Chapter 5: Connecting the dots

Both players played the same number of matches (4). Let us compare their totals and averages.

Shubman. Total runs $$= 0 + 17 + 21 + 90 = 128$$. Average per match $$= \dfrac{128}{4} = 32$$.

Yashasvi. Total runs $$= 67 + 55 + 18 + 35 = 175$$. Average per match $$= \dfrac{175}{4} = 43.75$$.

Yashasvi scored more runs in the series (175 vs 128) and a higher average per match (43.75 vs 32). Yashasvi also scored above 15 in every match, while Shubman scored 0 in Match 1 and got most of his runs from a single big score of 90.

So Yashasvi was both the higher scorer and the more consistent batter, and performed better in this series.

Vaishnavi's comparison is not fair, because the two players did not play the same number of matches. The dash ("–") in Yashasvi's Match 4 shows that he did not play that match.

So Shubman's 110 runs come from 5 matches while Yashasvi's 96 runs come from only 4 matches. To compare fairly we should look at the runs per match — the average.

Shubman: played 5 matches. $$\text{Average} = \dfrac{23 + 7 + 10 + 52 + 18}{5} = \dfrac{110}{5} = 22 \text{ runs per match}.$$

Yashasvi: played 4 matches. $$\text{Average} = \dfrac{26 + 53 + 2 + 15}{4} = \dfrac{96}{4} = 24 \text{ runs per match}.$$

Yashasvi's average (24) is higher than Shubman's (22). So on a per-match basis, Yashasvi actually performed slightly better in this series. Comparing only the totals — as Vaishnavi did — is misleading when the number of matches is different.

Yes, a single number can be used to summarise a whole group of numbers. Such a number is called a measure of central tendency. Two very common ones are:

• The average (mean) — the total divided by the number of matches. For Shubman it is $$110 \div 5 = 22$$ and for Yashasvi it is $$96 \div 4 = 24$$.
• The median — the middle value when the scores are arranged in order.

A single number is convenient because it lets us compare the two batters at a glance, and it works even if they played different numbers of matches. But it also has a limitation: it hides how the runs are distributed. For example, the mean 22 for Shubman doesn't tell us that he scored a huge 52 in one match and only 7 in another.

So a single number is a useful summary, but for a complete picture we should also look at the individual match scores (the variability).

An equal share for each member of a group means the total number of guavas divided by the number of members in the group. That is exactly the average (mean) for each group.

Shreyas' group has $$1 + 4 = 5$$ members (Shreyas and 4 friends).

Total guavas $$= 3 + 8 + 10 + 5 + 4 = 30$$.

$$\text{Share per member} = \dfrac{30}{5} = 6 \text{ guavas}.$$

Parag's group has $$1 + 5 = 6$$ members (Parag and 5 friends).

Total guavas $$= 5 + 4 + 6 + 3 + 4 + 8 = 30$$.

$$\text{Share per member} = \dfrac{30}{6} = 5 \text{ guavas}.$$

Even though both groups collected the same total of 30 guavas, Shreyas' group has fewer members, so each member gets more.

The average (mean) is the total divided by the number of days.

Total flowers over the 5 days $$= 2 + 7 + 9 + 4 + 3 = 25$$.

Number of days $$= 5$$.

$$\text{Average} = \dfrac{25}{5} = 5.$$

So on average, 5 Hibiscus flowers bloomed per day in Vaishnavi's garden.

The average (mean) $$=$$ total of all values $$\div$$ number of values.

Total number of bounces $$= 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40$$.

Number of attempts $$= 8$$.

$$\text{Average} = \dfrac{40}{8} = 5 \text{ bounces per attempt}.$$

This is an activity you should perform yourself. The steps are:

1. Take a bat and a ball.
2. Try to bounce the ball on the bat and count how many bounces you make before it falls. Record that count.
3. Repeat for 7 (or more) attempts, writing down each count.
4. Add all the counts and divide by the number of attempts. That gives the average.

Sample working. Suppose the 7 attempts give 4, 6, 3, 8, 5, 7, 2.

Total $$= 4 + 6 + 3 + 8 + 5 + 7 + 2 = 35$$.

Number of attempts $$= 7$$.

$$\text{Average} = \dfrac{35}{7} = 5.$$

So the average number of bounces for this sample data is 5. Your own average will depend on your data.

This is an observation activity. The steps are:

1. Choose one flowering plant near your home or school.
2. Every morning for 7 days, count the number of flowers that are open on the plant and note it in your notebook.
3. Add the seven daily counts.
4. Divide the total by 7 (the number of days) — that is the average number of flowers per day.

Sample working. Suppose on Monday through Sunday you count 3, 5, 4, 6, 8, 5, 4 flowers.

Total $$= 3 + 5 + 4 + 6 + 8 + 5 + 4 = 35$$.

$$\text{Average per day} = \dfrac{35}{7} = 5 \text{ flowers}.$$

So this plant produced on average 5 flowers per day during the observed week. Your own figures will depend on the plant and the week.

The one who is quicker takes less time. So we compare the average times.

Nikhil. Total time $$= 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122$$ seconds. Number of runs $$= 7$$.

$$\text{Average} = \dfrac{122}{7} \approx 17.43 \text{ seconds}.$$

Sunil. Total time $$= 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122$$ seconds. Number of runs $$= 7$$.

$$\text{Average} = \dfrac{122}{7} \approx 17.43 \text{ seconds}.$$

Both averages come out to be the same. So on average, Nikhil and Sunil ran at the same speed — neither was quicker on average.

Mean $$=$$ total enrolment $$\div$$ number of years.

Total enrolment: add the numbers in pairs to reduce mistakes.

$$1555 + 1670 = 3225$$

$$1750 + 2013 = 3763$$

$$2040 + 2126 = 4166$$

$$3225 + 3763 + 4166 = 11154$$

Number of years $$= 6$$.

$$\text{Mean enrolment} = \dfrac{11154}{6} = 1859.$$

So the mean enrolment during the six years was 1859 students.

The prices go up and down every month, so we cannot say straight away where onions are costlier. A fair way is to compare the two towns month by month, and also compare their average (mean) prices over the year.

Month-by-month comparison. In Yahapur, prices are higher than in Wahapur in Jan, Feb, Mar, Aug, Nov, Dec (6 months). In Wahapur, prices are higher in Apr, May, Jul, Sep, Oct (5 months). In June both towns are the same (₹35).

Average price for the year.

Yahapur: total $$= 25+24+26+28+30+35+39+43+49+56+59+44 = 458$$; over 12 months, $$\text{mean} = \dfrac{458}{12} \approx 38.17$$ rupees.

Wahapur: total $$= 19+17+23+30+38+35+42+39+53+60+52+42 = 450$$; over 12 months, $$\text{mean} = \dfrac{450}{12} = 37.5$$ rupees.

The two averages are very close (about ₹38), but Yahapur's average is slightly higher. Also Yahapur is more often the costlier of the two (6 months vs 5). So overall, onions are (a bit) costlier in Yahapur, but the difference is small.

Comparing two tables of numbers directly is tiring. There are several other ways that make the comparison easier:

• Dot plots — mark each month's price as a dot along a number line. We can quickly see which town's prices are usually higher or spread over a wider range.
• Double bar (column) graph — two bars side by side for every month, one for each town. Easy to see the month-wise gap.
• Line graphs — draw a line joining the monthly prices for each town on the same axes. This shows the trend across the year.
• Summary numbers — the mean, the median, the range (highest $$-$$ lowest), or the maximum and minimum price.

Each of these presents the same data in a different way and helps us see a different aspect (variability, trend, or overall level).

No, the dot plot does not capture all the information that was in the table.

A dot plot for onion prices shows each price value as a dot on a number line. From it we can see how many months had a particular price, the range of prices, and where the prices are clustered. But every dot loses its "identity" — we no longer know which month a particular dot came from. The table linked each price to a specific month; the dot plot does not.

So the dot plot captures the values of the prices and their distribution, but it drops the month information.

No, we cannot. As explained above, a dot plot only tells us which prices occurred and how many times. It does not remember the month a price came from.

Looking at Yahapur's dot plot, we can see that the price ₹25/kg occurred once during the year, but we cannot tell from the dot plot alone whether that occurred in January or some other month. For that we have to go back to the original table.

Add up the 12 monthly prices and divide by 12 for each town.

Yahapur.

$$25 + 24 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44 = 458.$$

$$\text{Average} = \dfrac{458}{12} \approx 38.17 \text{ rupees per kg}.$$

Wahapur.

$$19 + 17 + 23 + 30 + 38 + 35 + 42 + 39 + 53 + 60 + 52 + 42 = 450.$$

$$\text{Average} = \dfrac{450}{12} = 37.5 \text{ rupees per kg}.$$

The two averages are close, but Yahapur's average price is a little higher.

This is an open-ended question meant to make you think of further questions the data raises. Some things one might wonder about are:

• Why are onions the cheapest in the early months (Jan–Mar) and the most expensive around October–November?
• Is the price rise linked to the harvest season, monsoon damage or storage costs?
• Are Yahapur and Wahapur close to each other? If yes, why do their prices differ month to month?
• Are onions grown locally in these two towns, or are they brought in from elsewhere?
• Do the residents of these towns eat about the same amount of onions each month?
• Would other vegetables (potatoes, tomatoes) show the same pattern of monthly change?

Any similar question that helps us understand the reasons behind the price changes is a good answer.

Yaangba's family (6 members).

Total $$= 169 + 173 + 155 + 165 + 160 + 164 = 986$$ cm.

$$\text{Average} = \dfrac{986}{6} \approx 164.33 \text{ cm}.$$

Poovizhi's family (5 members).

Total $$= 170 + 173 + 165 + 118 + 175 = 801$$ cm.

$$\text{Average} = \dfrac{801}{5} = 160.20 \text{ cm}.$$

So Yaangba's family has a higher average height (about $$164.33$$ cm) than Poovizhi's family (about $$160.20$$ cm).

But is Yaangba's family really taller? Look at Poovizhi's family carefully. Four of the five heights (170, 173, 165, 175 cm) are actually bigger than every single height in Yaangba's family. The height 118 cm is very small compared to the others — most likely a young child. This one very small value pulls Poovizhi's average down.

So the higher average of Yaangba's family is misleading. We cannot honestly say that Yaangba's family is taller than Poovizhi's — in fact, most of Poovizhi's family members are taller than the members of Yaangba's family.

The problem with the average (mean) is that even a single very small (or very large) value called an outlier can shift it a lot. In Poovizhi's family the value 118 cm pulled the mean down.

A number that is not affected much by an outlier is the median — the middle value of the data when it is arranged in order.

Arrange the heights of Poovizhi's family in increasing order: $$118, 165, 170, 173, 175$$. The middle value (the 3rd number) is $$170$$ cm. This is much closer to the typical height in the family than the mean $$160.2$$ cm.

So a better representative number here is the median, which is $$170$$ cm for Poovizhi's family (and $$\dfrac{164 + 165}{2} = 164.5$$ cm for Yaangba's family).

Let us compute both measures for each family.

Yaangba's family (heights sorted): $$155, 160, 164, 165, 169, 173$$. There are 6 values, so the median is the average of the 3rd and 4th values: $$\dfrac{164 + 165}{2} = 164.5$$ cm. The mean is $$164.33$$ cm. Both are almost the same.

Poovizhi's family (heights sorted): $$118, 165, 170, 173, 175$$. There are 5 values, so the median is the middle (3rd) value: $$170$$ cm. The mean is $$160.20$$ cm. Here the two are quite different.

The mean 160.20 cm is well below almost every height in Poovizhi's family; it does not look like a typical member of that family. The median 170 cm sits comfortably in the middle of the four normal heights (165, 170, 173, 175) and is not dragged down by the outlier 118 cm. So the median is a much better representative of Poovizhi's family.

Also, comparing families by the median gives Yaangba 164.5 vs Poovizhi 170 — that matches our intuition that Poovizhi's family members are, on the whole, taller.

So yes — in this case, the median represents the family heights better than the average.

Removing 118 leaves the four heights $$165, 170, 173, 175$$ cm.

Mean. Total $$= 165 + 170 + 173 + 175 = 683$$ cm; number of values $$= 4$$.

$$\text{Mean} = \dfrac{683}{4} = 170.75 \text{ cm}.$$

Median. The values in order are $$165, 170, 173, 175$$. There are 4 values, so the median is the average of the middle two: $$\dfrac{170 + 173}{2} = 171.5$$ cm.

What changed. Earlier (with 118) the mean was $$160.20$$ cm and the median was $$170$$ cm.

• Mean jumped from $$160.20$$ to $$170.75$$ cm — a big change of about $$10.55$$ cm.
• Median moved only slightly from $$170$$ to $$171.5$$ cm — a change of $$1.5$$ cm.

This shows that the mean is very sensitive to outliers — removing (or including) one extreme value shifts it a lot. The median hardly changes — it only depends on the middle position and not on how far away the outlier lies.

Guess first. The number 40 in the list is far bigger than the rest of the values (most are between 0 and 15). A large value on the right side of the data usually pulls the mean up while the median stays near the middle. So we expect the mean to be greater than the median.

Mean. Total of all values: add them in convenient groups.

$$6 + 8 + 5 + 15 + 3 + 0 + 2 + 7 + 12 + 40 + 0 + 8 + 10 + 5 + 1$$

$$= (6 + 8 + 5 + 15) + (3 + 0 + 2 + 7 + 12) + (40 + 0 + 8 + 10 + 5 + 1)$$

$$= 34 + 24 + 64 = 122.$$

Number of students $$= 15$$.

$$\text{Mean} = \dfrac{122}{15} \approx 8.13.$$

Median. Arrange the 15 values in increasing order:

$$0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40.$$

There are 15 values (an odd number), so the median is the middle one — the 8th value. Counting from the left: 1st is 0, 2nd 0, 3rd 1, 4th 2, 5th 3, 6th 5, 7th 5, 8th 6.

So median $$= 6$$.

As guessed, the mean ($$8.13$$) is greater than the median ($$6$$) — the outlier 40 has pulled the mean up.

On the dot plot (0 to 40). Place one dot above each value: two dots at 0, one at 1, one at 2, one at 3, two at 5, one at 6, one at 7, two at 8, one at 10, one at 12, one at 15, one at 40. Mark the median with an arrow at 6, and the mean with an arrow slightly to the right at about 8.1. The single dot at 40 sits far away on the right — clearly an outlier.

An outlier is a value that lies very far away from the rest of the data.

Arrange the values in order: $$0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15, 40$$. All values other than $$40$$ lie between $$0$$ and $$15$$; the next-highest value is $$15$$, which is $$25$$ less than $$40$$. The value $$40$$ stands out clearly.

So the outlier is 40.

Remove the outlier $$40$$. The remaining 14 values are:

$$0, 0, 1, 2, 3, 5, 5, 6, 7, 8, 8, 10, 12, 15.$$

Mean. New total $$= 122 - 40 = 82$$. Number of values $$= 14$$.

$$\text{Mean} = \dfrac{82}{14} \approx 5.86.$$

Median. There are 14 values, an even number, so the median is the mean of the two middle values — the 7th and 8th.

7th value $$= 5$$, 8th value $$= 6$$.

$$\text{Median} = \dfrac{5 + 6}{2} = 5.5.$$

Comparison.

The mean fell by more than $$2$$ (from $$8.13$$ to $$5.86$$), while the median moved by only $$0.5$$ (from $$6$$ to $$5.5$$). Again, this shows that the mean reacts strongly to an outlier, but the median is barely affected.

The seven daily counts are $$16, 18, 20, 22, 26, 16, 10$$.

Mean. Total pages over 7 days:

$$16 + 18 + 20 + 22 + 26 + 16 + 10 = 128.$$

$$\text{Mean} = \dfrac{128}{7} \approx 18.29 \text{ pages}.$$

Median. Arrange the values in order: $$10, 16, 16, 18, 20, 22, 26$$. There are 7 values, so the median is the middle (4th) value.

$$\text{Median} = 18 \text{ pages}.$$

The mean and median are very close ($$18.29$$ vs $$18$$) — the data is fairly balanced with no strong outlier.

On the dot plot (0 to 25). Place one dot at 10, two dots at 16, one dot each at 18, 20, and 22. The value 26 lies outside the given range 0–25, so extend the scale or place it just past 25 with a note. Mark the median (18) with an arrow and the mean (about 18.3) with another arrow — they lie very close to each other. Most dots are clustered between 16 and 22, indicating a typical newspaper size of about 18 pages on any given day.

When outliers appear on both sides of the data — a very small value and a very large value — their effects on the mean tend to cancel each other out. Let us see this with an example.

Start with balanced data. Take $$4, 5, 5, 6, 6, 7, 7$$. Mean $$= \dfrac{4+5+5+6+6+7+7}{7} = \dfrac{40}{7} \approx 5.71$$ and median $$= 6$$.

Add a low outlier only. Data: $$0, 4, 5, 5, 6, 6, 7, 7$$. Mean $$= \dfrac{40}{8} = 5.00$$; median $$= \dfrac{5+6}{2} = 5.5$$. Mean fell by $$0.71$$.

Add a high outlier only. Data: $$4, 5, 5, 6, 6, 7, 7, 20$$. Mean $$= \dfrac{60}{8} = 7.50$$; median $$= \dfrac{5+6}{2} = 5.5$$. Mean rose by nearly 2.

Add outliers on both sides. Data: $$0, 4, 5, 5, 6, 6, 7, 7, 20$$. Mean $$= \dfrac{60}{9} \approx 6.67$$; median $$= 6$$ (the middle value). Compare with the original mean $$5.71$$ and median $$6$$: the median is unchanged, and the mean has only moved up a little (because the low outlier partly cancels the high outlier).

Conclusion.

• The median is barely affected by outliers, whether they are on one side or on both.
• The mean is pulled toward each outlier. If the outliers on the two sides are of roughly equal "pull", their effects cancel and the mean stays close to where it was. If one side has a stronger pull, the mean shifts in that direction.

So when outliers are present on both sides, the mean can appear misleadingly "okay" (close to the original) but the data itself is more spread out — the variability has increased even though the summary numbers may not show it. Looking at the dot plot alongside the mean and median is a good habit.

Let us compute the mean and median for the boys, the girls, and the whole class.

Boys (17 students). Total $$= 147+135+130+154+128+135+134+158+155+146+146+142+140+141+144+145+150 = 2430$$.

$$\text{Mean} = \dfrac{2430}{17} \approx 142.94 \text{ cm}.$$

Sorted: $$128, 130, 134, 135, 135, 140, 141, 142, 144, 145, 146, 146, 147, 150, 154, 155, 158$$. There are 17 values, so the median is the 9th value $$= 144$$ cm.

Girls (11 students). Total $$= 143+136+150+144+154+140+145+148+156+150+150 = 1616$$.

$$\text{Mean} = \dfrac{1616}{11} \approx 146.91 \text{ cm}.$$

Sorted: $$136, 140, 143, 144, 145, 148, 150, 150, 150, 154, 156$$. Median $$=$$ 6th value $$= 148$$ cm.

Whole class (28 students). Total $$= 2430 + 1616 = 4046$$. Mean $$= \dfrac{4046}{28} \approx 144.5$$ cm (rounded to about $$144.4$$ cm in the textbook).

Inferences.

• The girls' mean and median heights ($$\approx 146.9$$ and $$148$$ cm) are higher than the boys' ($$\approx 142.9$$ and $$144$$ cm) — in this class the girls are, on average, taller than the boys by about 4 cm. This is quite common at Grade 5, since girls typically start their growth spurt earlier.
• The boys' data has a wider spread ($$128$$ to $$158$$ cm, a range of $$30$$ cm), while the girls' data is more compact ($$136$$ to $$156$$ cm, range $$20$$ cm).
• For each group, the mean and median are close, so there are no strong outliers.

We list every value greater than $$144.4$$ cm.

Boys taller than 144.4 cm. From the sorted boys' list $$128, 130, 134, 135, 135, 140, 141, 142, 144, 145, 146, 146, 147, 150, 154, 155, 158$$, the values above 144.4 are: $$145, 146, 146, 147, 150, 154, 155, 158$$ — that is 8 boys.

Girls taller than 144.4 cm. From the sorted girls' list $$136, 140, 143, 144, 145, 148, 150, 150, 150, 154, 156$$, the values above 144.4 are: $$145, 148, 150, 150, 150, 154, 156$$ — that is 7 girls.

Total students taller than the class average $$= 8 + 7 = 15$$.

From the sorted list of boys' heights $$128, 130, 134, 135, 135, 140, 141, 142, 144, 145, 146, 146, 147, 150, 154, 155, 158$$, the boys whose height exceeds $$144.4$$ cm are $$145, 146, 146, 147, 150, 154, 155, 158$$.

Counting these gives $$8$$ boys.

The "correct" answer for the length of 1 minute is $$60$$ seconds. So the closer a group's mean and median are to $$60$$, the better that group estimated one minute.

Central tendency.

• Group A: mean $$58.21$$ s, median $$60$$ s. The median is exactly right, and the mean is a little below $$60$$ — a few children opened their eyes early.
• Group B: mean $$59.28$$ s, median $$59.5$$ s. Both are very close to $$60$$; the group's typical estimate is a shade under one minute.

Variability. In Group A the gap between mean and median is bigger ($$60 - 58.21 = 1.79$$), which suggests some low outliers pulling the mean down; the dots on that plot are more spread out. In Group B the mean and median are very close ($$59.5 - 59.28 = 0.22$$), which points to a symmetric, tight distribution — the dots are packed near $$60$$.

Overall. Both groups did well on average (mean/median within about $$2$$ seconds of $$60$$). Group B is more consistent — its estimates are more tightly clustered around one minute — while Group A had the exact median but a wider spread. So one could say Group B fared marginally better as a team, though the two are very close.

The 12 monthly prices for each town, arranged in increasing order.

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

Number of values $$= 12$$ (even). Median $$=$$ average of the two middle values (the 6th and 7th).

$$\text{Median}_{\text{Yahapur}} = \dfrac{35 + 39}{2} = 37 \text{ (₹ per kg)}.$$

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

Median $$=$$ average of the 6th and 7th values.

$$\text{Median}_{\text{Wahapur}} = \dfrac{38 + 39}{2} = 38.5 \text{ (₹ per kg)}.$$

So the median monthly onion price is ₹37 in Yahapur and ₹38.50 in Wahapur.

The two dashes ("—") stand for students who were absent, so they gave no data. We drop those and work with the remaining values.

The 20 valid values are:

$$0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4.$$

Mean.

$$\text{Total} = 0+1+0+4+8+0+0+2+1+1+5+3+4+0+0+10+25+2+2+4 = 72.$$

Number of students who gave data $$= 20$$.

$$\text{Mean} = \dfrac{72}{20} = 3.6 \text{ pets per student}.$$

Median. Arrange in increasing order:

$$0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25.$$

Number of values $$= 20$$ (even). Median $$=$$ mean of the 10th and 11th values.

10th value $$= 2$$, 11th value $$= 2$$.

$$\text{Median} = \dfrac{2 + 2}{2} = 2 \text{ pets}.$$

Describing the data. The median is 2 but the mean is 3.6, so mean is much larger than median. Looking at the sorted list, most students have very few pets (six students have zero, and most others 1–5), but two students have unusually large numbers — 10 and especially 25. The value 25 is a clear outlier; it single-handedly pulls the mean up. So the data is right-skewed: most values are small, with a long tail of a few very large values.

There are $$29$$ trees. Let us sort them first so the dot plot and median are easy.

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

Dot plot. On a horizontal number line from about $$43$$ to $$67$$ feet, place one dot above each value: one dot each at 43, 44, 45, 46, 49, 50, 51; two dots at 52; two at 54; two at 55; two at 56; one at 57, 58, 59; four dots at 60; two at 61; one each at 62, 63, 65, 66, 67. Most of the dots cluster between about 50 ft and 62 ft.

Mean. Total (add the sorted list):

$$43+44+45+46+49+50+51+52+52+54+54+55+55+56+56+57+58+59+60+60+60+60+61+61+62+63+65+66+67 = 1621.$$

$$\text{Mean} = \dfrac{1621}{29} \approx 55.90 \text{ feet}.$$

Median. $$29$$ is odd, so the median is the middle (15th) value of the sorted list. Counting: positions 1–14 use the values up to and including 56, and position 15 is $$56$$. So median $$= 56$$ ft.

Description. Mean $$\approx 55.9$$ and median $$= 56$$ are very close. The heights range from 43 to 67 ft, and most trees are 50–62 ft tall — a fairly symmetric cluster around 56 ft, with no outliers.

A quicker way to find the mean (assumed-mean method). Guess a round number close to the middle, say $$A = 55$$. Find how far each value is from $$A$$ (positive if above, negative if below), average these differences, and add back to $$A$$.

Differences from 55 (in order):
$$-12, -11, -10, -9, -6, -5, -4, -3, -3, -1, -1, 0, 0, 1, 1, 2, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8, 10, 11, 12$$.

Sum of differences $$= 26$$.

$$\text{Mean} = 55 + \dfrac{26}{29} \approx 55 + 0.90 = 55.90 \text{ ft}.$$

Same answer, but with smaller numbers to add.

How many trees are shorter than the mean (55.90 ft)? From the sorted list, the values less than $$55.9$$ are $$43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55$$ — that is 13 trees.

Dot plot. Use a horizontal number line from about $$2.4$$ to $$4.2$$ kg, with tick marks every $$0.1$$ kg. Place blue dots (say) for boys above 2.6, 3.2, 3.4, 3.5, 3.8, 4.1 (one each) and orange dots for girls above 2.5, 3.1, 3.4, 3.4, 3.7, 4.0.

Boys — 6 values. Sorted: $$2.6, 3.2, 3.4, 3.5, 3.8, 4.1$$.

Total $$= 2.6 + 3.2 + 3.4 + 3.5 + 3.8 + 4.1 = 20.6$$ kg.

$$\text{Mean}_{\text{boys}} = \dfrac{20.6}{6} \approx 3.43 \text{ kg}.$$

Median $$= \dfrac{3.4 + 3.5}{2} = 3.45$$ kg (average of 3rd and 4th values).

Girls — 6 values. Sorted: $$2.5, 3.1, 3.4, 3.4, 3.7, 4.0$$.

Total $$= 2.5 + 3.1 + 3.4 + 3.4 + 3.7 + 4.0 = 20.1$$ kg.

$$\text{Mean}_{\text{girls}} = \dfrac{20.1}{6} = 3.35 \text{ kg}.$$

Median $$= \dfrac{3.4 + 3.4}{2} = 3.4$$ kg.

Analysis and comparison.

• The mean weight of the boys ($$\approx 3.43$$ kg) is slightly greater than that of the girls ($$3.35$$ kg) — a difference of only about $$80$$ g.
• The medians ($$3.45$$ kg for boys, $$3.4$$ kg for girls) tell the same story.
• The boys' weights range from $$2.6$$ to $$4.1$$ kg (range $$1.5$$ kg); the girls' range from $$2.5$$ to $$4.0$$ kg (range $$1.5$$ kg). Very similar spread.
• For both groups the mean and median are close, so there is no strong outlier.

Overall, boys and girls have very similar weight distributions; boys are on average marginally heavier.

We are given the following summaries for this second section:

Observations.

• Boys vs girls in this section. The boys' mean and median (about $$142$$–$$143$$ cm) are a little higher than the girls' (about $$140$$ cm). So in this section, boys are on average slightly taller than girls — a difference of only about $$2$$ cm.
• Mean vs median for the whole class. Whole-class mean $$141.21$$ cm is a bit less than the median $$142.5$$ cm. Since mean $$<$$ median, there is a slight pull from a few short students (low-side outliers or a small tail on the left of the dot plot).
• Comparison with the earlier section (mean $$\approx 144.4$$ cm, girls taller than boys). This new section is shorter on average — by about 3 cm — and here the boys are taller than the girls, opposite to what we saw before.
• The gaps between mean and median in each group are small (at most $$1.3$$ cm), so there are no strong outliers in this section.

To compare two groups of different sizes, use the averages.

Average weight of a sumo wrestler.

Total $$= 295.2 + 250.7 + 234.1 + 221.0 + 200.9 = 1201.9$$ kg.

$$\text{Mean}_{\text{sumo}} = \dfrac{1201.9}{5} \approx 240.38 \text{ kg}.$$

Average weight of a ballet dancer.

Total $$= 40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 = 254.5$$ kg.

$$\text{Mean}_{\text{ballet}} = \dfrac{254.5}{6} \approx 42.42 \text{ kg}.$$

Ratio.

$$\dfrac{\text{sumo mean}}{\text{ballet mean}} = \dfrac{240.38}{42.42} \approx 5.67.$$

So on average a sumo wrestler weighs about $$5\tfrac{2}{3}$$ times — roughly $$5$$ to $$6$$ times — as much as a ballet dancer.

Bring together what we know about the two sections.

Observations.

• Section 1 is taller on average (mean $$\approx 144.4$$ cm) than Section 2 (mean $$\approx 141.2$$ cm) by about $$3$$ cm.
• In Section 1 the girls are noticeably taller than the boys (146.9 vs 142.9). In Section 2 the boys are slightly taller than the girls (142.1 vs 140.1). The pattern is reversed.
• The boys' mean is about the same in both sections ($$\approx 142$$ cm). The big difference is in the girls: much taller in Section 1 than in Section 2.
• In both sections the mean and median are close, so neither section has strong outliers.

Prices in the onion table range from a low of ₹17 per kg to a high of ₹60 per kg. For a bar graph to fit in one page, each unit on the vertical axis stands for a fixed number of rupees.

In the textbook's graph the vertical axis is marked at $$0, 10, 20, 30, 40, 50, 60$$ (rupees per kg). Every big gridline represents ₹10, so the scale is:

$$1 \text{ unit} = ₹10 \text{ per kg}.$$

The horizontal axis just lists the twelve months, with two coloured bars (one for Yahapur, one for Wahapur) side by side for each month.

Yes. In a clustered column graph the two bars for the same month are drawn right next to each other. Just by glancing at each pair we can immediately see:

• which town had the higher price that month (the taller bar),
• how big the gap between the two towns is (difference in bar heights), and
• how the prices in either town rise or fall over the year (the pattern of bar heights across months).

With the original table one has to read two numbers, subtract them and repeat that for every month; with the graph the comparison is visual and instant.

First identify the country each organisation belongs to.

Observations from the graph.

• SpaceX (USA) has by far the largest bars and is growing very fast each year — from around 30 launches in 2021 to about 60 in 2022 and nearly 100 in 2023.
• CASC (China) is the second biggest launcher and has also grown steadily (roughly 50 → 55 → 65).
• Roscosmos (Russia) has stayed roughly the same or even decreased over the three years.
• Arianespace (France) has very few launches (only a handful each year).
• ISRO (India) launches a small number every year, and the number has been going up (roughly 2, 5, 7).
• Chinese private companies (Galactic Energy, Expace) have small but slowly rising bars.
• Overall the total number of worldwide launches has grown sharply, mainly because of SpaceX and CASC.

How the graph is organised. The graph is a clustered horizontal bar graph. The organisations are listed one below the other on the vertical axis. For each organisation there are three adjacent bars — one each for 2021, 2022 and 2023, drawn in different shades of blue as shown in the legend. The horizontal axis measures the number of launches.

Scale. The horizontal axis is marked at 0, 20, 40, 60, 80, 100. Each big step covers 20 launches, so the scale is $$1 \text{ unit length} = 20 \text{ launches}$$.

Patterns.

• The three bars for SpaceX rise sharply from 2021 to 2023 — clearly the fastest growth.
• CASC's three bars are all tall and rise gently — steady growth in the Chinese program.
• Roscosmos's three bars are of similar height (a slight downward tendency).
• Most other organisations have short bars, and some (Arianespace, ULA) have tiny or even shrinking bars.
• The colour for 2023 is generally the tallest in each group, meaning most organisations launched more in 2023 than in the earlier years.

Take each observation and think about what it means in the real world.

• SpaceX's rapid rise. SpaceX uses reusable rockets, which makes each launch cheaper. Being able to reuse a rocket lets them launch many more times per year — this is the main reason for the sharp growth.
• CASC's steady rise. China's national space program has been expanding — more satellites, its own space station and Moon-mission tests. So CASC has more work each year.
• Roscosmos is flat or declining. Russia's space program has been slowed by economic and political factors, and by losing some foreign customers (many of whom now use SpaceX).
• Arianespace, ULA (small bars). These are older programs that are being replaced by newer rockets; also SpaceX has taken away a lot of their commercial launch business.
• ISRO growing gently. India's space program is also expanding, with more commercial and scientific missions each year, but the numbers are still small compared to the biggest launchers.
• 2023 tallest across most organisations. The overall demand for launches — for communication satellites, imaging satellites, science missions — is growing worldwide.

"Consistently launched more every year" means the 2022 bar is taller than the 2021 bar and the 2023 bar is taller than the 2022 bar. Scanning the graph:

• SpaceX — bars roughly $$31 \to 61 \to 96$$: clearly rising each year. ✓
• CASC — bars roughly $$48 \to 53 \to 65$$: rising each year. ✓
• ISRO — bars roughly $$2 \to 5 \to 7$$: rising each year. ✓

Other organisations do not show the same strictly-rising pattern (Roscosmos, Arianespace, ULA even drop in some years).

Add up the 2023 bars of every organisation. Reading approximate heights from the graph:

• SpaceX $$\approx 96$$
• CASC $$\approx 65$$
• Roscosmos $$\approx 19$$
• Rocket Lab $$\approx 10$$
• ULA $$\approx 3$$
• Arianespace $$\approx 3$$
• ISRO $$\approx 7$$
• Galactic Energy $$\approx 3$$
• Expace $$\approx 1$$
• Other $$\approx 15$$

Total $$\approx 96 + 65 + 19 + 10 + 3 + 3 + 7 + 3 + 1 + 15 = 222$$.

This is between $$200$$ and $$400$$, so the correct option is (b).

How the numbers are turned into daily averages. The table shows total monthly sunshine hours. To find the average per day for a month, divide the total by the number of days in that month.

For example, in January (31 days):

$$\text{City 1 average per day} = \dfrac{210}{31} \approx 6.8 \text{ hours/day},$$

$$\text{City 2 average per day} = \dfrac{459}{31} \approx 14.8 \text{ hours/day}.$$

(Note: the totals in the table are unusually large because they mix daylight-hours and sunshine differently in NCERT's example; the point is only to draw the daily-average bar graph.)

How the graph is organised. Along the horizontal axis are the 12 months (Jan to Dec). For each month there are two bars side by side — one for City 1 and one for City 2, in two different colours (see legend). The vertical axis shows average daylight hours per day.

Scale. Since the values range up to about $$19$$ hours per day, the vertical scale is marked in equal steps of $$5$$ hours (0, 5, 10, 15, 20). So the scale is $$1$$ unit $$= 5$$ hours per day.

Patterns.

• City 1: the bar heights grow from January, peak in June/July, and fall back down to December — the shape rises then falls.
• City 2: the pattern is the opposite. Bar heights are highest in December/January, drop through the middle of the year, hit their minimum in June/July, and rise again toward December.
• For every month, one city's bar is high while the other's is comparatively low — the two patterns are almost mirror-images of each other.

Every observation from the previous question can be interpreted in terms of the seasons.

• City 1 peaks in June/July. Long daylight in June and July is a summer pattern for places in the Northern Hemisphere. The days are longest in June (around the summer solstice) and shortest in December.
• City 2 peaks in December/January. Long daylight in December is a summer pattern for places in the Southern Hemisphere. Days are shortest around June.
• The mirror-image behaviour. When one hemisphere has summer, the other has winter. So City 1 and City 2 always "trade" — whichever has long days now, the other will six months later.
• The daily averages themselves tell us how far from the equator each city is: the larger the seasonal swing, the further from the equator.

Summary/conclusion. City 1 and City 2 are in opposite hemispheres. City 1 (peak in June) is a Northern-Hemisphere city; City 2 (peak in December) is a Southern-Hemisphere city. The seasons in the two cities are exactly six months apart.

Yes. The month in which daylight is longest immediately hints at which hemisphere a city belongs to.

• City 1's longest daylight is in June — this is characteristic of the Northern Hemisphere (e.g. cities like Delhi, London, or New York).
• City 2's longest daylight is in December — this is characteristic of the Southern Hemisphere (e.g. cities like Sydney, Cape Town, or Buenos Aires).

The size of the seasonal swing (difference between highest and lowest daylight) also tells us how far the city is from the equator — a larger swing means a location closer to the poles. From the numbers, both cities show fairly large swings, so both are probably at moderate to high latitudes in their respective hemispheres.

This is an open-ended "wondering" question. Some things one might want to explore are:

• Exactly which cities these numbers correspond to (guess names based on latitude).
• How the pattern would look for a city on the equator (should be nearly the same daylight all year round).
• How the pattern changes as we move closer to the poles (longer summer days, shorter winter days, until we reach the polar circles where there is 24-hour daylight or 24-hour night for part of the year).
• Whether the shortest-day (winter solstice) and longest-day (summer solstice) dates are the same for every city in a hemisphere.
• How this seasonal daylight pattern affects things like farming, birds' migration, our sleep, and electricity use.

Step 1: Make a frequency table by counting the letters.

Grade 5 (25 students): Going through the letters one by one — w a a a w n s a n w a a a a a w s s a a n w a a n — we count:

• Water (w): 5
• Air (a): 13
• Space (s): 3
• None (n): 4

Check: $$5 + 13 + 3 + 4 = 25$$. ✓

Grade 9 (25 students): n w s a s w s s a a w s s a s a n w s s a w a w a — count:

• Water (w): 6
• Air (a): 8
• Space (s): 9
• None (n): 2

Check: $$6 + 8 + 9 + 2 = 25$$. ✓

Step 2: Draw the double-bar graph. On the horizontal axis mark the four choices (Water, Air, Space, None). For each choice draw two adjacent bars — one for Grade 5 (say, blue) and one for Grade 9 (say, orange). The largest count is 13, so a convenient scale is $$1 \text{ unit} = 1 \text{ student}$$ (or 1 unit $$= 2$$ students to make the graph shorter).

Step 3: Observations.

• Grade 5 mostly chose Air (13 out of 25); Grade 9 mostly chose Space (9 out of 25).
• Grade 9 has three times as many "Space" choices as Grade 5 (9 vs 3).
• Grade 5 has slightly more "None" (undecided) than Grade 9.
• "Water" is roughly equally popular in both grades (5 vs 6).

Step 1: Draw the graph. On the horizontal axis mark the eight time slots (12 am, 3 am, 6 am, …, 9 pm). For each time draw two bars side by side — one for Day 1 and one for Day 2, in two different colours. On the vertical axis mark $$0, 4, 8, 12, \ldots, 44$$ °C so that $$1$$ unit $$= 4^{\circ}\mathrm{C}$$. Draw each Day 1 and Day 2 bar to the corresponding height. For example, $$20^{\circ}\mathrm{C}$$ has height $$\dfrac{20}{4} = 5$$ units, and $$43^{\circ}\mathrm{C}$$ has height $$\dfrac{43}{4} \approx 10.75$$ units.

Step 2: Observations from the graph.

• On both days the temperature is lowest around $$6$$ am and highest around $$3$$ pm — the daily cycle we expect.
• Every Day 2 bar is much taller than the corresponding Day 1 bar; on average Day 2 is about $$14$$–$$16^{\circ}\mathrm{C}$$ hotter.

Step 3: Guessing the months. Take the average of the eight readings for each day:

Day 1: $$\dfrac{20+18+16+20+26+34+30+24}{8} = \dfrac{188}{8} = 23.5^{\circ}\mathrm{C}$$. Minimum $$16^{\circ}\mathrm{C}$$, maximum $$34^{\circ}\mathrm{C}$$. Such moderate temperatures in Jodhpur match a spring or autumn month like February or March (or October–November).

Day 2: $$\dfrac{37+34+30+33+37+43+42+39}{8} = \dfrac{295}{8} \approx 36.9^{\circ}\mathrm{C}$$. Minimum $$30^{\circ}\mathrm{C}$$, peak $$43^{\circ}\mathrm{C}$$. Jodhpur reaches such extreme heat in peak summer — May or June.

Compare the mean heights across the two schools:

Observations.

• In each school, average height increases as grade goes up (older students are taller). This matches how children grow.
• For every grade, School B students are noticeably taller than School A students — the gap is about $$14$$–$$15$$ cm in Grade 6 and Grade 7. So School B seems to have taller children overall.
• Within each school and each grade, boys and girls have almost the same mean height (differences are less than 2 cm). No one gender is clearly taller than the other at this age band.
• The jump in height from Grade 6 to Grade 7 is about $$7$$ cm in School A and $$6$$ cm in School B. Between Grade 7 and Grade 8 the jump is smaller — for School B, boys' mean is almost the same in Grade 7 and Grade 8 ($$156.14$$ cm both times).

Some key findings from the table:

1. Change over 30 years (1989 → 2019). For nearly every age, both boys and girls became taller. For example, at age 19: boys went from $$163.5$$ cm (1989) to $$166.5$$ cm (2019), a gain of $$3$$ cm. Girls at age 19 went from $$151.9$$ to $$155.2$$ cm — also about $$3$$ cm gain. The rise is small at very young ages and clearer at older ages.

2. Boys vs girls in a single year. Take 2019.

• Ages 5–8: boys and girls have almost the same heights.
• Ages 9–11: boys and girls are very close (girls even a hair taller at 11: $$137.0$$ vs $$138.6$$).
• Ages 12–13: girls are still catching up; heights are similar.
• Ages 14–19: boys become clearly taller than girls, with the gap widening to about $$11$$ cm at age 19 ($$166.5$$ vs $$155.2$$).

So boys are not always taller — around ages 10–12 girls almost match or slightly exceed boys because girls start their growth spurt earlier.

3. Growth between successive ages in 2019.

Boys' fastest growth years are $$12 \to 14$$ (about $$6$$ cm each year). Girls' fastest growth years are $$10 \to 12$$ (about $$5$$–$$6$$ cm each year). After age $$16$$ or so, both slow down considerably.

For each successive pair of ages, subtract the earlier mean from the later mean (in the 2019 columns). The pair with the largest difference is the one where growth was fastest.

Boys (2019). The consecutive gains are approximately:

$$6.0, 5.5, 4.9, 4.6, 4.5, 4.4, 5.2, 6.2, 6.0, 4.6, 3.6, 2.0, 1.4, 0.5.$$

The biggest gain is $$6.2$$ cm — between age $$12$$ and age $$13$$ (heights $$142.2 \to 148.4$$).

Girls (2019). The consecutive gains are approximately:

$$5.7, 5.1, 4.7, 4.9, 5.2, 5.8, 5.2, 3.9, 2.7, 2.0, 1.4, 0.9, 0.5, 0.0.$$

The biggest gain is $$5.8$$ cm — between age $$10$$ and age $$11$$ (heights $$132.8 \to 138.6$$).

This is consistent with what we know biologically: girls hit their growth spurt earlier (around ages 10–12) than boys (around ages 12–14).

The table starts at age 5 with a height of about $$107$$ cm (2019 boys). Given a newborn is $$50$$ cm, the total gain from birth to age 5 is about $$107 - 50 = 57$$ cm over 5 years — an average of about $$11$$ cm per year.

However, growth is fastest in the first year of life (about $$25$$ cm), then slows down. A reasonable estimate is:

So a child grows rapidly in the first year (roughly to $$75$$ cm at age 1), then adds about $$10$$–$$12$$ cm in year 2, and about $$6$$–$$8$$ cm each year through ages 3 and 4, reaching about $$100$$–$$105$$ cm by age 4.

Over the 30 years from 1989 to 2019, average heights rose only a little — typically about $$1$$–$$3$$ cm per age. If the same trend continues, from 2019 to 2029 (10 more years, i.e. one-third of that period), we can expect roughly a further $$0.5$$–$$1$$ cm rise at each age.

Simple estimation rule. Add about $$1$$ cm (a bit more for older ages, a bit less for younger ages) to each 2019 value.

(These are only estimates — real data in 2029 could be affected by nutrition, health care and other factors.)

Organisation. The horizontal axis lists a set of countries — Timor-Leste, Yemen, Bangladesh, Liberia, Indonesia, India, Bhutan, …, up to Iceland and Netherlands. For every country there are four markers drawn as small symbols:

• B-1989 (blue "×") — 19-year-old boys' average height in 1989
• B-2019 (blue triangle) — 19-year-old boys' average height in 2019
• G-1989 (orange "×") — 19-year-old girls' average height in 1989
• G-2019 (orange triangle) — 19-year-old girls' average height in 2019

The vertical axis shows height in cm, starting from $$145$$ cm (a zoomed-in view) up to about $$185$$ cm.

Information presented. For every country we can see (i) how tall 19-year-old boys and girls were in 1989 and in 2019 and (ii) how much they grew (or shrank) between those two years. Sorting the countries left to right roughly by 2019 boys' heights lets us compare countries at a glance.

A few things stand out.

• The countries on the right (Netherlands, Iceland, Ukraine, Norway) have the tallest 19-year-olds — boys are around $$180$$ cm on average. The countries on the left (Timor-Leste, Yemen, Bangladesh) have the shortest — boys are only about $$160$$ cm on average.
• For almost every country the 2019 markers are higher than the 1989 markers — average heights have gone up worldwide.
• Some countries have had a very large increase: Indian and Chinese 19-year-olds are noticeably taller in 2019 than in 1989.
• In every country, 19-year-old boys are taller than 19-year-old girls (the two orange markers stay clearly below the two blue markers).
• The gap between the tallest and the shortest country is around $$20$$ cm — bigger than the average change over 30 years in any one country. Where you live matters a lot for average height (probably because of nutrition and health).
• The vertical axis starts at $$145$$ cm (not $$0$$) — the graph zooms in to show small differences clearly, but a hurried reader could mistake a moderate difference for a huge one.

Group 1 (10 students). Scores: $$85, 76, 90, 85, 39, 48, 56, 95, 81, 75$$.

Total $$= 85+76+90+85+39+48+56+95+81+75 = 730$$.

$$\text{Mean}_1 = \dfrac{730}{10} = 73.$$

Sorted: $$39, 48, 56, 75, 76, 81, 85, 85, 90, 95$$. There are 10 values, so median $$=$$ mean of 5th and 6th values:

$$\text{Median}_1 = \dfrac{76 + 81}{2} = 78.5.$$

Group 2 (9 students). Scores: $$68, 59, 73, 86, 47, 79, 90, 93, 86$$.

Total $$= 68+59+73+86+47+79+90+93+86 = 681$$.

$$\text{Mean}_2 = \dfrac{681}{9} \approx 75.67.$$

Sorted: $$47, 59, 68, 73, 79, 86, 86, 90, 93$$. 9 values, so median $$=$$ 5th value:

$$\text{Median}_2 = 79.$$

Comparison and description.

Both groups performed well. Group 2 has a slightly higher mean and median, so on the whole Group 2 did marginally better. In each group the mean is a fair bit below the median, meaning a few low scores (39, 48 for Group 1; 47 for Group 2) pulled the mean down. The typical (median) student in either group scored around $$78$$–$$79$$ out of 100.

Choosing a scale. The largest value is $$1240$$, and most values are multiples of about $$100$$. A convenient scale is $$1 \text{ unit} = 100 \text{ people}$$, so the vertical axis is marked at $$0, 100, 200, \ldots, 1300$$.

How to draw the graph. Along the horizontal axis mark the five sports (Cricket, Basketball, Swimming, Hockey, Athletics). For each sport draw two adjacent bars — a coloured one for "Watching" and another colour for "Participating". Bar heights (in units):

Observations.

• Cricket is by far the most popular — both for watching (1240) and for participating (620). Nothing else comes close.
• For every sport, the number who watch is roughly twice the number who participate. This is very clear for cricket ($$1240$$ vs $$620$$), basketball ($$470$$ vs $$320$$), hockey ($$430$$ vs $$250$$), athletics ($$250$$ vs $$105$$).
• Basketball, Swimming and Hockey are watched by roughly the same number of people (around 400–500).
• Athletics has the fewest people in both categories.

Step 1: Sort the 17 heights.

$$101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125.$$

Step 2: Splitting. The number of students is 17 (odd), so we cannot split them into two exactly equal halves; the best we can do is 8 vs 9 (or 9 vs 8). The natural cut-off is the median — the 9th value, which is $$115$$ cm.

Using $$115$$ cm as the threshold:

• Heights less than $$115$$ cm: 101, 102, 106, 109, 110, 110, 112 — that is $$7$$ students.
• Height equal to $$115$$ cm: 115, 115, 115, 115, 115 — $$5$$ students.
• Heights greater than $$115$$ cm: 117, 120, 120, 123, 125 — $$5$$ students.

To get an 8-and-9 split (equal in size to within 1), the teacher can put 1 of the five 115-cm students into the "shorter" group and the other 4 into the "taller" group. Then:

• Shorter group ($$\le 115$$ cm, one 115 included): $$7 + 1 = 8$$ students.
• Taller group ($$\ge 115$$ cm, four 115s included): $$4 + 5 = 9$$ students.

(Or swap the roles — the point is that the median $$115$$ cm is the natural dividing height.)

Step 3: Estimating the age from the Telling Tall Tales table. The 17 heights range from $$101$$ cm to $$125$$ cm, with a median of $$115$$ cm and a mean of $$\dfrac{101+102+106+109+110+110+112+115+115+115+115+115+117+120+120+123+125}{17} = \dfrac{1930}{17} \approx 113.5$$ cm.

Looking at the 2019 columns of the Telling Tall Tales table, the average height of $$107$$ cm corresponds to age $$5$$, $$113$$ cm to age $$6$$–$$7$$, $$118$$–$$120$$ cm to age $$7$$–$$8$$. So a group with an average height of about $$113$$–$$115$$ cm most closely matches age $$6$$ or $$7$$ (Class 1 or 2). These are young primary-school students.

This is an activity. Steps to do it:

1. Collect the heights (in cm) of every student in your class.
2. Add all the heights and divide by the number of students to get the mean.
3. Sort the heights in increasing order. If there are $$n$$ students, the median is the middle value (if $$n$$ is odd) or the average of the two middle values (if $$n$$ is even).
4. Draw a horizontal number line from the smallest height to the largest and put one dot above each height value — this is your dot plot.

Sample working. Suppose a Class 7 has 20 students with heights (in cm): $$138, 140, 141, 142, 142, 143, 144, 145, 146, 147, 147, 148, 149, 150, 151, 152, 153, 155, 157, 160$$.

Total $$= 2950$$; mean $$= \dfrac{2950}{20} = 147.5$$ cm.

Sorted median (average of 10th and 11th values) $$= \dfrac{147 + 147}{2} = 147$$ cm.

Mean and median are almost the same ($$147$$–$$147.5$$ cm), so the class's heights are fairly symmetric with no strong outliers.

Your own numbers will depend on your class — the key is to make both the mean and the median from the actual data you collect.

Both sections have the same number of students (30 each), but they have different children. Knowing the mean of one section tells us nothing about the actual heights of the students in the other section — those children could be taller, shorter, or about the same on average.

Only additional information (e.g. the other section's data, or an assumption that the two sections are drawn from an identical population) could pin down the mean. Without it, the mean of the other section is not determined.

So the correct option is (d).

This is a hands-on activity. Steps:

1. Before touching a ruler, write down your estimate (guess) of each length in cm.
2. Then use a ruler and write down the actual measurement in cm.
3. Positive difference $$=$$ |Estimate $$-$$ Measure|.
4. Draw a double-bar graph with the five objects on the x-axis, two bars (Estimate and Measure) per object.
5. Average difference $$= \dfrac{\text{sum of all 5 positive differences}}{5}$$.

Sample data (yours will differ).

Sum of differences $$= 1 + 1 + 2 + 3 + 1 = 8$$ cm. Average difference $$= \dfrac{8}{5} = 1.6$$ cm — the estimates were off by about $$1.6$$ cm on average, which is reasonably close.

For the double bar graph: choose scale $$1$$ unit $$= 2$$ cm. For each object draw two adjacent bars — one for the estimate, one for the measured value.