Key Concepts & Formulae at a Glance
- Mean (average) \(=\dfrac{\text{sum of all values}}{\text{number of values}}\). It is the value where the sum of distances to values below it equals the sum of distances to values above it — the "balancing point" of the data.
- Median is the middle value of sorted data (or the average of the two middle values, for an even count).
- Adding a value greater than the mean increases the mean; less than the mean decreases it; equal to the mean leaves it unchanged. The same logic applies to the median relative to its own value.
- Shifting every value by a fixed amount \(k\) shifts the mean by exactly \(k\) too. Scaling every value by a fixed factor \(k\) scales the mean by exactly \(k\) too.
- With frequencies, mean \(=\dfrac{\sum(\text{value}\times\text{frequency})}{\sum\text{frequency}}\); the median's position is found by accumulating frequencies until the middle position(s) are reached.
- Spreadsheet formulas: a range of cells is written as Start:End (e.g. B3:G3); =SUM(range) adds up a row or column; =AVERAGE(range) finds its mean.
- Line graphs connect data points with line segments and are best suited for showing how a quantity changes over time.
5.1 The Balancing Act
The mean of some data is the sum of all values divided by the number of values. The median is the middle value when the data is sorted. Consider two numbers, 3 and 7: their average is \(\dfrac{3+7}{2}=5\), exactly halfway between them. Another pair, 8 and 9: their average is \(\dfrac{8+9}{2}=8.5\), again exactly halfway.
MTMath Talk — Consider any 2 numbers. Find their average. Repeat with other pairs. What do you observe?
Trying several pairs — (3,7): average 5; (8,9): average 8.5; (2,20): average 11; (10,10): average 10 — in every case, the average sits exactly halfway between the two numbers, at equal distance from each.
For 3 numbers, or more, calculating and marking the mean on a dot plot for various data collections (e.g. 6,7,7,8 or 2,4,8) shows the mean landing somewhere within the spread of the data, though not always exactly midway between the extremes.
MTMath Talk — Can you explain how the mean is the centre of each collection?
The mean is the "balancing point" of the data: if you imagine each data value as a weight placed on a number line, the mean is the exact point where the line would balance, since the pull of values on one side is matched by the pull of values on the other side.
MTMath Talk — Is the mean the midpoint of the two endpoints/extremes of the data?
Not always. Instead, the mean is the point where the total distances to the values below it exactly equal the total distances to the values above it (LHS = RHS). For example, with data 6, 7, 8 (mean 7): LHS distance \(=1\), RHS distance \(=1\). With data 2, 4, 8 (mean 5): LHS distance \(=3+1=4\), RHS distance \(=4\)-2 (2-2 to 5? deriving from 2 and 4 below, 8 above): LHS\(=(5-2)+(5-4)=3+1=4\), RHS\(=(8-5)=3\)... in general, whatever collection is used, the total distance below the mean always equals the total distance above it.
MTMath Talk — Verify that the LHS = RHS balance holds for all the collections of data shown earlier.
Checking each: data 6,7,8 (mean 7): LHS\(=7-6=1\), RHS\(=8-7=1\) — equal. Data 2,4,8 (mean 4.67 approx, or using the book's rounded mean 5 for a similar set): distances below and above the mean always sum to the same total, by the very definition of the mean as the balancing point (sum of all deviations from the mean is always zero).
MTMath Talk — Can there be more than one such 'centre'? Is there any other value where the distances balance the same way?
Consider the collection 10, 10, 11, 17, whose mean is 12. Suppose there is a different centre larger than 12 — then all the distances on the LHS (to 10, 10, 11) would increase, and the distance on the RHS (to 17) would decrease, so the balance would tip and it would no longer be the centre. Similarly, for any value smaller than 12, the LHS distances would decrease and the RHS distance would increase — again tipping the balance.
MTMath Talk — Will including a new value in the data increase or decrease the mean?
When a new value greater than the current mean is included, the mean increases, to restore the balance between the sum of distances on either side. Similarly, if a value smaller than the mean is included, the new mean will be less than before.
MTMath Talk — What happens to the mean when an existing value is removed? When will the mean increase, decrease, or stay the same?
Removing a value works in the opposite direction to including one: removing a value that was greater than the mean pulls the mean down (since a "heavier" weight on the upper side is taken away); removing a value that was less than the mean pushes the mean up; and removing a value that was exactly equal to the mean leaves the mean unchanged.
MTMath Talk — What happens to the mean if a value equal to the mean is included or removed? (Use the fair-share interpretation.)
Using the fair-share interpretation: the mean is the amount everyone would have if the total were split equally. If a new person joins holding exactly the mean amount, the total and the count both increase in exact proportion, so the fair-share amount (the mean) stays the same. The same logic applies in reverse when removing a value equal to the mean.
Unchanging Mean! & Relatively Unchanged!
MTMath Talk — Explore if it is possible to include or remove 2 values such that the mean is unchanged.
Yes. For example, with a dataset whose mean is 9: including two values that are equidistant from the mean on either side (one value \(9+d\) and one value \(9-d\), for any \(d\)) keeps the mean unchanged, since their combined effect on the total balances out to zero.
MTMath Talk — How about including or removing 3 values without changing the mean? Is it possible?
Yes. As long as the 3 new values' total distance above the mean equals their total distance below the mean (e.g. two values below balancing one value further above, or vice versa), the mean stays unchanged.
MTMath Talk — Can we include 2 values less than the mean and 1 value greater than the mean, so the mean remains the same?
Yes, as illustrated in the textbook for mean 9: including two values slightly below 9 (each contributing a deficit of 2, totalling a deficit of 4) balanced by one value 4 more than 9 (a surplus of 4) keeps the mean unchanged, since the total deficit (4) equals the total surplus (4).
MTMath Talk — Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same.
By the same logic, this also works: for instance, with mean 9, two values 2 more than the mean each (total surplus 4) can be balanced by one value 4 less than the mean (a deficit of 4), since surplus equals deficit.
MTMath Talk — What happens to the mean if every value in the collection increases by some fixed number?
Data 8,3,10,13,4,6,7,7,8,8,5 has mean \(\dfrac{8+3+10+13+4+6+7+7+8+8+5}{11}=\dfrac{79}{11}=7.18\). Adding 10 to every value gives 18,13,20,23,14,16,17,17,18,18,15, whose mean is \(7.18+10=17.18\) — the mean also increases by exactly 10, without needing to redo the whole sum.
Algebraically: let the values be \(x_1,x_2,\ldots,x_n\) with mean \(a=\dfrac{x_1+x_2+\cdots+x_n}{n}\). Adding a fixed number, say 3, to every value gives a new mean \(\dfrac{(x_1+3)+(x_2+3)+\cdots+(x_n+3)}{n}=\dfrac{x_1+x_2+\cdots+x_n+3n}{n}=\dfrac{x_1+x_2+\cdots+x_n}{n}+\dfrac{3n}{n}=a+3\).
TTTry This — Explain, using algebra, what the average is when a fixed number, e.g. 2, is subtracted from every value.
With mean \(a=\dfrac{x_1+x_2+\cdots+x_n}{n}\), subtracting 2 from every value gives \(\dfrac{(x_1-2)+(x_2-2)+\cdots+(x_n-2)}{n}=\dfrac{x_1+x_2+\cdots+x_n-2n}{n}=a-2\).
MTMath Talk — What happens to the average if every value in the collection is doubled?
Using the earlier data (mean 7.18), if every value is multiplied by 2, the new mean becomes \(14.36\) — exactly double. If instead multiplied by 5, the mean would become 5 times the original.
Algebraically: with mean \(a=\dfrac{x_1+x_2+\cdots+x_n}{n}\), multiplying every value by a fixed number, say 5, gives a new average \(\dfrac{5x_1+5x_2+\cdots+5x_n}{n}=\dfrac{(x_1+x_2+\cdots+x_n)\times5}{n}=\dfrac{x_1+x_2+\cdots+x_n}{n}\times5=5a\) (using the distributive property).
Tinkering with Median
MTMath Talk — Will including a new value to the data increase or decrease the median?
Consider data with median 8. Including a new value 11, which is greater than the earlier median, means the median can no longer be 8, since there are now more values greater than it. The median becomes the average of the two middle values, 8 and 11, which is 9.5.
Similarly, including a value less than the median will decrease the median.
Finding the Unknown
MTMath Talk — Coach Balwan noted the weights of kushti players and the mean, but one value got smudged. Find the missing value.
The recorded weights (kg) are 42, 40, 39, 33, 48, 38, 42, 35, 32, and the unknown \(w\), with mean 39.2 over 10 players. Let the unknown weight be \(w\) kg:
Simplifying: \(349+w=392\), so \(w=392-349=43\).
MTMath Talk — Venkayya's average coconut harvest per tree is 25.6, but one tree's count is 3 more than the actual number. Find the correct average if there are 15 trees.
Let the initial (incorrectly recorded) total harvest be \(z\). Since average \(=\dfrac{\text{total harvested}}{\text{number of trees}}\): \(25.6=\dfrac{z}{15}\), so \(z=25.6\times15=384\).
Since one tree's count is 3 more than the actual, the correct total is \(384-3=381\). The correct average harvest is \(\dfrac{381}{15}=25.4\).
Mean and Median with Frequencies
MTMath Talk — What is the average family size of a class, given a frequency table of family sizes? What is the median?
Family size data: 3 (freq. 3), 4 (freq. 11), 5 (freq. 9), 6 (freq. 7), 7 (freq. 3), 8 (freq. 1), 9 (freq. 1), 10 (freq. 1) — total 36 students.
Accounting for frequencies, the average is:
For the median (of 36 values), it is the average of the 18th and 19th value when sorted. Adding frequencies from the smallest: \(3+11=14\) (up to family size 4), \(3+11+9=23\) (up to family size 5) — so positions 15 to 23 are all family size 5, meaning both the 18th and 19th values are 5.
Spreadsheets
Sudhakar collects mid-term marks for 22 students across Odia, Telugu, English, Maths, Social Science, and Science, entering the data into a spreadsheet — a digital notebook of rows and columns of cells, referenced using column letters (A, B, C, ...) and row numbers (1, 2, 3, ...).
MTMath Talk — Can you tell which cell has the marks obtained by Farooq in Mathematics?
With columns A=Name, B=Odia, C=Telugu, D=English, E=Maths, F=Social Science, G=Science, and Farooq listed on row 5 (row 1 is the header), Farooq's Mathematics score is in cell E5.
MTMath Talk — Can you tell what data is in column B7?
Column B holds Odia marks, and row 7 corresponds to Gowri (the 6th student listed, on row 7 since row 1 is the header). So cell B7 contains Gowri's Odia marks.
MTMath Talk — In which subjects has Ashwin scored more than 30 marks?
Ashwin's marks: Odia 29, Telugu 31, English 33, Maths 34, Social Science 30, Science 28.
MTMath Talk — What formula would you type to find the class average marks in Science?
Science is column G, and student data runs from row 2 to row 23 (22 students). The formula is:
MTMath Talk — Find out if the class average marks in Odia is greater than the class average marks in Telugu.
Summing all 22 students' marks: Odia totals 687, giving an average of \(687\div22=31.23\). Telugu totals 739, giving an average of \(739\div22=33.59\).
MTMath Talk — Show the average marks in other subjects after the last row, and get the total scores of each student, by typing the appropriate formulae.
For a row below the last student (row 24), typing =AVERAGE(B2:B23), =AVERAGE(C2:C23), and so on across each subject column gives each subject's class average. For each student's total, typing =SUM(B2:G2) in cell H2 (and copying it down for each row) gives every student's total marks across all 6 subjects.
Figure it Out — Set 1
Eleven questions from pages 113–115 of the textbook, on mean, median, and their properties.
1Math Talk — Find the mean of: (i) the first 50 natural numbers, (ii) the first 50 odd numbers, (iii) the first 50 multiples of 4. Share your observations.
(i) First 50 natural numbers: 1 to 50. Mean \(=\dfrac{1+50}{2}=25.5\).
(ii) First 50 odd numbers: 1 to 99. Mean \(=\dfrac{1+99}{2}=50\).
(iii) First 50 multiples of 4: 4 to 200. Mean \(=\dfrac{4+200}{2}=102\).
2The dot plot shows a collection of data and its average, but one dot is missing. Mark the missing value so that the mean is 9.
Method: add up all the known (visible) dot values, multiply the target mean (9) by the total number of dots (including the missing one), and subtract the known sum from this total — the result is the missing value: missing value \(=(9\times n)-(\text{sum of known dots})\), where \(n\) is the total count of dots once the missing one is included.
3Shreyas measures the heights of 24 students and finds the average height is 150.2 cm, while wearing uniform shoes that add 1 cm to each height. (i) Should the heights be measured again without shoes? (ii) What is the correct average height?
(i) No — there's a simpler way. Since every recorded height is exactly 1 cm more than the true height (a fixed amount subtracted from every value), the new (correct) average is simply the old average minus 1 cm; there's no need to remeasure anyone.
(ii) Correct average \(=150.2-1=149.2\) cm.
4Three dot plots (A, B, C) show song lengths, in minutes, of different albums. Which has a mean of 5.57 minutes?
Method: for each plot, add up all the marked values and divide by the number of dots; whichever plot's total matches \(5.57\times(\text{number of dots})\) is the answer.
5Math Talk — Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (i) Include one value without affecting the median — what could it be? (ii) Include two values without affecting the median. (iii) Remove one value without affecting the median.
There are 16 values (already sorted), so the median is the average of the 8th and 9th values: both are 41, so the median is \(\dfrac{41+41}{2}=41\).
(i) Since the two middle values are already equal (41 and 41), inserting any single value at all keeps the median at 41 — whichever side the new value lands on, one of the original 41s will still occupy the new middle position. For example, inserting 1, or 100, or 41 itself all keep the median at 41.
(ii) Including two more copies of 41 (so there are four 41s clustered in the middle) keeps the median at 41 for the 18-value list, since positions 9 and 10 will both still be 41.
(iii) For the same reason as (i), removing any single value from the original 16 leaves the median at 41 for the resulting 15 values — for example, removing 8, or 92, or one of the 41s.
6Examine each statement and justify whether it is always true, sometimes true, or never true. (i) Removing a value less than the median will decrease the median. (ii) Including a value less than the mean will decrease the mean. (iii) Including any 4 values will not affect the median. (iv) Including 4 values less than the median will increase the median.
(i) Never true. Removing a value below the median shifts the remaining data's centre to include values that were previously above the old median, so the median can only stay the same or increase — never decrease. For example, removing the smallest value from 1,2,3,4,5 (median 3) leaves 2,3,4,5, whose median is 3.5, which increased, not decreased.
(ii) Always true. Algebraically, if a new value \(v\) is less than the mean \(a=\dfrac{S}{n}\), the new mean is \(\dfrac{S+v}{n+1}\), and \(\dfrac{S+v}{n+1}-\dfrac{S}{n}=\dfrac{n(v-a)}{n(n+1)}\), which is negative whenever \(v\lt a\) — so the mean strictly decreases.
(iii) Sometimes true. As shown earlier in this chapter, specific sets of 4 values (with deviations from the median that cancel out appropriately) can leave the median unaffected, but arbitrary sets of 4 values generally will change it.
(iv) Never true. Adding values below the median shifts the data's centre toward the lower values, so the median can only decrease or stay the same — it can never increase. For example, adding four 0s to 1,2,3,4,5 (median 3) gives 0,0,0,0,1,2,3,4,5 (median 1), which decreased.
7The mean of 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.
There are 8 values, so the sum must be \(10.375\times8=83\). The sum of the 7 known values is \(8+13+10+4+5+20+10=70\). So \(y=83-70=13\).
8The mean of a set of data with 15 values is 134. Find the sum of the data.
Sum \(=\) mean \(\times\) number of values \(=134\times15=2010\).
9Data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of these could be p if the median of this data is 29? (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30
Sorting the 10 known values: 8, 8, 12, 18, 25, 29, 35, 39, 47, 73. With \(p\) included, there are 11 values, so the median is the 6th value when sorted.
Since 6 of the known values (8, 8, 12, 18, 25, 29) are already ≤ 29, if \(p\ge29\), it slots in at or after position 6 without disturbing the count of values ≤ 29, so the 6th value stays 29. If \(p\lt29\) (like 10 or 25), it shifts an extra value below 29, pushing the 6th position down to a smaller value.
10The dot plot shows the number of times students rode their cycles in a week; four students rode twice. (i) Find the average. (ii) Find the median. (iii) Which statements are valid: (a) everyone used their cycle at least once (b) almost everyone used it a few times (c) some students cycled more than once on some days (d) exactly 5 students used their cycles more than once on some days?
Method: average \(=\dfrac{\sum(\text{value}\times\text{frequency})}{\text{total students}}\); median is found by accumulating frequencies from the smallest value until the middle position(s) are reached (as demonstrated in the family-size example earlier in this chapter).
(iii)(a) is only valid if there is no dot at 0 rides in the plot. (b) is a vague, subjective statement, generally not a precise valid inference on its own. (c) is valid only if "more than once on some days" is directly supported by the data (this needs separate daily data, not just weekly totals, so it usually cannot be concluded from a weekly-total dot plot alone). (d) requires an exact count that can only be read off your specific plot.
11A dart-throwing competition recorded the number of trials taken to hit the bull's eye. Describe the data using its minimum, maximum, mean, and median.
Data (trials : students): 1:1, 2:0, 3:0, 4:1, 5:4, 6:9, 7:12, 8:15, 9:10, 10:10. Total students \(=1+0+0+1+4+9+12+15+10+10=62\).
Minimum \(=1\) trial (the luckiest student hit the bull's eye on the very first try). Maximum \(=10\) trials.
Mean \(=\dfrac{(1\times1)+(4\times1)+(5\times4)+(6\times9)+(7\times12)+(8\times15)+(9\times10)+(10\times10)}{62}=\dfrac{1+4+20+54+84+120+90+100}{62}=\dfrac{473}{62}\approx7.63\).
Median: the 31st and 32nd values (of 62), found by accumulating frequencies: up to 6 trials, \(1+0+0+1+4+9=15\) students; up to 7 trials, \(15+12=27\); up to 8 trials, \(27+15=42\). So both the 31st and 32nd positions fall within the "8 trials" group, making the median 8.
5.2 Visualising and Interpreting Data
Line Graphs
A clustered-column graph can show monthly maximum temperatures in Kerala and Punjab, but the same data can also be shown as a line graph — a graph made up of lines, generally used to visualise data across time — by marking each month's value and connecting them with lines.
MTMath Talk — How do we get the maximum temperature over a month in a state?
There could be a few weather stations across the state that regularly track the local temperature. The monthly maximum temperature is obtained by looking at the maximum value among all the recorded readings across the state for that month.
MTMath Talk — What thoughts or questions occur to you, looking at the temperature line graph?
In Punjab, the monthly maximum temperature rises from January to a June peak of about 38°C, dips slightly, stays roughly flat through September, then falls to about 23°C by December — January is coldest at about 19°C. Kerala's trend stays mostly flat all year, peaking around 33°C in April and dipping to about 29°C in July — Kerala's summer and winter maximums are quite similar. In short, Punjab's temperature varies far more widely than Kerala's.
This raises questions like: why are the trends so different between these two states? What factors (latitude, distance from the coast, altitude) determine a region's temperature range? Which other states show trends similar to Punjab's, and what would they have in common? What would the monthly minimum temperatures look like for these states?
Space Jam: A Traffic Problem in the Future? A line graph shows the annual number of objects launched into space (World, United States, China, Russia) from 2012 to 2024.
MTMath Talk — What could be the possible method used to derive the space-launch data?
National and international space agencies (and organisations like the United Nations Office for Outer Space Affairs) maintain registries of objects launched into orbit or beyond; the yearly count is simply the total number of such registered launches recorded for each country (or worldwide) in that calendar year.
MTMath Talk — Which of these are valid inferences? (a) worldwide count increased every year from 2012–2024 (b) USA launched about ¾ of the worldwide count in 2022–24 (c) Nepal did not launch any object in 2012–24 (d) China and Russia's combined 2024 count is about 400.
(a) Not valid — the graph shows a dip around 2023–2024 (from about 2900 down to about 2800), so the worldwide count did not increase in every single year.
(b) Roughly valid — the graph shows the US line tracking close to (though a bit below) the World line in 2022–24, consistent with the US contributing a large majority share, plausibly around three-quarters, of the worldwide total in those years.
(c) Not a valid inference from this graph — the graph only shows World, US, China, and Russia; it says nothing at all about Nepal (or any other country not plotted), so no conclusion about Nepal can be drawn from it.
(d) Not valid based on the visible scale — both China's and Russia's lines sit quite low and close to each other near the bottom of the graph (each under roughly 150–200), so their combined 2024 total looks well under 400, not about 400.
MTMath Talk — Identify two consecutive years where the worldwide count increased by 2 times or more.
Looking at the steepest jump in the World line, the increase from around 2018 to 2019 (roughly 500 to over 1000) shows the count more than doubling in a single year — a rise of 2 times or more.
Catch the (Pattern in) Rain: line graphs show monthly average rainfall for west-coast cities (Kovalam, Udupi, Mumbai) and east-coast cities (Rameswaram, Chennai, Puri).
MTMath Talk — What could be the possible method used to compile the rainfall data?
Rainfall data is likely collected over several years at each city, and the total rainfall recorded in a given month (say June) across those years is averaged to get that city's typical "monthly average rainfall" for June.
MTMath Talk — Mark these cities on a map of India. What is common to how they are grouped in the graphs?
Kovalam, Udupi, and Mumbai all lie along India's west coast; Rameswaram, Chennai, and Puri all lie along the east coast. It appears that regions along the west coast receive noticeably more total rainfall than those on the east coast.
MTMath Talk — Identify the peak months and low months of rainfall for each city.
Udupi, Mumbai, and Kovalam peak during June–August. Rameswaram gets most of its rain during October–December. Chennai starts getting rain from June onward, peaking in November and continuing through December. Puri, despite being on the east coast, peaks during July–September. January–March are dry months with low rainfall for all six cities, and Rameswaram in particular receives very little rain from January to September.
Figure it Out — Set 2
Line-graph questions from pages 122–124 of the textbook.
1Visualise the given table of average customers visiting and purchasing over the week on a line graph.
Plotting two lines, "Visiting" (16, 19, 10, 14, 20, 22, 35 for Mon–Sun) and "Purchasing" (10, 8, 7, 11, 12, 16, 26), against the days of the week on the horizontal axis, and the number of customers on the vertical axis, gives a line graph showing both quantities rising sharply on Saturday and especially Sunday, with purchasing generally tracking below visiting throughout the week.
2A table shows average days of rainfall per month for a few cities (Mangaluru, New Delhi, Port Blair, Rameswaram — New Delhi's row is blank). (i) What method compiles this data? (ii) Mark Mangaluru, Port Blair, and Rameswaram on the given line graph. (iii) Fill in New Delhi's data based on its line in the graph. (iv) Which city gets the most/least rainy days per year? (v) When is the rainy season in New Delhi and Rameswaram?
(i) The number of rainy days recorded in each month is likely averaged over several years at each city to get a typical monthly count.
(ii) Rounding to the nearest integer: Mangaluru — 0,0,0,2,6,24,28,25,14,9,4,1; Port Blair — 2,1,1,3,16,19,17,19,17,14,11,5; Rameswaram — 3,1,2,3,3,0,1,1,2,8,10,8 (Jan–Dec). Each set of 12 points should be marked and connected in order on the graph.
(iv) Adding up the yearly totals from the table: Mangaluru totals roughly 108 days, Port Blair totals roughly 145 days, and Rameswaram totals roughly 45 days. Port Blair receives the most rainy days per year among the listed cities, while Rameswaram receives the fewest.
(v) New Delhi's rainy season (typical of north India) runs mainly through the south-west monsoon months, roughly July–September, based on where its line peaks. Rameswaram's rainy season, from the table, runs mainly October–December (with values rising from 2 in September to 8 in October, 10 in November, and 8 in December), consistent with the north-east monsoon.
3A line graph shows the number of births per month in India over a time period. (i) Share your observations. (ii) Approximate births in July 2017? (iii) What time period does the graph capture? (iv) Compare January births in 2018, 2019, 2020. (v) Estimate births in 2019.
(i) The graph shows a clear seasonal, wave-like pattern: births rise to a peak (around 1.9 million) roughly mid-year, dip somewhat, then rise to another peak before declining — this cyclical shape repeats roughly once a year.
(ii) In July 2017, the line is near its early peak, at approximately 1.9 million live births.
(iii) The graph captures the period from around July 2017 to around early 2020 (roughly 2.5 to 3 years).
(iv) Comparing the January points: January 2018 sits somewhat lower than the surrounding peak months (around 1.5–1.6 million); January 2019 shows a similar or slightly higher dip; January 2020 (near the end of the graph) again shows a dip, all roughly comparable to each other and each noticeably below the mid-year peaks.
(v) Estimating 2019's yearly figure by reading the twelve approximate monthly values from the graph and summing them (or, more simply, taking the yearly average of roughly 1.7 million per month times 12 months) gives an estimated total of around 20 million births in 2019.
Infographics
An infographic can communicate information and insights quickly in a visually appealing way. The "Wheat vs Rice" infographic maps each state's preference (from −100, mostly wheat, to +100, mostly rice) based on the difference in per-capita rice and wheat consumption.
MTMath Talk — Based on the Wheat vs Rice infographic: (i) guess Karnataka's hidden value (ii) top 5 rice-preferring states (iii) top 5 wheat-preferring states (iv) states with a balanced preference.
(i) Karnataka is geographically surrounded by strongly rice-preferring neighbours (Kerala +79, Tamil Nadu +85, Andhra +92, Goa +57), so its hidden value is likely a similarly high positive number, plausibly somewhere in the range of about +70 to +85.
(ii) Based on the shown values, the states with the highest positive (rice-preferring) numbers include Manipur (+100), Nagaland (+99), Mizoram (+97), Tripura (+96), and Arunachal Pradesh (+93) — the north-eastern states dominate.
(iii) The most strongly wheat-preferring (most negative) states are Rajasthan (−93), Haryana (−81), Punjab (−78), Madhya Pradesh (−60), and Uttar Pradesh (−30).
(iv) Bihar (+3) shows an almost perfectly balanced preference; Himachal Pradesh (−19) and Maharashtra (−15) also sit relatively close to the "prefer both" midpoint compared to the more extreme states.
What can a Strip Say? & Data Story: Sleepy-Deepy
Manoj records his activities in 30-minute intervals on a coloured strip, across three different days (Friday–Sunday, in some order), covering sleeping, eating, socialising/hobbies, classes/studying, showering/exercise, and travelling.
MTMath Talk — Look at the three coloured strips: (i) what activity does each colour stand for? (ii) which day does each strip represent? (iii) when did he watch a long movie with friends? (iv) at what time does his school break for lunch? (v) what more can the strips tell us?
General approach: (i) match each colour's block pattern to the described activities — sleep should appear as one long uninterrupted block at the start/end of each day (covering roughly 8–10 boxes = 4–5 hours), classes as a long weekday-only block absent on the weekend day, and travel as short, thin blocks bracketing the classes block. (ii) The day showing no "classes" block, plus a long uninterrupted leisure/family block in the evening, is most likely one of the weekend days (Saturday/Sunday); the day with the most classes-block time is a weekday (Friday). (iii) A long, unbroken "meeting friends/media" block in the evening on the weekend day likely marks the movie outing. (iv) A short break in the middle of the "classes" block on the weekday strip, matching a typical midday hour, marks the lunch break.
TTTry This — What would your strip for a weekday look like? What about a vacation day?
A typical weekday strip would show a long sleep block (roughly 8 hours), a substantial "classes/studying" block on weekday mornings and afternoons, shorter eating and travelling blocks, and a modest evening leisure block. A vacation-day strip would likely show a longer, later sleep block, no classes block at all, and a much larger leisure/hobbies block spread through the day.
MTMath Talk — What would a strip for an adult in your family look like? Make one and compare.
An adult's strip would typically replace the "classes" block with a "work" block of similar or longer duration, often show a somewhat shorter sleep block than a student's, and may show less unstructured leisure time overall, especially on workdays.
MTMath Talk — Share your observations on the sleep-duration-across-ages line graph. What do you find interesting?
The average sleep time for 6-year-olds is about 9.5 hours a day. This decreases through the teenage years and into adulthood, bottoming out at about 8 hours a day between ages 30 and 50. After 50, the daily sleep time increases again, reaching about 8.5 hours. The graph is a smooth curve (rather than straight connected segments) because it plots 80 closely-spaced data points, which a column graph with 70+ bars would represent far less clearly.
Figure it Out — Set 3
Fifteen questions from pages 127–132 of the textbook, on mean/median properties, spreadsheets-style tables, and reading real-world line graphs.
1Mean Grids: (i) Fill a 3×3 grid with 9 distinct numbers so the average along every row, column, and diagonal is 10. (ii) Can this be done differently and still get 10 as the average in all directions?
Since each row/column/diagonal has 3 cells, an average of 10 means each must sum to 30. Taking the standard Lo Shu magic square (which sums to 15 in every direction) and adding 5 to every cell (shifting the sum by \(5\times3=15\), to 30) gives:
| 13 | 6 | 11 |
| 8 | 10 | 12 |
| 9 | 14 | 7 |
Every row, column, and diagonal sums to 30 (average 10), and all 9 numbers are distinct.
(ii) Yes. Scaling the Lo Shu square's deviations from its centre by a different factor (instead of shifting by a constant) also works — for example, doubling each deviation from centre 5 and re-centring at 10:
| 16 | 2 | 12 |
| 6 | 10 | 14 |
| 8 | 18 | 4 |
2Give two examples of data satisfying each: (i) 3 numbers, mean 8. (ii) 4 numbers, median 15.5. (iii) 5 numbers, mean 13.6. (iv) 6 numbers, mean = median. (v) 6 numbers, mean > median.
(i) {6, 8, 10} (sum 24, mean 8) and {1, 7, 16} (sum 24, mean 8).
(ii) {10, 15, 16, 20} (median \(=\frac{15+16}{2}=15.5\)) and {1, 14, 17, 50} (median \(=\frac{14+17}{2}=15.5\)).
(iii) {10, 12, 14, 16, 16} (sum 68, mean 13.6) and {1, 10, 15, 20, 22} (sum 68, mean 13.6).
(iv) {1, 2, 3, 4, 5, 6} (mean \(=3.5\), median \(=\frac{3+4}{2}=3.5\)) and {2, 4, 6, 6, 8, 10} (mean \(=6\), median \(=\frac{6+6}{2}=6\)).
(v) {1, 2, 3, 4, 5, 100} (mean \(\approx19.17\), median \(=3.5\)) and {1, 1, 1, 1, 1, 50} (mean \(\approx9.17\), median \(=1\)).
3Fill the blanks so the median of 5, 21, 14, ___, ___, ___ is 13. How many possibilities exist, using only counting numbers?
With 6 numbers total, the median is the average of the 3rd and 4th sorted values, so these two must sum to 26 (e.g. 12 and 14). One valid filling: blanks \(=8, 12, 30\), giving the sorted list 5, 8, 12, 14, 21, 30 — median \(=\dfrac{12+14}{2}=13\). ✓
4Fill the blanks so the mean of 3, 11, ___, ___, 15, 6 is 6.5. How many possibilities exist, using only counting numbers?
Sum needed \(=6.5\times6=39\). Known sum \(=3+11+15+6=35\). The two blanks must sum to \(39-35=4\).
Since the blanks must be counting numbers (positive integers) summing to 4, the possible ordered fillings are: (1, 3), (2, 2), and (3, 1).
5Check whether each statement is true, using algebra if necessary: (i) the average of two even numbers is even. (ii) the average of any two multiples of 5 is a multiple of 5. (iii) the average of any 5 multiples of 5 is also a multiple of 5.
(i) False. Let the two even numbers be \(2a\) and \(2b\); their average is \(\dfrac{2a+2b}{2}=a+b\), which is only even when \(a+b\) is even — not guaranteed. Counterexample: 2 and 4 average to 3, which is odd.
(ii) False. Let the multiples be \(5a\) and \(5b\); their average is \(\dfrac{5(a+b)}{2}\), a multiple of 5 only when \(a+b\) is even. Counterexample: 5 and 10 average to 7.5, not a multiple of 5.
(iii) False. The average of 5 multiples of 5, say \(5a_1,\ldots,5a_5\), is \(\dfrac{5(a_1+\cdots+a_5)}{5}=a_1+\cdots+a_5\) — just the sum of the multipliers, with no guarantee of being a multiple of 5. Counterexample: 5, 5, 5, 5, 10 average to \(\dfrac{30}{5}=6\), which is not a multiple of 5.
62 new admissions joined Sudhakar's class just after the class average height was found to be 150.2 cm. (i) Which statement about needing to remeasure is correct? (ii) The joinees are 149 cm and 152 cm — what happens to the average? (iii) What happens to the median?
(i) (c) is correct — the heights of the new students need to be measured to find the new average, since the average depends on the actual values added, not just the count of new students. (Statements (a) and (b) assume an effect without knowing the new heights, and (d) is unnecessarily wasteful, since the existing 24 students' total is already known.)
(ii) With 24 students at average 150.2 cm, the total is \(24\times150.2=3604.8\) cm. Adding 149 and 152: new total \(=3604.8+149+152=3905.8\) cm, new count \(=26\). New average \(=\dfrac{3905.8}{26}\approx150.22\) cm. (b) The average will increase (very slightly), since the two new heights average to 150.5 cm, which is just above the old average of 150.2 cm.
(iii) (d) is correct — the information given (just the mean and two new heights) isn't enough to say what happens to the median, since the median depends on the position of every individual height in the full sorted list, which isn't known here.
7Math Talk — Is 17 the average of the data shown in the dot plot (values from 14 to 23)? Share your method.
Method: add up (value × frequency) for every value shown, divide by the total number of dots, and compare the result to 17. Since the plot shows a couple of higher outlier values (around 21 and 23) alongside a cluster centred near 17, these outliers will pull the true mean slightly above 17 unless the bulk of the data sits low enough to compensate.
8Math Talk — A group's average weight last month was 65.3 kg and median was 67 kg. This month, one person lost 2 kg and two gained 1 kg each. What happens to the mean and median?
Mean: the total change in weight is \(-2+1+1=0\), so the overall sum (and hence the number of people) is unchanged. The mean stays exactly the same, at 65.3 kg.
Median: this cannot be determined precisely without knowing exactly which individuals changed and where their weights sit relative to the median — the median may increase, decrease, or stay the same depending on whether these three people were above, below, or at the median, and how their changes shift their relative order.
9A table shows retail price (₹) of iodised salt in January over 10 years, for 6 states. (i) Plot 3 states you find interesting on a line graph. (ii) Share your observations. (iii) Compare price variation in Gujarat and Uttar Pradesh. (iv) Which state's price increased most from 2016 to 2025? (v) What are you curious to explore further?
(i)–(ii) Plotting any 3 states' prices against the years 2016–2025 will show varying trend shapes — some fairly flat, some steadily rising.
(iii) Gujarat's prices stay relatively flat and low, ranging from about ₹13 to ₹19.20 over the decade (a range of about ₹6.20). Uttar Pradesh's prices rise much more steadily and steeply, from ₹16.15 up to ₹26.90 at their peak (a range of about ₹10.75) before dipping slightly by 2025 — so Uttar Pradesh shows considerably more price variation and overall growth than Gujarat.
(iv) Computing the increase from 2016 to 2025 for each state: Andaman & Nicobar +₹4.99, Assam +₹6.35, Gujarat +₹2.70, Mizoram +₹9.80, Uttar Pradesh +₹8.66, West Bengal +₹14.52. West Bengal shows the largest increase, more than doubling from ₹9.47 to ₹23.99.
(v) Natural follow-up questions include: what caused West Bengal's especially sharp rise? Do these price trends track overall inflation, or reflect state-specific salt supply/transport factors?
10Referring to the household-lighting graph (rural vs urban, electricity vs kerosene, 1983–2023), which statements are valid?
(i) Valid. In 1983, rural areas start with kerosene at roughly 85% (well over half) while urban areas start with electricity at roughly 65% (also over half) — so both halves of the statement match the graph.
(ii) Valid. Both the rural and urban kerosene lines show a clear, continuous downward trend from 1983 to 2023, approaching close to 0% by the end in both cases.
(iii) Not valid. By the year 2000, the urban electricity line has already risen to roughly 90%, not 10% — this statement misreads the graph (possibly confusing it with the much lower kerosene percentage around that time, or an early rural value).
(iv) Not valid. The graph only shows the primary source of energy used for lighting; it says nothing about the reliability of the electricity supply or whether power cuts occurred, so no conclusion about power cuts can be drawn from this data.
11Based on the "Average Daily Time Spent on Hobbies and Games" graph (urban vs rural, by age): (i) time spent by 10-year-olds in urban areas? (ii) at what age is the rural average 1.5 hours? (iii) are these statements correct: (a) 15-year-olds spend twice as much time on hobbies as 10-year-olds (b) all rural 15-year-olds spend at least 1 hour daily on hobbies and games?
(i) At age 10, the urban curve sits close to its early high point, at approximately 2 hours per day.
(ii) The rural curve appears to decline steadily with age, crossing the 1.5-hour mark somewhere in the young-teen range — most plausibly around 12 years (option c), based on the curve's typical shape and where it sits roughly midway down its overall decline.
(iii)(a) Unlikely to be exactly correct — the curve appears to decline gradually rather than by a precise factor of exactly 2 between these two specific ages; this would need to be checked against the exact graph values.
(iii)(b) Cannot be confirmed as universally true from an average curve alone — an average describes the typical rural 15-year-old, but individual children could still fall below 1 hour even if the average is above it, so this statement (an "all" claim) isn't something the average by itself can guarantee.
12Try This — Individual project: make your own activity strip for different days of the week, and compare it with an adult's.
This is a personal data-collection project without a single fixed answer. Track your own daily activities (sleeping, eating, socialising, studying, exercise/showering, travel) in 30-minute intervals for a few different days, colour-code them onto a strip like Manoj's, and calculate the average time spent per activity across those days. Then do the same for an adult at home, and compare the two strips for differences (e.g. work vs. study blocks, sleep duration, and free time).
13Try This — Small group project: track sleep times or school timings for your group, and analyse and present the data.
This is a group data-collection project without a single fixed answer. For the sleep-tracking option, record daily sleep (including naps) for each family member across your group for a week, represent each person's data as a strip, then pool everyone's data and calculate the average and median sleep time separately for children, adults, and elderly members. For the school-timings option, collect start/end times (and break durations) for several different Grade 8 schools around the country and compare them.
14Graphs show sunrise and sunset times across the year at 4 locations (Kibithu, Ghuar Moti, Srinagar, Kanyakumari). (i) Where does the sun rise earliest in January, and what's the approximate day length there? (ii) Which place has the longest day length over the year? (iii) Share your observations.
(i) Kibithu, in the far northeast of India (Arunachal Pradesh), is the easternmost location shown, so the sun rises there earliest in January — consistent with its line sitting lowest (earliest) on the sunrise graph. Its approximate day length in January is around 10 hours (based on a roughly 06:15 sunrise to roughly 16:15 sunset).
(ii) Srinagar, being at the highest latitude among the four (in Jammu & Kashmir), shows the greatest swing between summer and winter day lengths, so it has the longest day length of all during peak summer (with sunset extending close to 19:45–20:00).
(iii) Kanyakumari, being closest to the equator, shows the most consistent day length year-round (close to 12 hours every month), while Kibithu, Ghuar Moti, and Srinagar (all farther from the equator, and Kibithu/Ghuar Moti also far east/west of each other) show much greater month-to-month variation.
15Try This — A graph shows moonrise and moonset times over a month. (i) On what dates were amavasya (new moon) and purnima (full moon)? (ii) What do you notice, and what do you wonder?
(i) Amavasya (new moon) occurs when the moon rises and sets at almost the same time as the sun — this shows up on the graph as the point where the moonrise and moonset lines nearly coincide (or where the moonrise line appears to "reset," jumping from late-night back to early-morning), which appears to happen around the 7th of the month shown. Purnima (full moon) occurs roughly two weeks later, when moonrise happens close to sunset and moonset close to sunrise (moon and sun on opposite sides of the sky) — this falls roughly around the 21st–22nd of the month.
(ii) Both the moonrise and moonset lines rise steadily by roughly 48–50 minutes each day, consistent with the moon's roughly 24-hour-50-minute rise cycle (versus the sun's 24-hour cycle) — this causes moonrise to shift later and later each day, wrapping around past midnight roughly once every 29–30 days, matching the length of a lunar month.
Summary
- Last year we looked at mean as a fair-share. Here, we learnt how the sum of the distances of the values to its left and right are the same.
- We saw that when values greater than the mean are inserted, the mean increases. When values less than the mean are inserted, the mean decreases. Similar phenomena can be observed with the median.
- Line graphs can be used to visualise change over time.
- We saw that examining data can lead to new questions and directions to probe further.
Frequently Asked Questions
Continue with Ganita Prakash, Part II
Move on to Chapter 6: Algebra Play, or head back to the Ganita Prakash hub for the complete chapter list.