Class 9 Mathematics Notes – Chapter 12: Information Handling

Punjab Board | 2026

EXERCISE 12.1

Question 1

Given frequency distribution:

Scores	Number of Students
24–28	3
29–33	6
34–38	12
39–43	23
44–48	15
49–53	6
Total	65

(i) What is the upper limit of the last class?

The last class is 49–53.

Answer: 53

(ii) What is the lower limit of the class 39–43?

Answer: 39

(iii) What is the midpoint of the class 34–38?

Midpoint (Class Mark):

\[ \begin{aligned} \text{Midpoint} &=\frac{\text{Lower Limit}+\text{Upper Limit}}{2}\\ &=\frac{34+38}{2}\\ &=\frac{72}{2}\\ &=\boxed{36} \end{aligned} \]

(iv) What are the class frequencies of 29–33 and 44–48?

Frequency of 29–33 = 6
Frequency of 44–48 = 15

(v) What is the class size?

\[ \begin{aligned} \text{Class Size}&=(28-24)+1\\ &=\boxed{5} \end{aligned} \]

Alternatively, the difference between consecutive lower limits is:

\[ 29-24=5 \]

(vi) In which class does the minimum number of students fall?

The minimum frequency is 3.

Answer: 24–28

(vii) What is the lower limit of the class having frequency 15?

The class with frequency 15 is 44–48.

Answer: 44

(viii) How many students have scores between 24 and 43?

\[ 3+6+12+23=\boxed{44} \]

Answer: 44 students

Question 2

Given data:

145, 152, 153, 156, 158, 160, 146, 152, 155, 159, 161, 163, 165, 147, 148, 151, 154, 156, 158, 160, 144, 167, 151, 150, 152

Minimum value = 144
Maximum value = 167
Class size = 3

| Class Interval | Tally Marks | Frequency (\(f\)) |
|:—:|:—:|—:|
| 144–146 | |||| | 4 |
| 147–149 | ||| | 3 |
| 150–152 | ||||| ||| | 8 |
| 153–155 | ||||| | | 6 |
| 156–158 | |||| | 4 |
| 159–161 | |||| | 4 |
| 162–164 | || | 2 |
| 165–167 | || | 2 |
| Total | — | 33 |

Frequencies of the last three classes:

Class Interval	Frequency
159–161	4
162–164	2
165–167	2

Note: The supplied data contains 25 observations, but the listed frequencies total 33. The frequency table should be checked against the original textbook question.

Question 3

Weights of students (in kg):

30, 33, 24, 21, 15, 39, 37, 44, 42, 33, 33, 28, 29, 32, 31, 28, 26, 32, 34, 35, 38, 36, 41, 30, 35, 41, 23, 26, 18, 34

Class size = 5

Class Interval	Frequency (\(f\))
15–19	2
20–24	3
25–29	5
30–34	10
35–39	6
40–44	4
Total	30

To construct a frequency polygon:

Find the midpoint of each class interval.
Plot the class midpoints on the \(x\)-axis and the corresponding frequencies on the \(y\)-axis.
Join the plotted points consecutively using straight line segments.

Note: The supplied table lists a frequency of 7 for class 35–39. Counting the given data gives 6; verify the original question.

Question 4

Marks obtained out of 100:

58, 59, 58, 33, 40, 58, 45, 46, 43, 45, 45, 50, 52, 49, 50, 57, 52, 55, 49, 50, 62, 49, 48, 44, 42, 47, 46, 47, 46, 53, 40, 44

Minimum value = 33
Maximum value = 62
Class size = 6

Class Interval	Frequency
33–38	2
39–44	6
45–50	14
51–56	5
57–62	5
Total	32

Least-frequency class: 33–38, with frequency 2.

Note: The supplied frequencies (2, 8, 12, 6, 4) do not match the provided data. The table above reflects a count of the observations as pasted.

Question 5

Given frequency distribution:

Weight (kg)	Frequency (\(f\))
10–14	6
15–19	17
20–24	23
25–29	30
30–34	22
35–39	13

Steps to draw a frequency polygon on a histogram:

Convert the class intervals into continuous class boundaries: 9.5–14.5, 14.5–19.5, 19.5–24.5, and so on.
Draw the histogram using the class boundaries on the \(x\)-axis and the frequencies on the \(y\)-axis.
Mark the midpoints of the tops of the rectangles.
Join the midpoints with straight line segments to form the frequency polygon.

Question 6

The number of heads obtained in 50 sets of tossing a coin five times is recorded.

To form a discrete frequency distribution:

List the possible numbers of heads: 0, 1, 2, 3, 4, and 5.
Count how many times each result occurs in the 50 trials.
Record the respective frequencies in a table.

Note: The individual results or frequency counts are not included in the supplied text.

Question 7

Marks of Grade 10 students in Mathematics:

Class Interval	Frequency
35–37	2
38–44	12
45–54	16
55–61	13
62–67	9
68–72	3

Steps to construct a histogram:

Convert the class intervals into continuous class boundaries.
Since the class widths are unequal, use frequency density (frequency divided by class width) for the bar heights.
Draw adjacent rectangular bars using the class boundaries on the \(x\)-axis and the appropriate heights on the \(y\)-axis.

Question 8

Given grouped data:

Class Interval	Frequency (\(f\))
5–8	2
8–12	12
12–20	25
20–25	32
25–27	14
27–32	5

To construct a histogram and frequency polygon:

Identify the class boundaries and widths.
Draw the histogram. For unequal class widths, use frequency density to determine the heights.
Find the midpoint of each class.
Plot the midpoints against the corresponding frequencies (or frequency densities, consistently with the histogram) and join them with straight line segments.

EXERCISE 12.2

Question 1

(i) Data: 4, 6, 10, 12, 15, 20, 25, 30

Arithmetic Mean:

\[ \begin{aligned} \bar{x}&=\frac{\sum x}{n}\\ &=\frac{4+6+10+12+15+20+25+30}{8}\\ &=\frac{122}{8}\\ &=\boxed{15.25} \end{aligned} \]

(ii) Data: 12, 18, 19, 0, −19, −18, −12

\[ \begin{aligned} \bar{x}&=\frac{12+18+19+0-19-18-12}{7}\\ &=\frac{0}{7}\\ &=\boxed{0} \end{aligned} \]

(iii) Data: 6.5, 11, 12.3, 9, 8.1, 16, 18, 20.5, 25

\[ \bar{x}=\frac{126.4}{9}\approx\boxed{14.04} \]

(iv) Data: 8, 10, 12, 14, 16, 20, 22

\[ \bar{x}=\frac{112}{7}=\boxed{16} \]

Question 2

Heights (in inches):

55, 53, 54, 58, 60, 61, 62, 56, 57, 52, 51, 63

Number of observations: \(n=12\)

Arranging in ascending order:

51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63

Since \(n=12\) is even, the median is the average of the 6th and 7th observations.

\[ \text{Median}=\frac{56+57}{2}=\boxed{56.5\text{ inches}} \]

Question 3

Earnings (Rs.):

88, 70, 72, 125, 115, 95, 81, 90, 95, 90

(i) Arithmetic Mean

\[ \bar{x}=\frac{921}{10}=\boxed{\text{Rs. }92.1} \]

(ii) Median

Arranged data:

70, 72, 81, 88, 90, 90, 95, 95, 115, 125

\[ \text{Median}=\frac{90+90}{2}=\boxed{90} \]

(iii) Mode

Both 90 and 95 occur twice.

Answer: The data is bimodal. Modes = 90 and 95.

Question 4

Frequency distribution of marks:

Marks	Frequency (\(f\))	Midpoint (\(x\))	\(fx\)
15–19	9	17	153
20–24	18	22	396
25–29	35	27	945
30–34	17	32	544
35–39	5	37	185
Total	84	—	2223

(i) Arithmetic Mean

Using the supplied data:

\[ \begin{aligned} \bar{x}&=\frac{\sum fx}{\sum f}\\ &=\frac{2223}{84}\\ &\approx\boxed{26.46} \end{aligned} \]

Note: The supplied text gives \(\sum fx=2248\), but the listed frequencies and midpoints give \(\sum fx=2223\). Check the original textbook table.

(ii) Median

\[ \frac{N}{2}=\frac{84}{2}=42 \]

The median class is 25–29. Using continuous boundaries:

\(L=24.5\)
\(N=84\)
\(c=27\), the cumulative frequency before the median class
\(f=35\)
\(h=5\)

\[ \text{Median}=L+\frac{N/2-c}{f}\times h \]

\[ \begin{aligned} &=24.5+\frac{42-27}{35}\times5\\ &\approx\boxed{26.64} \end{aligned} \]

Question 5

Frequency distribution:

Class Interval	Frequency
5–9	1
10–14	8
15–19	18
20–24	11
25–29	2

The modal class is 15–19 because it has the highest frequency.

Lower class boundary, \(L=14.5\)
Modal frequency, \(f_m=18\)
Previous frequency, \(f_1=8\)
Next frequency, \(f_2=11\)
Class width, \(h=5\)

\[ \text{Mode}=L+\frac{f_m-f_1}{(f_m-f_1)+(f_m-f_2)}\times h \]

\[ \begin{aligned} &=14.5+\frac{18-8}{(18-8)+(18-11)}\times5\\ &=14.5+\frac{50}{17}\\ &\approx\boxed{17.44} \end{aligned} \]

Question 6

Wages (Rs.):

4250, 4350, 4400, 4250, 4350, 4410, 4500, 4300, 4500, 4390

Mean using the short method

Assumed mean: \(A=4350\)

\[ \sum(x-A)=-100+0+50-100+0+60+150-50+150+40=200 \]

\[ \bar{x}=A+\frac{\sum(x-A)}{n} \]

\[ \bar{x}=4350+\frac{200}{10}=\boxed{\text{Rs. }4370} \]

Median

Arranged data:

4250, 4250, 4300, 4350, 4350, 4390, 4400, 4410, 4500, 4500

\[ \text{Median}=\frac{4350+4390}{2}=\boxed{\text{Rs. }4370} \]

Mode

4250, 4350, and 4500 each occur twice.

Answer: The data is multimodal, with modes 4250, 4350, and 4500.

Question 7

The arithmetic mean of 45 numbers is 80.

\[ \text{Sum}=\text{Mean}\times n \]

\[ 80\times45=\boxed{3600} \]

Question 8

Data: 1, 4, 0, 7, 9

Mean

\[ \bar{x}=\frac{1+4+0+7+9}{5}=\frac{21}{5}=\boxed{4.2} \]

Median

Arranged data: 0, 1, 4, 7, 9

Median = 4

Mode

Each value occurs once.

Answer: There is no mode.

Question 9

Data: 148, 145, 160, 157, 156, 160

Arranged data: 145, 148, 156, 157, 160, 160

Mode

\[ \boxed{160} \]

Median

\[ \text{Median}=\frac{156+157}{2}=\boxed{156.5} \]

Mean

\[ \begin{aligned} \bar{x}&=\frac{148+145+160+157+156+160}{6}\\ &=\frac{926}{6}\\ &\approx\boxed{154.33} \end{aligned} \]

Conclusion:

\[ \text{Mode}>\text{Median}>\text{Mean} \]

Question 10

Attendance:

21, 15, 16, 18, 14, 17, 15, 12, 13, 11

Arranged data:

11, 12, 13, 14, 15, 15, 16, 17, 18, 21

Median

\[ \text{Median}=\frac{15+15}{2}=\boxed{15} \]

Mode

\[ \boxed{15} \]

Mean using the assumed mean method

Take \(A=15\). The deviations are:

−4, −3, −2, −1, 0, 0, 1, 2, 3, 6

\[ \sum D=2 \]

\[ \begin{aligned} \bar{x}&=A+\frac{\sum D}{n}\\ &=15+\frac{2}{10}\\ &=\boxed{15.2} \end{aligned} \]

Question 11

Pocket money of 50 students:

Pocket Money	Frequency
5–10	12
10–15	9
15–20	18
20–25	7
25–30	4
Total	50

(i) Median

\[ \frac{N}{2}=\frac{50}{2}=25 \]

The median class is 15–20.

Using the supplied method and values:

\(L=14.5\)
\(c=21\)
\(f=18\)
\(h=5\)

\[ \begin{aligned} \text{Median}&=14.5+\frac{25-21}{18}\times5\\ &\approx\boxed{15.61} \end{aligned} \]

Mode

The modal class is 15–20.

\[ \text{Mode}=14.5+\frac{18-9}{(18-9)+(18-7)}\times5 \]

\[ \begin{aligned} &=14.5+\frac{45}{20}\\ &=\boxed{16.75} \end{aligned} \]

Note: The supplied text gives 16.73. The displayed formula evaluates to 16.75.

(ii) Arithmetic Mean

The supplied text gives the coding-method result:

\[ \boxed{\bar{x}=16.0} \]

Question 12

The mean age of 20 boys is 13 years, 4 months, and 5 days.

Sum of ages

\[ 20\times(13\text{ years }4\text{ months }5\text{ days}) \]

\[ =266\text{ years }8\text{ months }100\text{ days} \]

Using 30 days per month:

\[ =266\text{ years }11\text{ months }10\text{ days} \]

If one boy is exactly 15 years old, the sum of the remaining 19 boys is:

\[ 266\text{ years }11\text{ months }10\text{ days}-15\text{ years} \]

\[ =251\text{ years }11\text{ months }10\text{ days} \]

Note: The supplied answer gives the average age of the remaining boys as 13 years, 2 months, and 16 days. This result should be checked against the textbook’s conventions for converting months and days.

Question 13

(i) Given \(D=X-140\), \(\sum D=500\), and \(n=10\):

\[ \bar{x}=140+\frac{500}{10}=\boxed{190} \]

(ii) Given \(U=\frac{X-130}{6}\), \(\sum U=-150\), and \(n=15\):

\[ \begin{aligned} \bar{x}&=130+6\left(\frac{-150}{15}\right)\\ &=\boxed{70} \end{aligned} \]

(iii) Given \(D=X-25\), \(\sum D=300\), and \(\sum f=20\):

\[ \bar{x}=25+\frac{300}{20}=\boxed{40} \]

(iv) Given \(U=\frac{X-120}{5}\), \(\sum U=60\), and \(\sum f=100\):

\[ \bar{x}=120+5\left(\frac{60}{100}\right)=\boxed{123} \]

Question 14

The total marks and means are:

Student	Sum of Marks	Mean
Haris	350	\(350/5=70\)
Maham	293	\(293/5=58.6\)
Miral	313	\(313/5=62.6\)

Answer: Haris has the highest average, 70, and will be awarded Rs. 1000.

Question 15

Given \(D=X-20\):

\(D\)	Frequency (\(f\))	\(fD\)
−6	1	−6
−4	3	−12
−2	6	−12
0	20	0
2	26	52
4	12	48
6	2	12
Total	70	82

Using the listed values:

\[ \bar{x}=A+\frac{\sum fD}{\sum f} \]

\[ \bar{x}=20+\frac{82}{70}\approx\boxed{21.17} \]

Note: The supplied text gives \(\sum fD=84\) and a mean of 21.2. The listed frequencies and deviations give \(\sum fD=82\).

Question 16

Data: 45, 51, 58, 61, 74, 48, 46, 50

Assumed mean: \(A=58\)

Deviations (\(D=X-58\)):

−13, −7, 0, 3, 16, −10, −12, −8

\[ \sum D=-31,\qquad n=8 \]

\[ \begin{aligned} \bar{x}&=58+\frac{-31}{8}\\ &=58-3.875\\ &=\boxed{54.125} \end{aligned} \]

Question 17

Weighted Mean of Food Items

Food Item	Weight (\(w\))	Price (\(x\))	\(wx\)
Rice	10	96	960
Flour	12	48	576
Ghee	4	190	760
Sugar	3	49	147
Mutton	2	650	1300
Total	31	—	3743

\[ \begin{aligned} \text{Weighted Mean}&=\frac{\sum wx}{\sum w}\\ &=\frac{3743}{31}\\ &\approx\boxed{120.74} \end{aligned} \]

Question 18

Weighted Mean for Household Items

Item	Weight (\(w\))	Value (\(x\))	\(wx\)
Washing Machine	5	35	175
Heater	3	5	15
Stove	2	13	26
Dispenser	6	18	108
Total	16	—	324

\[ \text{Weighted Mean}=\frac{324}{16}=\boxed{20.25} \]

Question 19

Budgets (in millions): 5, 7, 8, 6, 7

\[ \begin{aligned} \text{Average Budget}&=\frac{5+7+8+6+7}{5}\\ &=\frac{33}{5}\\ &=\boxed{6.6\text{ million}} \end{aligned} \]

Question 20

Weighted Marks

Subject	Marks (\(x\))	Weight (\(w\))	\(wx\)
Urdu	78	5	390
English	65	4	260
Science	80	2	160
Mathematics	90	3	270
Islamiyat	85	2	170
Computer	72	4	288
Total	—	20	1538

\[ \text{Weighted Mean}=\frac{1538}{20}=\boxed{76.9} \]

REVIEW EXERCISE 12

Question 1: Multiple Choice Questions (MCQs)

No.	Question	Correct Answer
(i)	Which data takes only some specific values?	(b) Discrete data
(ii)	The number of times a value occurs in data is called:	(a) Frequency
(iii)	Midpoint is also known as:	(d) Class mark
(iv)	A frequency polygon is also constructed by using:	(c) Class boundaries
(v)	The difference between the greatest and smallest values is called:	(d) Range
(vi)	A measure of central tendency is used to find the _____ of a data set.	(c) Middle or centre value
(vii)	If the mean of 5, 7, 8, 9, and \(x\) is 7.5, find \(x\).	(c) 8.5
(viii)	Find the mode: 2, 5, 8, 9, 0, 1, 3, 7, 10.	(d) No mode
(ix)	The value that occurs most often is called:	(b) Mode
(x)	Find the median: 110, 125, 122, 130, 124, 127, 120.	(a) 124

Question 2: Definitions

(i) Frequency Distribution

A tabular arrangement of data showing the frequency of observations in different classes or groups.

(ii) Histogram with Unequal Class Intervals

A graphical representation of a frequency distribution. When class widths vary, the bar heights should be proportional to frequency density.

(iii) Mean

The sum of all observations divided by the total number of observations.

\[ \bar{x}=\frac{\sum x}{n} \]

(iv) Median

The middle value that divides an ordered data set into two equal halves.

Question 3

For the weights of 40 students:

Construct a frequency table using a class size of 10.
Draw the corresponding histogram.
Draw the frequency polygon.

Note: The individual weights are not included in the supplied text, so the frequency table and graphs cannot be completed from the provided information alone.

Question 4

Frequency Polygon on a Histogram

Convert the class intervals to continuous class boundaries.
Draw the histogram using the appropriate bar heights.
Find the midpoint of each class.
Mark the midpoint at the top of each bar.
Join the marked points with straight line segments.

For unequal class widths, use frequency density consistently when constructing the histogram and polygon.

Question 5

Biology marks of 45 students:

Marks	Frequency
20–24	5
25–29	8
30–34	12
35–39	15
40–44	3
45–49	2
Total	45

(i) Upper class boundary of the fifth class (40–44)

\[ \boxed{44.5} \]

(ii) Lower class boundaries

\[ 19.5,\ 24.5,\ 29.5,\ 34.5,\ 39.5,\ 44.5 \]

(iii) Midpoints of all classes

\[ 22,\ 27,\ 32,\ 37,\ 42,\ 47 \]

(iv) Class interval with the least frequency

Answer: 45–49, with frequency 2.

Question 6

Frequency Polygon and Histogram

Plot the classes 5–9, 10–14, 15–19, 20–24, 25–29, and 30–34 with the respective frequencies:

1, 8, 18, 11, 2, 5.

Use continuous class boundaries for the histogram. Then plot the class midpoints and join them to form the frequency polygon.

Question 7

Weighted Mean for Items

Item	Weight (\(w\))	Price (\(x\))	\(wx\)
Chair	20	500	10000
Table	20	400	8000
Blackboard	10	750	7500
Tube Light	25	230	5750
Cupboard	9	950	8550
Total	84	—	39800

\[ \begin{aligned} \text{Weighted Mean}&=\frac{39800}{84}\\ &\approx\boxed{473.81} \end{aligned} \]

Question 8

A total of Rs. 50,000 is allocated across five sectors.

\[ \begin{aligned} \text{Average Allocation}&=\frac{50000}{5}\\ &=\boxed{\text{Rs. }10000} \end{aligned} \]

Question 9

Saad’s marks in six tests:

84, 91, 72, 68, 87, 78

\[ \begin{aligned} \text{Mean}&=\frac{84+91+72+68+87+78}{6}\\ &=\frac{480}{6}\\ &=\boxed{80} \end{aligned} \]

Question 10

Maximum-load capacities of ropes:

Maximum Load (kg)	Frequency
93–97	2
98–102	5
103–107	8
108–112	12
113–117	6
118–122	2
Total	35

Using the short or coding method, with assumed mean \(A=110\) and class width \(h=5\), the supplied text gives:

\[ \bar{x}=110+\left(\frac{\sum fU}{\sum f}\right)\times5 \]

\[ \boxed{\bar{x}=108.44} \]

Note: The coding table or \(\sum fU\) is not provided, so the reported mean cannot be independently verified from the supplied details.

Question 11

Sums obtained when rolling a die eight times:

8, 5, 6, 6, 9, 4, 3, 11

Arranged data:

3, 4, 5, 6, 6, 8, 9, 11

Median

\[ \text{Median}=\frac{6+6}{2}=\boxed{6} \]

Mode

\[ \boxed{6} \]

Question 12

Weekly wages of employees:

Weekly Wages (Rs.)	Frequency (\(f\))	Midpoint (\(x\))	\(fx\)
600–700	3	650	1950
700–800	5	750	3750
800–900	7	850	5950
900–1000	21	950	19950
1000–1100	11	1050	11550
Total	47	—	43150

Arithmetic Mean

\[ \begin{aligned} \bar{x}&=\frac{\sum fx}{\sum f}\\ &=\frac{43150}{47}\\ &\approx\boxed{\text{Rs. }917.02} \end{aligned} \]

Note: The supplied text gives \(\sum fx=43050\) and a mean of Rs. 915.96. Multiplying the listed midpoints by the listed frequencies gives \(\sum fx=43150\); check the original table.

Notes by Sir Dilawar

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