Tales by Dots and Lines

www.akankshaclasses.com
CLASS VIII Mathematics ~4–5 marks Ch 12 of 14
Tales by Dots and Lines

Class 8 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • The mean is the balance point of data on a dot plot — the total distance of values on its left equals the total distance on its right. There is only one such centre.
  • Add a value above the mean $\Rightarrow$ mean rises; add one below $\Rightarrow$ mean falls; add one equal to the mean $\Rightarrow$ mean unchanged. The median shifts in the same direction.
  • If every value is increased by $k$, the mean increases by $k$. If every value is multiplied by $k$, the mean is multiplied by $k$.
  • With a frequency table: $\text{Mean}=\dfrac{\sum(\text{value}\times\text{frequency})}{\sum\text{frequency}}$, and the median is found by adding up frequencies (cumulative) to reach the middle position.
  • Line graphs join points with line segments to show how a quantity changes over time; spreadsheets use =SUM() and =AVERAGE() to crunch big tables.
  • Board weightage: ~4–5 marks — usually a "find the missing value / corrected mean" sum, a frequency-table mean & median, and a line-graph reading/interpretation question.
Detailed notes

1. What this chapter is about

The title "Tales by Dots and Lines" says it all: data can be told as a story using dots (dot plots, points) and lines (line graphs). Last year you met the mean and median; this chapter looks at them from a fresh angle — as a balance point — and then moves to visualising data with line graphs, infographics and activity strips. Quick recall:

  • Mean (average) $=\dfrac{\text{sum of all values}}{\text{number of values}}$ — a measure of central tendency that represents the "centre".
  • Median $=$ the middle value when the data is arranged in order. With an even count, it is the average of the two middle values.

2. The Balancing Act — mean as the centre

For just two numbers the mean sits exactly halfway between them: the mean of $3$ and $7$ is $\dfrac{3+7}{2}=5$, and the mean of $8$ and $9$ is $8.5$. But for three or more values the mean is not the midpoint of the smallest and largest. Instead:

The mean is the point where the total distance to all the values on its left equals the total distance to all the values on its right. It is the balance point (like a see-saw).

For the collection $10,10,11,17$, the mean is $\dfrac{10+10+11+17}{4}=\dfrac{48}{4}=12$. Check the balance:

  • Left of $12$: distances $12-10=2,\ 12-10=2,\ 12-11=1$ — total $=5$.
  • Right of $12$: distance $17-12=5$ — total $=5$. Balanced!

Only one centre exists. If we tried a centre larger than $12$, every left-distance grows and every right-distance shrinks, so the two sides can no longer be equal. A centre smaller than $12$ does the opposite. So the balancing value is unique.

3. What happens to the mean when data changes?

Think of the see-saw — the mean moves to keep it balanced.

  • Include a value greater than the mean $\Rightarrow$ the mean increases.
  • Include a value less than the mean $\Rightarrow$ the mean decreases.
  • Include a value equal to the mean $\Rightarrow$ the mean stays the same (and removing such a value also leaves it unchanged).

You can even add or remove several values without changing the mean, as long as the extra distance above is balanced by the extra distance below — for example, including two values below the mean and one above can keep the mean fixed if the pulls cancel out.

Worked example — keep the mean fixed

Data $4,6,8$ has mean $6$. Suppose we add the value $6$ (equal to the mean). New data $4,6,6,8$ has mean $\dfrac{24}{4}=6$ — unchanged, exactly as predicted. Adding $6$ and $6$ again still gives mean $6$.

4. Shifting and scaling — "Relatively Unchanged"

Suppose the mean of $n$ values $x_1,x_2,\dots,x_n$ is $a$, i.e. $\dfrac{x_1+x_2+\dots+x_n}{n}=a$.

Adding a fixed number to every value. If we add $k$ to each value:

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

So the new mean $=a+k$ — it just shifts by $k$. (Subtracting $k$ from each value lowers the mean by $k$.) The relative position of the mean within the data is unchanged.

Multiplying every value by a fixed number. If we multiply each value by $k$:

$$\dfrac{kx_1+kx_2+\dots+kx_n}{n}=\dfrac{k(x_1+\dots+x_n)}{n}=k\cdot a.$$

So the new mean $=ka$ — doubling every value doubles the mean.

NCERT example — shoes and heights

The average height of $24$ students measured with shoes is $150.2$ cm. The shoes add $1$ cm to each student. We do not need to re-measure everyone! Since every value is $1$ cm too big, the mean is $1$ cm too big. Correct average height $=150.2-1=\textbf{149.2 cm}$.

5. Finding an unknown value from the mean

A favourite exam trick: the mean equation is just $\text{Sum}=\text{Mean}\times\text{Count}$. If one value is missing, call it $w$ and solve.

NCERT example — Coach Balwan's wrestlers

Ten players' weights (kg): $42,40,39,33,48,38,42,35,32,w$, with mean $39.2$.

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

The known nine sum to $349$, so $349+w=39.2\times10=392\Rightarrow w=392-349=\textbf{43 kg}$.

NCERT example — Venkayya's coconuts (corrected mean)

Average harvest per tree is $25.6$ over $15$ trees, so total recorded $=25.6\times15=384$. One tree's count was written $3$ too high, so the true total $=384-3=381$. Correct average $=\dfrac{381}{15}=\textbf{25.4}$.

6. Tinkering with the median

The median is the middle value, so including a new value moves it in the natural direction:

  • Include a value greater than the median $\Rightarrow$ median increases (or stays, depending on position).
  • Include a value less than the median $\Rightarrow$ median decreases.
NCERT example — median jumps

Suppose some data has median $8$. Including the value $11$ (greater than $8$) makes the count even and creates two middle values $8$ and $11$, so the new median $=\dfrac{8+11}{2}=\textbf{9.5}$. The median rose because we added a larger value.

Median is robust: you can often add a value between the two middle numbers (or remove one) without changing the median at all — handy for "fill in the blank so the median stays the same" questions.

7. Mean and median with frequencies

When a value repeats, listing it once and noting how often it appears (its frequency) saves work. The mean must weight each value by its frequency:

$$\text{Mean}=\dfrac{\sum(\text{value}\times\text{frequency})}{\sum\text{frequency}}=\dfrac{\sum f_i x_i}{\sum f_i}.$$
NCERT example — average family size

Family-size data: value $3,4,5,6,7,8,9,10$ with frequencies $3,11,9,7,3,1,1,1$ (total $36$ students).

Sum $=(3\times3)+(4\times11)+(5\times9)+(6\times7)+(7\times3)+(8\times1)+(9\times1)+(10\times1)$

$=9+44+45+42+21+8+9+10=188$. So mean $=\dfrac{188}{36}\approx\textbf{5.22}$.

Common error: averaging the distinct values $\dfrac{3+4+\dots+10}{8}=6.5$ is wrong — it ignores how often each value occurs.

Median from frequencies: with $36$ values, the median is the average of the $18$th and $19$th values. Add frequencies cumulatively: $3$ (up to value 3), $3+11=14$ (up to value 4), $3+11+9=23$ (up to value 5). So positions $15$ to $23$ are all $5$ — both the $18$th and $19$th values are $5$. Median $=5$.

8. Spreadsheets — letting the computer do the sums

A spreadsheet is a digital grid of cells in rows (numbered $1,2,3,\dots$) and columns (lettered $A,B,C,\dots$). A cell is named by its column then row — e.g. E5 is column $E$, row $5$. A block of cells is written Start:End, e.g. B3:G3 is one row, D2:D6 is part of a column.

  • =SUM(B3:G3) adds all the marks in row 3 (one student's total).
  • =AVERAGE(B7:D7) finds the average of three subjects for one student.
  • =AVERAGE(G2:G23) would give the class average in the column $G$ subject.

Spreadsheets (Excel, Google Sheets, LibreOffice Calc) make computing dozens of totals and averages instant — far faster than a calculator for big tables.

9. Line graphs — showing change over time

A line graph plots data points and joins consecutive points with straight line segments. It is the best choice for showing how a quantity changes over time (months, years).

  • The horizontal axis usually shows time; the vertical axis shows the measured quantity (temperature, count, price).
  • Different colours and marker shapes distinguish multiple data series (e.g. Kerala in blue circles, Punjab in red) — shapes help when printed in greyscale.
  • The steepness (slope) of a segment shows how fast the value is changing — a steeper line means a bigger increase or decrease.

Two-step reading method: (1) Identify what is given — the axes, scale, and which line is which; (2) Infer — describe the trends (rising, falling, flat) and draw conclusions, while noting limits like missing data.

NCERT example — temperature trends

Punjab's monthly maximum temperature climbs from about $19^{\circ}$C in January to a peak near $38^{\circ}$C in June, then falls to about $23^{\circ}$C by December — it varies a lot. Kerala stays nearly flat all year (about $29^{\circ}$C to $33^{\circ}$C). Reading the lines lets us compare two places at a glance.

Why a line graph over columns? The "space objects launched" data had $13$ years $\times\ 4$ countries $=52$ bars in a clustered-column graph — crowded and hard to read. A line graph connects each country's points so trends pop out clearly.

10. Other "dots and lines" stories

  • Smooth-looking line graphs: the sleep-vs-age graph uses about $80$ closely spaced points, so it looks like a smooth curve — lighter and clearer than $70$ columns. Sleep falls from ~$9.5$ h at age $6$ to ~$8$ h around $30$–$50$, then rises slightly.
  • Infographics: combine maps, colours and numbers to tell a story quickly — e.g. a rice-vs-wheat preference map where a colour scale from $-100$ to $+100$ shows the geographical split across India.
  • Activity strips: a strip of $48$ boxes (each $=30$ minutes from midnight to midnight) coloured by activity shows how a person spends a whole day — a compact "dots and lines" data story.

The big idea of the chapter: examining data carefully always leads to new questions worth exploring.

11. Worked exam-style problems

Solved — find $y$ from the mean

The mean of $8,13,10,4,5,20,y,10$ is $10.375$. There are $8$ values, so sum $=10.375\times8=83$. The seven known values sum to $8+13+10+4+5+20+10=70$. Hence $70+y=83\Rightarrow y=\textbf{13}$.

Solved — sum from the mean

A set of $15$ values has mean $134$. Sum $=\text{Mean}\times\text{Count}=134\times15=\textbf{2010}$.

Solved — median of an even list

Median of $8,10,19,23,26,34,40,41,41,48,51,55,70,84,91,92$ ($16$ values): the $8$th and $9$th values are $41$ and $41$, so median $=\dfrac{41+41}{2}=\textbf{41}$.

12. Common mistakes to avoid

  • Treating the mean as the midpoint of the smallest and largest value — it is the balance point, not the midpoint (true only for two numbers).
  • Forgetting frequencies: averaging only the distinct values instead of weighting each by how often it occurs.
  • Re-doing all the work after a uniform change — if every value shifts by $k$ or scales by $k$, just shift/scale the mean.
  • For an even number of values, taking one middle value as the median instead of the average of the two middle values.
  • On a line graph, misreading the scale on the axes or confusing which coloured line is which series.

13. Quick revision checklist

  • Mean $=$ balance point; total left-distance $=$ total right-distance; only one centre.
  • Add above mean $\to$ up; below $\to$ down; equal $\to$ no change. Median moves the same way.
  • Add $k$ to all $\to$ mean $+k$; multiply all by $k$ $\to$ mean $\times k$.
  • Sum $=$ Mean $\times$ Count — the key for missing-value and corrected-mean sums.
  • Frequency mean $=\dfrac{\sum f_i x_i}{\sum f_i}$; median via cumulative frequency to the middle position.
  • Line graphs $=$ change over time; steeper line $=$ faster change; identify, then infer.
Practice MCQs
1. The mean of the data $10,10,11,17$ is:
  1. $13$
  2. $12$
  3. $11.5$
  4. $14$
Answer: (B) $\dfrac{10+10+11+17}{4}=\dfrac{48}{4}=12$ — the balance point.
2. If a value greater than the current mean is added to the data, the mean will:
  1. decrease
  2. stay the same
  3. increase
  4. become zero
Answer: (C) the balance point shifts towards the larger value, so the mean increases.
3. The mean of some data is $25$. If $4$ is added to every value, the new mean is:
  1. $25$
  2. $100$
  3. $29$
  4. $21$
Answer: (C) adding $k$ to every value increases the mean by $k$: $25+4=29$.
4. The mean of some data is $12$. If every value is multiplied by $3$, the new mean is:
  1. $15$
  2. $36$
  3. $4$
  4. $12$
Answer: (B) multiplying every value by $k$ multiplies the mean by $k$: $12\times3=36$.
5. The average height with shoes (adding $1$ cm each) is $150.2$ cm. The correct average without shoes is:
  1. $151.2$ cm
  2. $150.2$ cm
  3. $149.2$ cm
  4. need to re-measure
Answer: (C) every value is $1$ cm too big, so the mean is $1$ cm too big: $150.2-1=149.2$ cm.
6. The median of $8,10,19,23,26,34,40,41$ (8 values) is:
  1. $23$
  2. $24.5$
  3. $26$
  4. $25$
Answer: (B) the $4$th and $5$th values are $23$ and $26$; median $=\dfrac{23+26}{2}=24.5$.
7. In a frequency table, the mean is found by:
  1. averaging the distinct values
  2. $\dfrac{\sum(\text{value}\times\text{frequency})}{\sum\text{frequency}}$
  3. the most frequent value
  4. the middle frequency
Answer: (B) each value must be weighted by its frequency.
8. In a spreadsheet, the cell in column $E$, row $5$ is named:
  1. $5E$
  2. $E5$
  3. $EE5$
  4. $5,E$
Answer: (B) column letter then row number: $E5$.
9. Which spreadsheet formula gives the total of the cells $B3$ to $G3$?
  1. =AVERAGE(B3:G3)
  2. =TOTAL(B3,G3)
  3. =SUM(B3:G3)
  4. =ADD(B3-G3)
Answer: (C) =SUM(B3:G3) adds every cell from $B3$ to $G3$.
10. A graph that joins data points with line segments to show change over time is a:
  1. pie chart
  2. bar graph
  3. line graph
  4. pictograph
Answer: (C) a line graph is best for trends over time.
11. On a line graph, a steeper line segment means:
  1. no change in value
  2. a smaller change
  3. a greater change
  4. missing data
Answer: (C) the steeper the segment, the larger the increase (or decrease) in that period.
12. The mean of $8,13,10,4,5,20,y,10$ is $10.375$. Then $y=$
  1. $10$
  2. $13$
  3. $15$
  4. $8$
Answer: (B) sum $=10.375\times8=83$; known values $=70$, so $y=83-70=13$.
13. How many "centres" (balance points) does a set of data have?
  1. exactly one
  2. two
  3. as many as there are values
  4. none
Answer: (A) the balancing argument shows the mean is the unique centre.
Assertion–Reason
A: Adding $5$ to every value of a data set increases its mean by $5$.   R: The sum increases by $5n$ for $n$ values, and dividing by $n$ adds $5$ to the mean.
Answer: Both A and R are true, and R correctly explains A — see the shifting rule in §4.
A: The mean of three numbers is always the midpoint of the smallest and largest.   R: The mean is the balance point of the data.
Answer: A is false (true only for two numbers), R is true — the mean balances total distances, it is not generally the midpoint.
Exam-style questions
Q1. The weights (kg) of $10$ players are $42,40,39,33,48,38,42,35,32,w$ and the mean is $39.2$. Find $w$. (3 marks)
Answer: Sum $=39.2\times10=392$. Known nine sum to $349$, so $w=392-349=43$ kg.
Q2. Family-size data has values $3,4,5,6,7,8,9,10$ with frequencies $3,11,9,7,3,1,1,1$. Find the mean and the median. (4 marks)
Answer: $\sum f=36$; $\sum fx=9+44+45+42+21+8+9+10=188$, so mean $=\dfrac{188}{36}\approx5.22$. Cumulative frequencies $3,14,23,\dots$ show the $18$th and $19$th values are both $5$, so median $=5$.
Q3. Venkayya's $15$ trees gave an average of $25.6$ coconuts, but one tree's count was recorded $3$ too high. Find the correct average. (3 marks)
Answer: Recorded total $=25.6\times15=384$; true total $=384-3=381$; correct average $=\dfrac{381}{15}=25.4$.
Q4. The line graph of monthly maximum temperature shows Punjab rising from $19^{\circ}$C (Jan) to $38^{\circ}$C (Jun) then falling to $23^{\circ}$C (Dec), while Kerala stays near $29$–$33^{\circ}$C all year. State two valid inferences. (2 marks)
Answer: (i) Punjab's temperature varies much more across the year than Kerala's. (ii) Kerala is mostly flat, with similar temperatures in summer and winter; June is Punjab's hottest month, January its coldest.
Want personal coaching in Dwarka?
Book a free demo class
More Class 8 Mathematics chapters
Chat with us