visited destination in 2021-22? (A) Majuli (B) Manas National Park (C) Kamakhya Temple (D) Historical places of Sivsagar
Answer: (A) Majuli (0 tourists in 2021-22).
4. Which place had the highest increase in 2023-24 over previous years? (A) Srimanta Sankaradev Kalakshetra (B) Kaziranga National Park (C) Historical places of Sivasagar (D) Kamakhya Temple
Answer: (B) Kaziranga National Park (9183 to 13904, the largest rise).
5. Which place had more visitors in 2022-23 than in 2021-22? (A) Majuli (B) Historical places of Sivasagar (C) Manas National Park (D) Srimanta Sankaradev Kalakshetra
Answer: In fact every listed place had more visitors in 2022-23 than in 2021-22; the most striking rise is (C) Manas National Park (24 → 808).
6. The 2022 tea production (Million Kg) of some districts of Assam is shown below.
(a) Write the estimated tea production of Sivasagar, Sonitpur and Cachar.
Answer: Reading the bar lengths — Sivasagar ≈ 118 M. Kg, Sonitpur ≈ 60 M. Kg, Cachar ≈ 20 M. Kg.
(b) State whether correct or incorrect — (i) Only two districts produce more tea than Golaghat. (ii) Udalguri produces more than Biswanath. (iii) Tinsukia is the largest producer among the listed districts. (iv) Nagaon’s production is less than half of Udalguri’s.
Answer: (i) Districts producing more than Golaghat (72) are three — Sivasagar (118), Dibrugarh (122), Tinsukia (138); so “only two” is incorrect. (ii) Udalguri (36) < Biswanath (38) — incorrect. (iii) Tinsukia (138) is the largest — correct. (iv) Half of Udalguri = 18; Nagaon (13) < 18 — correct.
Data Detective
Well-organised data raises new questions, and finding their answers reveals fresh facts — this process is Data Investigation. The three-year tea production (Million Kg) of States/Regions of India is given in Table 6.
| State/Region | 2020-21 | 2021-22 | 2022-23 |
|---|---|---|---|
| Assam | 626 | 672 | 698 |
| West Bengal | 396 | 409 | 418 |
| South India (Tamil Nadu, Kerala, Karnataka) | 232 | 231 | 225 |
Table 6 shows that tea production in Assam and West Bengal is rising steadily, while South India’s is falling from 2020-21 to 2022-23. Assam produces more than half of India’s total tea, which is why it is called the “Tea Garden of India”.
Work it Out 13.6
1. Draw the dot plot and clustered column graph of the tea production data of Table 6.
Answer: On a dot plot, put each value (Assam 626/672/698, West Bengal 396/409/418, South India 232/231/225) as a dot on a line from 200 to 700. Showing the three years side by side with three colours gives the clustered column graph below —
2. The male and female literacy rates of Assam (per 100) — Year 1951, 1971, 1991, 2011; Male 28, 44, 72, 78; Female 8, 24, 58, 66. Draw a double bar graph and answer the questions.
Answer: The double bar graph is shown below —
(a) What are the ranges of male and female literacy? Whose growth is more remarkable?
Answer: Male range = 78 − 28 = 50; Female range = 66 − 8 = 58. Female literacy rose from 8 to 66 (a gain of 58), which is comparatively faster, so female literacy growth is more remarkable.
(b) Find the mean and median of both data sets.
Answer: Male: mean = $\frac{28+44+72+78}{4}$ = $\frac{222}{4}$ = 55.5; median = $\frac{44+72}{2}$ = 58. Female: mean = $\frac{8+24+58+66}{4}$ = $\frac{156}{4}$ = 39; median = $\frac{24+58}{2}$ = 41.
(c) Why might female literacy have been low in 1951?
Answer: Social restrictions, lack of schools for girls, early marriage and gender bias meant girls had little access to education at that time.
(d) Who started the women’s literacy movement in Assam, and when?
Answer: Nineteenth-century social reformers and later women’s organisations — such as the Asom Mahila Samiti (founded 1926) — played an important role in spreading women’s education in Assam. (Consult your teacher and local history for exact details.)
(e) What steps would improve the literacy rate further?
Answer: More schools for girls, scholarships, adult-literacy programmes, awareness campaigns, free education and incentive schemes should be implemented.
3. Be a data detective for the Summer freshness (Table 3) and the tourist destinations (Fig. 13.6) you have studied.
Answer: This is open exploration — for example, does weather, market size or transport cost affect sugarcane-juice sales; did COVID-19 or transport facilities affect tourist numbers? Frame such questions, collect data and seek answers.
Project
Record the daily study time (hours) of you and a friend for one week; find each person’s average study time and range, and draw a double bar graph to compare.
Answer: This is a project. Add the seven days’ study time and divide by 7 for the average, and maximum − minimum gives the range. Example: if your times are 3, 4, 2, 5, 3, 4, 3 hours, average = $\frac{24}{7}$ ≈ 3.4 hours and range = 5 − 2 = 3 hours. Show your values and your friend’s side by side in two colours as a double bar graph.
Exercise 13
1. Choose the correct answer. (a) The average of the first six natural numbers — (A) 3 (B) 4 (C) 3.5 (D) 6
Answer: First six natural numbers are 1, 2, 3, 4, 5, 6; average = $\frac{21}{6}$ = (C) 3.5.
(b) The median of the first eight even natural numbers — (A) 8 (B) 9 (C) 10 (D) 12
Answer: The numbers are 2, 4, 6, 8, 10, 12, 14, 16; the middle two are 8 and 10; median = $\frac{8+10}{2}$ = (B) 9.
(c) The difference between the mean and median of 8, 3, 0, 5, 2, 5, 3, 5, 9, 0 — (A) 0 (B) 1 (C) 2 (D) 4
Answer: Mean = $\frac{40}{10}$ = 4. Ascending: 0, 0, 2, 3, 3, 5, 5, 5, 8, 9; median = $\frac{3+5}{2}$ = 4. Difference = 4 − 4 = (A) 0.
2. True or false.
Answer: (a) “Roses are more beautiful than Hibiscus” is a statistical statement — False (an opinion, not data-based). (b) The mean is always one of the values in the data set — False. (c) The mean and median can sometimes be equal — True. (d) “Were book sales at the Assam Book Fair less or more in 2025 than 2024?” is a statistical question — True.
3. Statement (I): the median of 25, 16, 26, 32, 19, 35, 28 is 31. Statement (II): to find the median, arrange the values in ascending or descending order. (A) both true (B) I true, II false (C) I false, II true (D) both false
Answer: Ascending: 16, 19, 25, 26, 28, 32, 35 — the middle (4th) value is 26, so the median is 26, not 31 — Statement (I) is false. Statement (II) is true. So (C).
4. Match Column A with Column B — (a) Average of 5, 7, 11, 6, 1; (b) Median of 3, 1, 5, 2, 4; (c) Difference of mean and median of 5, 0, 1, 4, 2, 3, 5, 0. B: 1. 0 2. 6 3. 3
Answer: (a) Average = $\frac{30}{5}$ = 6 → 2. (b) Ascending 1, 2, 3, 4, 5; median = 3 → 3. (c) Mean = $\frac{20}{8}$ = 2.5; ascending 0, 0, 1, 2, 3, 4, 5, 5, median = $\frac{2+3}{2}$ = 2.5; difference = 0 → 1. So (a)-2, (b)-3, (c)-1.
5. Sunil and Niyar’s five-day 200 m times (seconds) — Sunil: 42, 39, 41, 43, 40; Niyar: 44, 38, 39, 42, 37. By average, who runs faster? Show it on a dot plot.
Answer: Sunil’s average time = $\frac{205}{5}$ = 41 s; Niyar’s = $\frac{200}{5}$ = 40 s. Less time means faster, so Niyar runs faster. On a dot plot from 37 to 44, show each runner’s times with dots of a different colour.
6. Readers on five days — 12, 16, 20, 10, 22. Find the mean and median.
Answer: Mean = $\frac{80}{5}$ = 16. Ascending: 10, 12, 16, 20, 22; median = 16.
7. Rohit Sharma’s runs in 10 ODI innings — 121, 73, 8, 76, 28, 15, 20, 41, 1, 119. Find the average and median.
Answer: Total = 502; average = $\frac{502}{10}$ = 50.2. Ascending: 1, 8, 15, 20, 28, 41, 73, 76, 119, 121; median = $\frac{28+41}{2}$ = 34.5.
8. The average age of 15 students is 14 years. Adding the class teacher’s age makes the average 16 years. Find the teacher’s age.
Answer: Total age of 15 students = 14 × 15 = 210 years. Total of 16 people = 16 × 16 = 256 years. Teacher’s age = 256 − 210 = 46 years.
9. Favourite activities — Reading (R), Playing (P), Singing (S). Seventh: R, R, P, S, P, P, P, R, S, S, R, R, R, P, S, R, S; Eighth: S, S, R, R, R, R, P, P, R, P, S, R, R, P, R, P, R. Draw a bar graph and answer.
Answer: Counting — Seventh: R = 7, P = 5, S = 5; Eighth: R = 9, P = 5, S = 3.
Class R (Reading) P (Playing) S (Singing) Seventh 7 5 5 Eighth 9 5 3
10. The clustered column graph below shows the runs of Chinmay (C), Rupam (R) and Biraj (B) in six inter-school matches. Complete the table and answer.
Complete the table and answer — (a) Which player has the highest average? (b) Who scored the highest in each match?
Answer: Completed from the graph —
Match 1 2 3 4 5 6 Average Chinmay (C) 20 16 50 25 14 50 29.2 Rupam (R) 28 23 13 60 5 4 22.2 Biraj (B) 60 14 33 9 16 45 29.5
11. Goals by three players in four matches — Bipul: 1, 2, 3, 2; Anthony: 0, 2, 1, not played; Abdul: 2, 3, 2, 2. (a) Anthony’s average goals per match? (b) By average, who performed best?
Answer: (a) Anthony played only three matches; total = 0 + 2 + 1 = 3; average = $\frac{3}{3}$ = 1. (b) Bipul’s average = $\frac{8}{4}$ = 2; Anthony’s = 1; Abdul’s = $\frac{9}{4}$ = 2.25. So Abdul performed best.
12. Science marks of 12 students — 82, 91, 74, 85, 59, 93, 67, 85, 78, 90, 62, 80; Mathematics — 72, 65, 84, 91, 76, 55, 88, 69, 95, 73, 81, 77. Find the mean and median of each and compare.
Answer: Science: total = 946; mean = $\frac{946}{12}$ ≈ 78.8; the middle two (ascending) are 80 and 82, median = 81. Mathematics: total = 926; mean = $\frac{926}{12}$ ≈ 77.2; the middle two are 76 and 77, median = 76.5. Both the mean (78.8) and median (81) of Science are higher, so students performed slightly better in Science.
13. Modes of transport to school — Class Six: walking 48, cycling 12, bus 5, auto 28, guardian’s vehicle 7; Class Seven: walking 36, cycling 23, bus 7, auto 30, guardian’s vehicle 4. Draw a double bar graph and note your observation.
Answer: The double bar graph is shown below —
Observation:
Answer: In both classes walking is the most common mode. Class Six has more students walking (48) and using autos (28), while Class Seven has more cyclists (23) than Class Six. The fewest students in both classes come by bus.
14. The figure shows the geographical area and forest area (hectares) of some districts of Assam.
| District | Geographical area (ha) | Forest area (ha) |
|---|---|---|
| Tinsukia | 379000 | 126468 |
| Nalbari | 104644 | 0 |
| Kamrup Metro | 87150 | 29590 |
| Karbi Anglong | 739900 | 199689 |
| Nagaon | 276605 | 51093 |
| Dima Hasao | 488800 | 67277 |
| Sonitpur | 340900 | 115554 |
(a) Which district has the largest geographical area? (b) Which has the largest forest area? (c) Which district has no forest? (d) Does the district with the largest land area also have the largest forest area?
Answer: (a) Karbi Anglong (739900 ha). (b) Karbi Anglong (199689 ha). (c) Nalbari (forest area 0). (d) Yes — Karbi Anglong, with the largest land area, also has the largest forest area.
Puzzle 1: The average age of Jonti and Sonti is 10, of Sonti and Dhanti is 11, of Dhanti and Jonti is 12. What is the sum of their ages? How old are they?
Answer: Jonti + Sonti = 20, Sonti + Dhanti = 22, Dhanti + Jonti = 24. Adding, 2 × (Jonti + Sonti + Dhanti) = 66, so the sum of ages = 33 years. Jonti = 33 − 22 = 11, Sonti = 33 − 24 = 9, Dhanti = 33 − 20 = 13 years.
Puzzle 2: The median of 6 natural numbers is 12. Five are given: 24, 5, 2, 16, 10. Find the sixth number.
Answer: For 6 numbers the median is the average of the 3rd and 4th values = 12, so 3rd + 4th = 24. Sorting the five: 2, 5, 10, 16, 24. Taking the sixth number x = 14 gives 2, 5, 10, 14, 16, 24 — median = $\frac{10+14}{2}$ = 12. So the sixth number is 14.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The sum of all observations divided by their number is the — (a) median (b) range (c) average (d) outlier
Answer: (c) average (arithmetic mean).
2. The middle value of data arranged in order is the — (a) mean (b) median (c) range (d) total
Answer: (b) median.
3. The difference between the maximum and minimum value is the — (a) mean (b) median (c) range (d) density
Answer: (c) range.
4. A value lying far from the rest is called an — (a) median (b) outlier (c) mean (d) range
Answer: (b) outlier.
5. The average of the first five natural numbers is — (a) 2 (b) 2.5 (c) 3 (d) 5
Answer: (c) 3 [$\frac{1+2+3+4+5}{5}=3$].
6. For an even number of values the median is the ___ of the middle two — (a) sum (b) product (c) average (d) difference
Answer: (c) average.
7. An outlier most affects the — (a) median (b) range (c) mean (d) repetition
Answer: (c) mean.
8. Two related data sets shown side by side as bars form a — (a) dot plot (b) double bar graph (c) line graph (d) pie chart
Answer: (b) double bar graph.
9. The average of 4, 6, 8, 10, 12 is — (a) 8 (b) 9 (c) 10 (d) 40
Answer: (a) 8 [$\frac{40}{5}=8$].
10. The median of 3, 5, 9, 11, 13 is — (a) 5 (b) 9 (c) 11 (d) 8.2
Answer: (b) 9 (the middle value).
Fill in the Blanks
1. Average = sum of all observations ÷ ______ number of observations.
Answer: total.
2. Data shown as dots on a line is called a ______.
Answer: dot plot.
3. The median divides the data into ______ equal parts.
Answer: two.
4. An outlier does not affect the ______.
Answer: median.
5. The range of 2, 4, 4, 6, 9 is ______.
Answer: 7 (9 − 2).
True or False
1. The mean is always equal to one of the values in the data.
Answer: False.
2. To find the median the data must be arranged in order.
Answer: True.
3. An outlier affects the median more than the mean.
Answer: False (it affects the mean more).
4. A double bar graph can compare two data sets.
Answer: True.
5. Range = maximum value + minimum value.
Answer: False (Range = maximum − minimum).
Short Answer Questions
1. What is the difference between a statistical question and a statistical statement?
Answer: A statistical question needs data to be collected and analysed and has varying answers; a statistical statement is a claim or summary based on the collected data and expressed in numbers.
2. When is the median more suitable than the mean?
Answer: When the data contains outliers (values far from the rest), the mean gets distorted; so the median is more suitable for representing data with large deviations.
3. What can a dot plot tell us easily?
Answer: It shows the maximum and minimum values, range, repetition of values, density, variability and consistency of the data at a glance.
4. How does a double bar graph differ from a clustered column graph?
Answer: A double bar graph shows two related data sets as adjacent bars; a clustered column graph groups two or more bars together (for example, three years of data shown side by side).
Key Terms
| Term | Meaning |
|---|---|
| Statistical question | A question answered only by collecting and analysing data |
| Statistical statement | A data-based claim expressed in numbers |
| Average / Arithmetic mean | Sum of observations ÷ number of observations |
| Median | The middle value of ordered data |
| Outlier | A value lying far from the rest of the data |
| Range | Maximum value − minimum value |
| Dot plot | Data shown as dots on a line |
| Bar graph | Data shown using bars |
| Double bar graph | Two data sets shown as adjacent bars |
| Clustered column graph | A graph with two or more bars grouped together |
| Density | How closely the dots are packed together |
| Data detective | One who raises fresh questions from data |