Chapter 5: Tales by Dots and Lines

Let us pick a few pairs and find their means:

Pair $$3, 7$$: mean $$= \dfrac{3+7}{2} = 5$$.

Pair $$8, 9$$: mean $$= \dfrac{8+9}{2} = 8.5$$.

Pair $$4, 20$$: mean $$= \dfrac{4+20}{2} = 12$$.

Pair $$11, 15$$: mean $$= \dfrac{11+15}{2} = 13$$.

Observation: In every case the mean lies exactly halfway between the two numbers. If the two numbers are $$a$$ and $$b$$, then the distance of the mean $$\dfrac{a+b}{2}$$ from $$a$$ equals its distance from $$b$$ (both equal $$\dfrac{|b-a|}{2}$$). So the mean of two numbers is the midpoint of the pair.

(a) Values $$6, 7, 8$$: $$\text{mean} = \dfrac{6+7+8}{3} = \dfrac{21}{3} = 7$$. Mark the mean at $$7$$.

(b) Values $$3, 6, 9$$: $$\text{mean} = \dfrac{3+6+9}{3} = \dfrac{18}{3} = 6$$. Mark the mean at $$6$$.

(c) Values $$2, 4, 8$$: $$\text{mean} = \dfrac{2+4+8}{3} = \dfrac{14}{3} \approx 4.67$$. Mark the mean at $$\dfrac{14}{3}$$.

(d) Values $$3, 9, 15$$: $$\text{mean} = \dfrac{3+9+15}{3} = \dfrac{27}{3} = 9$$. Mark the mean at $$9$$.

The mean can be viewed as a balance point of the collection. For each collection, if we look at the distances of every value from the mean, the total distance of values that are less than the mean equals the total distance of values that are greater than the mean.

(a) Values $$6, 7, 8$$, mean $$= 7$$. Distance below: $$|6-7| = 1$$. Distance above: $$|8-7| = 1$$. Both sides balance ($$1 = 1$$).

(b) Values $$3, 6, 9$$, mean $$= 6$$. Below: $$|3-6| = 3$$. Above: $$|9-6| = 3$$. Balanced.

(c) Values $$2, 4, 8$$, mean $$= \dfrac{14}{3}$$. Below: $$\left(\dfrac{14}{3}-2\right) + \left(\dfrac{14}{3}-4\right) = \dfrac{8}{3} + \dfrac{2}{3} = \dfrac{10}{3}$$. Above: $$8 - \dfrac{14}{3} = \dfrac{10}{3}$$. Balanced.

(d) Values $$3, 9, 15$$, mean $$= 9$$. Below: $$|3-9| = 6$$. Above: $$|15-9| = 6$$. Balanced.

So in each collection the mean is a point where the total distance of smaller values matches the total distance of larger values — a natural centre.

(a) Values $$10, 12, 14, 16, 18$$: $$\text{mean} = \dfrac{10+12+14+16+18}{5} = \dfrac{70}{5} = 14$$. Mark the mean at $$14$$.

(b) Values $$4, 5, 15, 16$$: $$\text{mean} = \dfrac{4+5+15+16}{4} = \dfrac{40}{4} = 10$$. Mark the mean at $$10$$.

(c) Values $$9, 10, 17$$: $$\text{mean} = \dfrac{9+10+17}{3} = \dfrac{36}{3} = 12$$. Mark the mean at $$12$$.

(d) Values $$2, 3, 10, 12$$: $$\text{mean} = \dfrac{2+3+10+12}{4} = \dfrac{27}{4} = 6.75$$. Mark the mean at $$6.75$$.

As before, the mean acts as a balance point: the total distance from the mean to the smaller values equals the total distance from the mean to the larger values.

(a) $$10, 12, 14, 16, 18$$, mean $$= 14$$. LHS: $$(14-10)+(14-12)=4+2=6$$. RHS: $$(16-14)+(18-14)=2+4=6$$. Balanced.

(b) $$4, 5, 15, 16$$, mean $$= 10$$. LHS: $$(10-4)+(10-5)=6+5=11$$. RHS: $$(15-10)+(16-10)=5+6=11$$. Balanced.

(c) $$9, 10, 17$$, mean $$= 12$$. LHS: $$(12-9)+(12-10)=3+2=5$$. RHS: $$(17-12)=5$$. Balanced.

(d) $$2, 3, 10, 12$$, mean $$= 6.75$$. LHS: $$(6.75-2)+(6.75-3)=4.75+3.75=8.5$$. RHS: $$(10-6.75)+(12-6.75)=3.25+5.25=8.5$$. Balanced.

Notice that the mean need not be the midpoint of the extreme values — it is the point where the total distances on either side are equal.

For every collection encountered so far we compute LHS $$=$$ sum of (mean $$-$$ each smaller value) and RHS $$=$$ sum of (each larger value $$-$$ mean) and check LHS $$=$$ RHS.

In every case LHS equals RHS, so the mean is indeed the balancing centre for each collection.

No — there is only one such balance point, and it is the mean.

Take any value $$c$$ that is larger than the mean $$a$$. Compared with the balance at $$a$$, every distance from a smaller value increases by $$(c-a)$$ and every distance from a larger value decreases by $$(c-a)$$ (values that were larger than $$a$$ but smaller than $$c$$ actually flip sides — but even then the LHS grows and RHS shrinks overall). So LHS becomes bigger than RHS and the balance is lost.

Symmetrically, if we take $$c$$ smaller than $$a$$, then LHS shrinks and RHS grows, again breaking the balance.

Hence the only value that keeps the two sums of distances equal is the mean itself.

At the mean $$12$$, LHS $$= (12-10)+(12-10)+(12-11) = 2+2+1 = 5$$ and RHS $$= 17-12 = 5$$; the two sides balance.

Now try a value $$c > 12$$. Since $$10, 10, 11$$ are all less than $$c$$, each of these distances increases compared to the distances from $$12$$. Meanwhile the value $$17$$, which is still larger than $$c$$ (as long as $$c<17$$), contributes a smaller distance $$17-c$$. So:

LHS increases while RHS decreases $$\Longrightarrow$$ LHS $$>$$ RHS.

For example, at $$c = 13$$: LHS $$= 3+3+2 = 8$$, RHS $$= 17-13 = 4$$ — clearly unequal. So no value larger than $$12$$ can serve as the centre. By the same reasoning, no value smaller than $$12$$ can either. The mean $$12$$ is the unique centre.

Suppose the current data has $$n$$ values with mean $$a$$, so the total is $$na$$. Include a new value $$v$$. The new total is $$na + v$$ and the new mean is:

$$a' = \dfrac{na + v}{n+1}.$$

Compare with $$a$$: $$a' - a = \dfrac{na+v}{n+1} - a = \dfrac{na + v - a(n+1)}{n+1} = \dfrac{v - a}{n+1}.$$

So the change in mean has the same sign as $$(v-a)$$:

• If $$v > a$$ (new value greater than the old mean), the mean increases.
• If $$v < a$$, the mean decreases.
• If $$v = a$$, the mean stays the same.

Suppose there are $$n$$ values with mean $$a$$, so the total is $$na$$. Remove a value $$v$$. The new total is $$na - v$$ and the new mean is:

$$a' = \dfrac{na - v}{n-1}.$$

Compare with $$a$$: $$a' - a = \dfrac{na - v}{n-1} - a = \dfrac{na - v - a(n-1)}{n-1} = \dfrac{a - v}{n-1}.$$

So the change has the same sign as $$(a - v)$$:

• If $$v < a$$ (we removed a value smaller than the mean), then $$a - v > 0$$, so the mean increases.
• If $$v > a$$, the mean decreases.
• If $$v = a$$, the mean stays the same.

Removing a value pulls the mean away from that value.

The mean does not change.

Algebra: If the current mean is $$a$$ with $$n$$ values (total $$na$$), then including a value equal to $$a$$ gives new mean $$\dfrac{na + a}{n+1} = \dfrac{a(n+1)}{n+1} = a$$. Removing a value equal to $$a$$ gives $$\dfrac{na - a}{n-1} = \dfrac{a(n-1)}{n-1} = a$$.

Fair-share interpretation: The mean is what each person would get if the total were shared equally. If everyone already has $$a$$ units and one more person joins with exactly $$a$$ units, no one has to give up or receive anything — each person still ends up with $$a$$. Similarly, if a person who had exactly $$a$$ units leaves, the remaining people still have $$a$$ each. So the fair share does not change.

Yes, it is possible. If we add two values $$u$$ and $$v$$ to a data of $$n$$ values with mean $$a$$, the new mean is:

$$a' = \dfrac{na + u + v}{n+2}.$$

For $$a' = a$$ we need $$na + u + v = a(n+2)$$, i.e. $$u + v = 2a$$, so the two new values must have average $$a$$.

For the shown data (mean $$= 9$$), any pair whose sum is $$18$$ works. Examples:

• Add $$4$$ and $$14$$ (both new values average to $$9$$). Sum increases by $$18$$, count by $$2$$, and mean stays $$9$$.
• Add $$6$$ and $$12$$, or $$5$$ and $$13$$, or $$9$$ and $$9$$ itself.

The same rule applies to removal: remove any two values whose average equals the mean. From the shown data one could remove two dots whose values sum to $$18$$ (say a $$7$$ and an $$11$$) and the mean would still be $$9$$.

Yes. Adding three new values $$u, v, w$$ to a data of $$n$$ values with mean $$a$$ gives new mean:

$$a' = \dfrac{na + u + v + w}{n+3}.$$

For $$a' = a$$ we need $$u + v + w = 3a$$, i.e., the three new values must themselves have mean $$a$$.

For the shown data (mean $$= 9$$), any triplet whose sum is $$27$$ works. Examples:

• Add $$3, 9, 15$$ (sum $$= 27$$, mean $$= 9$$).
• Add $$5, 8, 14$$ (sum $$= 27$$).
• Add $$9, 9, 9$$ (three copies of the mean).

The same idea applies to removal — remove any three values whose sum is $$3a$$ and the mean stays the same.

Yes. From the previous exploration, the three added values must sum to $$3 \times \text{mean}$$. For the shown data (mean $$= 9$$), the sum must be $$27$$.

We now insist that two of the three values are less than $$9$$ and one is greater than $$9$$.

Example: pick $$5$$ and $$7$$ (both less than $$9$$). Their sum is $$12$$, so the third value must be $$27 - 12 = 15$$, which is indeed greater than $$9$$. So we can add $$5, 7$$ and $$15$$.

Another example: $$3, 8, 16$$ (both $$3, 8 < 9$$ and $$16 > 9$$, sum $$= 27$$).

So it is definitely possible — as long as the total of the two smaller additions and one larger addition equals $$27$$.

Again the three added values must sum to $$27$$ (since the current mean is $$9$$).

Choose two values greater than $$9$$, say $$10$$ and $$14$$; their sum is $$24$$, so the third value (which must be less than $$9$$) has to be $$27 - 24 = 3$$. So we can add $$10, 14$$ and $$3$$.

Another possibility: $$11$$ and $$12$$ (sum $$23$$) and $$4$$ ($$< 9$$) — sum $$= 27$$.

Or $$10, 13$$ and $$4$$.

Suppose there are $$n$$ values $$x_1, x_2, \ldots, x_n$$ with mean $$a = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$$.

Add the same fixed number $$k$$ to every value. The new values are $$x_1 + k,\ x_2 + k,\ \ldots,\ x_n + k$$, and their mean is:

$$\dfrac{(x_1+k) + (x_2+k) + \cdots + (x_n+k)}{n} = \dfrac{(x_1 + x_2 + \cdots + x_n) + nk}{n} = a + k.$$

So the mean also increases by exactly $$k$$. Every dot on the dot plot shifts right by $$k$$, and so does the balance point.

Add up all $$11$$ values:

$$8+3+10+13+4+6+7+7+8+8+5 = 79.$$

Divide by the number of values ($$n = 11$$):

$$\text{Mean} = \dfrac{79}{11} \approx 7.18.$$

The new data is obtained from the previous one by adding $$10$$ to every value. By the property we just proved (question 16), the mean also increases by $$10$$:

$$\text{New mean} = \dfrac{79}{11} + 10 = \dfrac{79 + 110}{11} = \dfrac{189}{11} \approx 17.18.$$

Direct check: $$18+13+20+23+14+16+17+17+18+18+15 = 189$$ and $$189 \div 11 \approx 17.18$$. ✓

The quicker way: since every value shifted right by $$10$$, the mean shifts right by $$10$$ too — no need to recompute the whole sum.

Let the collection be $$x_1, x_2, \ldots, x_n$$ with mean $$a = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$$.

Subtract $$2$$ from every value. The new values are $$x_1 - 2, x_2 - 2, \ldots, x_n - 2$$. Their mean is:

$$\dfrac{(x_1-2) + (x_2-2) + \cdots + (x_n-2)}{n} = \dfrac{(x_1 + x_2 + \cdots + x_n) - 2n}{n} = a - 2.$$

So subtracting the fixed number $$2$$ from every value decreases the mean by $$2$$. More generally, subtracting any fixed $$k$$ decreases the mean by $$k$$.

Recall the fair-share picture: the mean is the amount each person would end up with if the total were shared equally.

Suppose there are $$n$$ people currently holding a fair-share of $$a$$ units each. If every person now gives away the same fixed amount (say $$2$$ units), each person is left with $$a - 2$$ units, which is still an equal share. So the new fair share — that is, the new average — is exactly $$a - 2$$.

The same reasoning explains why adding a fixed amount $$k$$ to every value raises the average by $$k$$: everyone gains the same $$k$$, so the fair share also rises by $$k$$.

Let the data be $$x_1, x_2, \ldots, x_n$$ with mean $$a = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$$.

Doubling every value gives $$2x_1, 2x_2, \ldots, 2x_n$$. Their mean is:

$$\dfrac{2x_1 + 2x_2 + \cdots + 2x_n}{n} = \dfrac{2(x_1 + x_2 + \cdots + x_n)}{n} = 2a.$$

So the average is also doubled. More generally, if every value is multiplied by a constant $$c$$, the average is multiplied by $$c$$.

The median of $$5, 7, 8, 11, 13$$ is the middle value $$8$$ (there are $$5$$ values, so the $$3$$rd is the median).

Case 1: new value greater than the median. Include $$11$$. New sorted data: $$5, 7, 8, 11, 11, 13$$ — $$6$$ values. The median is the average of the $$3$$rd and $$4$$th values: $$\dfrac{8+11}{2} = 9.5$$. The median increased from $$8$$ to $$9.5$$.

Case 2: new value less than the median. Include $$4$$. New sorted data: $$4, 5, 7, 8, 11, 13$$. Median $$= \dfrac{7+8}{2} = 7.5$$. The median decreased.

Case 3: new value equal to the median. Include $$8$$. New sorted: $$5, 7, 8, 8, 11, 13$$. Median $$= \dfrac{8+8}{2} = 8$$. The median stays the same.

So the effect on the median depends on how the new value compares with the current median.

Let the missing (smudged) weight be $$w$$ kg. There are $$10$$ players in all, and the average weight is $$39.2$$ kg. Using

$$\text{Average} = \dfrac{\text{Total weight}}{\text{Number of players}},$$

we get:

$$\dfrac{42+40+39+33+48+38+42+35+32+w}{10} = 39.2.$$

Add up the nine known weights: $$42+40+39+33+48+38+42+35+32 = 349$$.

So $$\dfrac{349 + w}{10} = 39.2$$, giving $$349 + w = 392$$, and therefore $$w = 392 - 349 = 43$$.

The missing weight is $$43$$ kg.

Let the (incorrect) total number of coconuts be $$z$$. Since the incorrect average is $$25.6$$ with $$15$$ trees:

$$\dfrac{z}{15} = 25.6 \implies z = 25.6 \times 15 = 384.$$

One tree's count was noted as $$3$$ more than the actual, so the actual total is

$$384 - 3 = 381.$$

The correct average harvest per tree is:

$$\dfrac{381}{15} = 25.4.$$

Yes. Even without the individual tree-wise data, we can find the total using

$$\text{Total} = \text{Average} \times \text{Number of trees}.$$

So the (originally recorded) total number of coconuts $$= 25.6 \times 15 = 384$$. After subtracting the over-count of $$3$$ on one tree, the actual number of coconuts harvested $$= 384 - 3 = 381$$.

To find the average family size, we would:

1. Ask each student in the class how many members are there in their family (including themselves).
2. Add up all these numbers to get the total family-member count across the class.
3. Divide the total by the number of students in the class.

Formally, $$\text{Average family size} = \dfrac{\text{Sum of family sizes of all students}}{\text{Number of students}}$$.

The exact numerical value depends on your particular class. In many Indian classrooms today the answer typically lies between about $$4$$ and $$6$$.

The family size $$3$$ occurs $$3$$ times, size $$4$$ occurs $$11$$ times, and so on. So we must multiply each value by its frequency, add these products, and divide by the total number of students.

Total number of students $$= 3+11+9+7+3+1+1+1 = 36$$.

Sum of family sizes $$= (3\times 3)+(4\times 11)+(5\times 9)+(6\times 7)+(7\times 3)+(8\times 1)+(9\times 1)+(10\times 1)$$

$$= 9+44+45+42+21+8+9+10 = 188.$$

Average family size $$= \dfrac{188}{36} = \dfrac{47}{9} \approx 5.22$$.

There are $$36$$ students, so when the data is sorted, the median is the average of the $$18$$th and $$19$$th values.

Build the cumulative frequency:

Positions $$15$$ through $$23$$ (in sorted order) all correspond to family size $$5$$. So both the $$18$$th and $$19$$th values are $$5$$, giving:

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

No, we do not need to write all $$36$$ numbers. Because we already have the frequency table, we can use cumulative frequencies:

• Start with the smallest value and add up frequencies as we go along.
• The moment the cumulative count reaches (or exceeds) the median position (here, positions $$18$$ and $$19$$), the corresponding value is the median.

In this data, adding frequencies of family sizes $$3, 4, 5$$ gives $$3+11+9 = 23$$. Since $$14 < 18 \le 23$$ and $$14 < 19 \le 23$$, both the $$18$$th and $$19$$th sorted values are $$5$$. So the median is $$5$$.

This is much faster than listing all $$36$$ values in sorted order.

In the spreadsheet the first row is a header row and Row $$1$$ has the column titles Name (A), Odia (B), Telugu (C), English (D), Maths (E), Social Science (F), Science (G).

The student rows start from Row $$2$$. In the given order, Farooq is the $$4$$th student, so his data lies in Row $$5$$ (because Row $$1$$ is the header and Rows $$2$$–$$4$$ hold Ratna, Nagesh, Ashwin).

Farooq's Mathematics mark ($$42$$) is therefore in the cell where column E meets row 5 — cell E5.

Column A of the spreadsheet contains the student names, so column B contains the marks of the next header — which is Odia.

Specifically, cells B2 through B23 hold the Odia marks of Ratna, Nagesh, Ashwin, Farooq, ..., Jyothi (in that order).

Reading Ashwin's row of marks:

We need marks that are strictly greater than $$30$$. Comparing each:

• Odia $$= 29$$ (not $$> 30$$)
• Telugu $$= 31$$ ✓
• English $$= 33$$ ✓
• Maths $$= 34$$ ✓
• Social Science $$= 30$$ (equal, not greater)
• Science $$= 28$$ (not $$> 30$$)

So Ashwin scored more than $$30$$ in Telugu, English and Maths.

The Science marks lie in column G (the sixth subject column), starting from cell G2 (Ratna) to cell G23 (Jyothi), covering all $$22$$ students. The spreadsheet function that computes an average is AVERAGE.

So the formula to type in an empty cell is:

$$\texttt{=AVERAGE(G2:G23)}$$

In a spreadsheet, we would type $$\texttt{=AVERAGE(B2:B23)}$$ (Odia) in one cell and $$\texttt{=AVERAGE(C2:C23)}$$ (Telugu) in another, then compare.

Doing the arithmetic by hand:

Odia total: $$25+41+29+47+33+27+16+31+39+12+48+25+29+23+34+35+42+37+15+29+41+29 = 687$$.

Odia average $$= \dfrac{687}{22} \approx 31.23$$.

Telugu total: $$39+43+31+46+35+29+19+38+37+17+45+28+36+25+36+42+45+32+18+24+44+30 = 739$$.

Telugu average $$= \dfrac{739}{22} \approx 33.59$$.

Since $$31.23 < 33.59$$, the Odia class average is not greater than the Telugu class average — in fact, Telugu has the higher average.

The last student's row is Row $$23$$. In Row $$24$$, type the following formulae in the cells B24 to G24:

Instead of typing each formula, you can also type only the first (B24) and then drag the fill handle rightwards through G24 — the spreadsheet automatically adjusts the column letter, giving each subject its own average.

The six subject-marks of each student lie in columns B through G. Put a heading Total in cell H1. Then in cell H2 (for Ratna) type:

$$\texttt{=SUM(B2:G2)}$$

This adds up Ratna's six marks. Drag the fill handle downward through H23 (or copy the formula into H3, H4, ..., H23). The spreadsheet automatically changes the row number, giving:

• H3 becomes $$\texttt{=SUM(B3:G3)}$$ (Nagesh's total)
• H4 becomes $$\texttt{=SUM(B4:G4)}$$ (Ashwin's total)
• ...and so on down to H23 for Jyothi.

Now column H contains the total marks scored by every student.

Let the value of the missing dot be $$x$$. If the visible dots have sum $$S$$ and count $$n$$, then after adding the missing dot the total becomes $$S + x$$ across $$n + 1$$ values. Setting the mean equal to $$9$$:

$$\dfrac{S + x}{n + 1} = 9 \implies x = 9(n+1) - S.$$

Reading the dot plot, the visible dots are one at $$4$$, one at $$7$$, two dots at $$8$$, four dots at $$9$$, and one at $$11$$ — so $$n = 9$$ visible dots and

$$S = 4 + 7 + 2(8) + 4(9) + 11 = 4 + 7 + 16 + 36 + 11 = 74.$$

After adding the missing dot the total count becomes $$n+1 = 10$$, and we need $$9 \times 10 = 90$$ as the new sum. So

$$x = 90 - 74 = 16.$$

Mark the missing dot at position $$16$$ on the number line.

Rather than compute each mean exactly, we can use the balance-point idea: the mean lies near the ‘centre of mass’ of the dots. A mean of $$5.57$$ minutes is fairly high — most of the dots must be clustered towards the higher end of the scale (between $$5$$ and $$7$$ minutes), with only a few (if any) low values pulling the mean down.

Plot A has almost all of its dots between $$5$$ and $$6.5$$ minutes; its balance point is roughly $$5.5$$–$$5.6$$. This matches $$5.57$$ well.

Plot B has several dots between $$0.5$$ and $$1.5$$ minutes, along with a cluster between $$3.5$$ and $$5$$. These small values pull the mean well below $$5$$, so the mean cannot be $$5.57$$.

Plot C has all dots between $$3.5$$ and $$4.5$$ minutes; the balance point is roughly $$4$$. This cannot give a mean of $$5.57$$ either.

Therefore Plot A has the mean of $$5.57$$ minutes.

(A quick verification for A: adding up the dot values (approximately $$5, 5, 5, 5.5, 5.5, 5.5, 5.5, 5.5, 6, 6, 6.5$$ or similar) and dividing by the count gives a value very close to $$5.57$$.)

There are $$8$$ numbers with mean $$10.375$$, so the total sum must be:

$$\text{Sum} = 10.375 \times 8 = 83.$$

Adding up the seven known numbers:

$$8 + 13 + 10 + 4 + 5 + 20 + 10 = 70.$$

Therefore

$$70 + y = 83 \implies y = 13.$$

Using $$\text{Sum} = \text{Mean} \times \text{Number of values}$$:

$$\text{Sum} = 134 \times 15 = 2010.$$

Total students: $$1+0+0+1+4+9+12+15+10+10 = 62$$.

Minimum: Looking at the table, the smallest value with a non-zero frequency is $$1$$ (one student hit the bull's eye on the very first throw). So minimum $$= 1$$.

Maximum: The largest value with a non-zero frequency is $$10$$. Maximum $$= 10$$.

Mean: Sum $$= 1(1)+2(0)+3(0)+4(1)+5(4)+6(9)+7(12)+8(15)+9(10)+10(10)$$

$$= 1+0+0+4+20+54+84+120+90+100 = 473.$$

$$\text{Mean} = \dfrac{473}{62} \approx 7.63.$$

Median: There are $$62$$ students (even), so the median is the average of the $$31$$st and $$32$$nd sorted values. Build cumulative frequencies:

Positions $$28$$ through $$42$$ are the value $$8$$. Both the $$31$$st and $$32$$nd values are $$8$$. So median $$= 8$$.

Summary: min $$= 1$$, max $$= 10$$, mean $$\approx 7.63$$, median $$= 8$$. Most students needed $$6$$–$$10$$ trials to hit the bull's eye, and both the mean and median suggest the typical student needed around $$7$$–$$8$$ trials.

Yes — both graphs display the same data (the monthly maximum temperatures in Kerala and Punjab across the $$12$$ months of $$2023$$). What differs is the visualisation:

• The clustered-column graph uses side-by-side vertical bars for each month.
• The line graph uses points connected by straight-line segments — one line per state.

A line graph is especially useful when the data changes over time, because it makes trends (rises, falls, stability) visually obvious and lets us compare two series easily.

Weather stations spread across the state (in different cities and towns) record the daily temperature. For a given month, we:

1. Note the highest temperature recorded on each day at each station.
2. The state's monthly maximum is the largest value among all those daily maxima across the whole state and the whole month.

In practice, agencies such as the India Meteorological Department (IMD) collect data from many weather stations and publish the monthly maximum temperature for each state.

Organisation. The horizontal axis shows the twelve months (January to December) of $$2023$$. The vertical axis shows temperature in degrees Celsius. Two lines are drawn — blue circular markers for Kerala and red square markers for Punjab — with a legend at the top explaining the colours/markers.

Scale. Temperature axis goes from $$0^{\circ}\mathrm{C}$$ to about $$40^{\circ}\mathrm{C}$$ in equal steps of $$10^{\circ}\mathrm{C}$$.

Patterns.

• Kerala's line is nearly flat, hovering between $$29^{\circ}\mathrm{C}$$ and $$33^{\circ}\mathrm{C}$$ all year.
• Punjab's line is very variable: it starts low in January ($$\approx 19^{\circ}\mathrm{C}$$), rises steeply to a peak near $$38^{\circ}\mathrm{C}$$ in June, and then falls again to about $$23^{\circ}\mathrm{C}$$ by December.
• Kerala is warmer than Punjab in winter, but Punjab overtakes Kerala from April through October.

• In Punjab, the monthly maximum temperature rises steadily from January ($$\approx 19^{\circ}\mathrm{C}$$) to a peak of about $$38^{\circ}\mathrm{C}$$ in June, drops slightly in July, stays around $$34$$–$$35^{\circ}\mathrm{C}$$ until September, and then falls continuously to about $$23^{\circ}\mathrm{C}$$ in December. January is the coldest month.
• Kerala shows a very different picture — its line stays roughly flat between about $$29^{\circ}\mathrm{C}$$ and $$33^{\circ}\mathrm{C}$$ all year, with a mild peak of about $$33^{\circ}\mathrm{C}$$ in April and a mild dip to about $$29^{\circ}\mathrm{C}$$ in July.
• Punjab experiences distinct summers and winters, while Kerala's temperature is nearly constant year-round.

Conclusion: Punjab has a strongly seasonal climate (hot summers, cool winters), whereas Kerala has a moderate, uniform tropical climate. This is because Punjab lies in the northern plains, far from the sea, whereas Kerala is a coastal state near the equator; the sea moderates its temperature.

Some natural follow-up questions:

• Why is Punjab's climate so seasonal but Kerala's is not? What geographical factors (latitude, altitude, proximity to sea) are at play?
• What do the monthly minimum temperature graphs look like? Would Kerala still stay flat and Punjab still swing widely?
• How do other states behave — do coastal states (e.g., Tamil Nadu, Odisha) look like Kerala, and inland states (e.g., Rajasthan, Uttar Pradesh) look like Punjab?
• Has the pattern changed over the last few decades because of climate change?
• How does the temperature difference between the two states affect crops, clothing, festivals, tourism, and school schedules?

The source given on the graph is the United Nations Office for Outer Space Affairs ($$2025$$). Every country that launches a spacecraft is expected to register the launch (spacecraft name, launch date, orbit, purpose) with this UN body under the international Registration Convention. So the data are compiled by:

1. Collecting launch registrations from every country's space agency.
2. Counting the number of objects (satellites, probes, landers, crewed spacecraft, space station modules) launched each calendar year.
3. Adding up the country-wise counts to get the worldwide total for that year.

The graph then shows these counts for the World and the three biggest-contributing nations (USA, China, Russia) year by year from $$2012$$ to $$2024$$.

Organisation. Horizontal axis: years from $$2012$$ to $$2024$$ (marked every $$2$$ years). Vertical axis: number of space objects launched. Four labelled lines — World (green), United States (blue), China (grey) and Russia (red).

Scale. Y-axis runs from $$0$$ to $$3{,}000$$ in equal steps of $$500$$.

Patterns.

• From $$2012$$ to about $$2016$$, all four counts are very low ($$< 500$$).
• Starting around $$2017$$ the World and USA lines rise sharply, and both keep climbing very steeply after $$2020$$.
• By $$2024$$ the World count is roughly $$2900$$, the USA is close behind at about $$2300$$, China is around $$300$$, and Russia is under $$100$$.
• The USA's line rises fastest — a large chunk of the worldwide growth comes from the USA (largely driven by SpaceX's Starlink launches).

• The world launched relatively few objects each year up to about $$2016$$ (a couple of hundred per year). The pace has accelerated dramatically since then, largely due to constellations of small satellites (like Starlink, OneWeb) that require dozens of launches at once.
• The USA is by far the largest contributor to recent growth; its line runs parallel to and just below the World line from about $$2020$$ onward, indicating it accounts for roughly $$\dfrac{3}{4}$$ of worldwide launches by $$2024$$.
• China's launches have grown steadily but at a much slower rate; by $$2024$$ they are roughly $$300$$ per year.
• Russia's line has stayed low and roughly flat (under $$100$$ per year), suggesting a decline in its share of world launches compared with the Soviet-era peaks.
• The gap between the sum of USA + China + Russia and the World line represents launches by all other countries (Europe, India, Japan, etc.).

Conclusion: Space activity has entered a new era of rapid growth after $$2017$$, dominated primarily by the USA. Understanding this trend is important for issues such as satellite congestion, space debris, and international space policy.

(1) Not valid. Looking closely at the World line, we see that the count did not increase in every single year. For instance, between $$2023$$ and $$2024$$ the worldwide count actually dips slightly (from about $$2900$$ to about $$2800$$). So the statement "increased every year" is false.

(2) Valid. In $$2022$$–$$24$$, the USA line lies very close to the World line — the USA count ($$\sim 2300$$) is roughly $$\dfrac{3}{4}$$ of the World count ($$\sim 2900$$) in $$2024$$. So the USA is indeed a major contributor.

(3) Not valid. The graph only shows data for the World, USA, China and Russia — Nepal is not shown at all. We cannot make any inference about Nepal from this graph.

(4) Not valid. Reading off the graph, in $$2024$$ China is around $$300$$ and Russia is under $$100$$, so their combined total is roughly $$300$$–$$400$$. The value $$400$$ is close but not quite exact. The graph itself shows the combined count is about $$300$$, so "about $$400$$" is a slight over-estimate — the statement is questionable at best. (Some students may accept it as valid; a careful reading suggests it is not accurate.)

Looking at the World line, we compare each pair of consecutive years to see where the count roughly doubled (or more) from one year to the next.

• $$2019 \to 2020$$: the World count jumped from about $$500$$ to about $$1000$$ — approximately a doubling.
• $$2020 \to 2021$$: rose from about $$1000$$ to about $$2000$$ — again nearly a doubling.

So one such consecutive-year pair is $$\mathbf{2019 \to 2020}$$ (about $$500 \to 1000$$, roughly $$2\times$$); another is $$\mathbf{2020 \to 2021}$$ (about $$1000 \to 2000$$).

Rainfall in each city is measured daily by rain-gauges maintained by the meteorological department. For each month:

1. Add up the daily rainfall over all the days of that month in that city — this gives the monthly rainfall for one year.
2. Repeat this for several years (for example, the last $$10$$ or $$30$$ years).
3. Compute the average of the monthly rainfall values across all those years. That is the "monthly average rainfall" plotted on the graph.

Averaging over several years smooths out year-to-year variations and shows the typical pattern of monsoon and dry periods.

Locating the six cities on a map of India:

• West-coast group — Kovalam (Kerala, southern tip), Udupi (coastal Karnataka) and Mumbai (Maharashtra). All three lie along the Arabian Sea on India's western coastline.
• East-coast group — Rameswaram (Tamil Nadu), Chennai (Tamil Nadu) and Puri (Odisha). All three lie along the Bay of Bengal on India's eastern coastline.

What is common: the three cities in each graph are grouped by the coast (sea) they face.

Observations from the graphs:

• The west coast gets very heavy rainfall between June and September (all three cities peak in this window). Udupi has the highest peak (about $$900$$ mm in July), Mumbai next (about $$600$$ mm), Kovalam moderate (about $$300$$ mm).
• The east coast cities show a different pattern: rainfall is generally lower and the peak comes later in the year (October–December). Chennai and Rameswaram peak in November, while Puri (further north) peaks in July.

Inference: The west coast is dominated by the South-West Monsoon (June–Sep), whereas most of the Tamil Nadu east coast receives its rain from the North-East Monsoon (Oct–Dec).

Reading the peaks (highest points) and lows (near-zero) from each city's line:

West-coast cities all peak in the middle of the year (June–August). Chennai and Rameswaram peak in November; Puri (further north on the east coast) peaks in July like the west coast.

South-West (SW) Monsoon. From around June to September, moisture-laden winds blow from the Arabian Sea and the Bay of Bengal towards the Indian sub-continent, driven from the south-west. These winds first hit the west coast (Kerala, Karnataka, Konkan, Goa, Maharashtra), then move north and east across most of India. So the SW monsoon brings the main rainy season to the west coast, central India, north India, most of the northeast, and to some extent Odisha.

North-East (NE) Monsoon. From around October to December, the winds reverse and blow from the north-east, picking up moisture over the Bay of Bengal. These winds bring the rainy season to the south-eastern coast of India — particularly Tamil Nadu, coastal Andhra Pradesh, southern Karnataka, and parts of Kerala. That is why cities like Chennai and Rameswaram get most of their rain in October, November and December, rather than in the SW monsoon months.

These two monsoons together define India's overall rainfall pattern and explain the different peak months seen on the west-coast and east-coast rainfall graphs.

How to draw the line graph:

1. Draw two perpendicular axes. On the horizontal axis, mark the seven days (Mon, Tue, ..., Sun) at equal intervals.
2. On the vertical axis, mark "Number of customers" using an even scale. Since the largest value is $$35$$, a scale from $$0$$ to $$40$$ in steps of $$5$$ works well.
3. For each day, plot a point for the Visiting value at that day. Join these seven points in order with straight-line segments — this gives the Visiting line (use one colour, say blue).
4. Repeat for the Purchasing row using a different colour (say red). Add a small legend on the graph identifying each colour.

Approximate shape of the graph (values from the table):

What the graph shows: The Visiting line lies above the Purchasing line every day. Both lines dip on Wednesday and rise sharply to a peak on Sunday. Both visitors and purchasers are highest on Sunday and lowest on Wednesday.

A typical weekday strip for a Grade $$8$$ student in India would look like this (roughly, per $$30$$-minute box):

• $$12$$:$$00$$ am – $$6$$:$$30$$ am: Sleeping (about $$6.5$$ hours of night sleep).
• $$6$$:$$30$$ – $$7$$:$$00$$ am: Getting ready, showering.
• $$7$$:$$00$$ – $$7$$:$$30$$ am: Breakfast.
• $$7$$:$$30$$ – $$8$$:$$00$$ am: Travel to school.
• $$8$$:$$00$$ am – $$3$$:$$00$$ pm: School (with a $$30$$-minute lunch break around $$1$$ pm).
• $$3$$:$$00$$ – $$3$$:$$30$$ pm: Travel back home.
• $$3$$:$$30$$ – $$5$$:$$00$$ pm: Snack, rest, hobbies.
• $$5$$:$$00$$ – $$7$$:$$00$$ pm: Studies / homework.
• $$7$$:$$00$$ – $$8$$:$$00$$ pm: Free time / family / media.
• $$8$$:$$00$$ – $$9$$:$$00$$ pm: Dinner.
• $$9$$:$$00$$ – $$10$$:$$00$$ pm: Reading, family time.
• $$10$$:$$00$$ pm onwards: Sleep.

Comparison with Manoj's weekday: Both students have long class/study blocks during the day and dedicated sleeping and eating times. Differences may include amount of screen/media time, whether one takes tuition classes, sports practice, or hobby time in the evening. Your exact strip depends on your personal schedule.

A vacation-day strip typically has:

• A longer sleep block (waking up later — perhaps $$8$$ am instead of $$6$$:$$30$$ am).
• No school/class block during the daytime.
• Much more time spent on hobbies, friends, media, family and travel (outings).
• Meal times may become slightly irregular.
• Some students may still keep a small study block for revision or reading; others drop it entirely.

Comparison with a school day: The school block (the biggest block on a weekday) is replaced by leisure and family time. The wake-up and bed-times shift later. Overall, the strip becomes more colourful/varied and less repetitive across days.

A typical adult's weekday strip (e.g., a working parent) might look like:

• $$11$$:$$00$$ pm – $$6$$:$$00$$ am: Sleeping ($$7$$ hours).
• $$6$$:$$00$$ – $$7$$:$$00$$ am: Getting ready, exercise, tea/breakfast.
• $$7$$:$$00$$ – $$8$$:$$00$$ am: Household chores or preparing kids for school.
• $$8$$:$$00$$ – $$9$$:$$00$$ am: Travel to work.
• $$9$$:$$00$$ am – $$1$$:$$00$$ pm: Work.
• $$1$$:$$00$$ – $$2$$:$$00$$ pm: Lunch.
• $$2$$:$$00$$ – $$6$$:$$00$$ pm: Work (or business tasks).
• $$6$$:$$00$$ – $$7$$:$$00$$ pm: Travel back / errands.
• $$7$$:$$00$$ – $$9$$:$$00$$ pm: Cooking/family time/dinner.
• $$9$$:$$00$$ – $$11$$:$$00$$ pm: Media/relaxation/reading.

Interesting comparisons with a student's day:

• Adults typically sleep less (about $$7$$ hours) than school-going children ($$8$$–$$9$$ hours).
• The "class/school" block is replaced by a longer "work" block, and it often extends into the evening.
• Adults spend more time on household chores and cooking.
• Free/hobby time is usually shorter for adults.

Answers will vary depending on which adult is chosen.

Observations from the sleep-across-ages line graph:

• Children of about age $$6$$ sleep the longest — around $$9.5$$ hours per night.
• The sleep duration decreases steadily as we move from childhood through the teen years into adulthood, reaching a minimum of about $$8$$ hours somewhere between ages $$30$$ and $$50$$.
• After age $$50$$, the sleep duration starts increasing again, reaching about $$8.5$$ hours by age $$75$$.
• So the curve is U-shaped (long in childhood, minimum in middle age, longer again in old age).
• The zoomed-in version of the same data makes the small dip in middle age easier to see (from $$\sim 8.0$$ to $$\sim 9.5$$ hours vertically).

Interesting points: The graph is smooth because it has $$70+$$ closely-packed data points; a bar chart would look cluttered. Middle-aged adults consistently get the least sleep — perhaps because of work and family responsibilities.

We have $$6$$ values in all, so the median is the average of the $$3$$rd and $$4$$th sorted values. For the median to be $$13$$, we need $$a_3 + a_4 = 26$$.

Given values already: $$5, 14, 21$$. Three blanks (say $$a, b, c$$) must be filled with counting numbers to make the sorted median $$13$$.

The two structural cases:

1. $$14$$ is the $$4$$th sorted value. Then the $$3$$rd value must be $$26 - 14 = 12$$, so exactly one blank is $$12$$; one other blank must be a counting number $$\le 12$$ (goes into positions $$1$$ or $$2$$); the third blank must be a counting number $$\ge 14$$ (goes into position $$5$$).
2. $$14$$ is the $$5$$th sorted value. Then $$a_3 + a_4 = 26$$ with both $$\le 14$$, forcing $$a_3 = a_4 = 13$$. So two blanks must equal $$13$$, and the third blank is either a counting number $$\le 13$$ (in positions $$1$$ or $$2$$) or $$\ge 21$$ (in position $$6$$).

Example fillings that work:

• $$5, 21, 14, 8, 12, 20$$ — sorted: $$5, 8, 12, 14, 20, 21$$; median $$= \dfrac{12+14}{2} = 13$$. ✓
• $$5, 21, 14, 6, 13, 13$$ — sorted: $$5, 6, 13, 13, 14, 21$$; median $$= 13$$. ✓
• $$5, 21, 14, 12, 15, 3$$ — sorted: $$3, 5, 12, 14, 15, 21$$; median $$= 13$$. ✓

How many possibilities? In case 1, one blank is fixed at $$12$$, a second blank can be any counting number in $$\{1, 2, \ldots, 12\}$$, and a third can be any counting number in $$\{14, 15, 16, \ldots\}$$ — this last range is unbounded. In case 2, two blanks are fixed at $$13$$ and the third can be any counting number in $$\{1, \ldots, 13\} \cup \{21, 22, 23, \ldots\}$$ — again unbounded.

So there are infinitely many possibilities.

There are $$6$$ values in all, so the sum must be $$6 \times 6.5 = 39$$.

Sum of the four known values: $$3 + 11 + 15 + 6 = 35$$. Therefore the two blanks $$a, b$$ must satisfy:

$$a + b = 39 - 35 = 4.$$

Since $$a, b$$ are counting numbers (positive integers $$\ge 1$$), the pairs $$(a, b)$$ that satisfy $$a + b = 4$$ are:

• $$(1, 3)$$ — sum $$= 4$$
• $$(2, 2)$$ — sum $$= 4$$
• $$(3, 1)$$ — sum $$= 4$$

Treating the two blank positions as distinguishable (they are different slots in the list), there are $$3$$ possibilities. If instead we treat $$(1,3)$$ and $$(3,1)$$ as the same unordered pair, there are $$2$$ distinct unordered fillings.

Verification for $$(1, 3)$$: collection is $$3, 11, 1, 3, 15, 6$$; sum $$= 39$$; mean $$= 39/6 = 6.5$$. ✓

No — $$17$$ is not the average.

Method 1 (balance point): If $$17$$ were the average, the total distance of dots below $$17$$ should equal the total distance of dots above $$17$$. Looking at the plot, only a few dots lie to the left of $$17$$ (at $$14, 15, 16$$), each contributing a small distance ($$1$$–$$3$$). But many dots lie to the right ($$18, 19, 20, 21, 22, 23$$), several at large distances up to $$6$$. So the right-side distances clearly outweigh the left-side distances, meaning the true balance point (the mean) lies to the right of $$17$$.

Method 2 (direct calculation): Reading approximate frequencies from the plot — $$14$$: $$2$$, $$15$$: $$2$$, $$16$$: $$3$$, $$17$$: $$5$$, $$18$$: $$5$$, $$19$$: $$4$$, $$20$$: $$3$$, $$21$$: $$2$$, $$22$$: $$1$$, $$23$$: $$1$$ — gives:

Total dots $$= 28$$. Sum $$= 2(14)+2(15)+3(16)+5(17)+5(18)+4(19)+3(20)+2(21)+22+23 = 504$$. Mean $$= \dfrac{504}{28} = 18$$.

So the true average is about $$18$$, not $$17$$.

Let $$n$$ be the number of people in the group. The old total weight is $$65.3n$$ kg.

Change in mean: The total weight changes by $$(-2) + 1 + 1 = 0$$ kg. So the new total is still $$65.3n$$ kg, and the mean is still

$$\dfrac{65.3n}{n} = 65.3 \text{ kg}.$$

The mean is unchanged at $$65.3$$ kg.

Change in median: Individual weights have moved by small amounts (by $$\pm 1$$ or $$-2$$). Whether the median shifts depends on which people gained/lost weight and how close they were to the middle. If none of the three affected people were among the middle values, the median stays exactly the same. If they were near the middle, the median could shift slightly (by at most $$\pm 2$$ kg in principle).

Without knowing individual weights we cannot pin down the new median exactly, but it can change by at most a small amount.