Q: Using the salary ogive (points plotted at 2k, 4k, 6k, 8k, 10k, 12k with cf 10, 20, 35, 55, 70, 75 up to N=80), find: (a) total number of workers. (b) median class. (c) lower-quartile class. (d) workers earning ≥ ₹6,000 but < ₹10,000.

(a) Total number of workers = 80 workers . [1 Mark] (b) Median = N/2 = 40th worker ⇒ lies in ₹6,000 – ₹8,000 class . [1 Mark] (c) Lower quartile Q1 = N/4 = 20th worker ⇒ ₹2,000 – ₹4,000 class . [1 Mark] (d) Number of workers = cf(10,000) − cf(6,000) = 70 − 35 = 35 workers . [1 Mark]

Q: Histogram of plant heights: 30–40 (4), 40–50 (2), 50–60 (8), 60–70 (12), 70–80 (6), 80–90 (3), 90–100 (4). (a) Frequency table. (b) State the modal class. (c) Identify mode of distribution. (d) Number of plants with height 80 cm to 90 cm.

(a) Table completed. [2 Marks] (b) Modal class = 60 − 70 cm . [1 Mark] (c) From intersection of diagonals: Mode = 64 cm . [1 Mark] (d) Reading bar 80–90 cm: 3 plants . [1 Mark]

Q: Ages of 100 policy holders: 20–25 (2), 25–30 (4), 30–35 (12), 35–40 (20), 40–45 (28), 45–50 (22), 50–55 (8), 55–60 (4). Draw ogive to find: (a) Median age (b) Policy holders above 52 years.

Plot at upper limits: (20,0), (25,2), (30,6), (35,18), (40,38), (45,66), (50,88), (55,96), (60,100). [2 Marks] (a) N /2 = 50 ⇒ Median age ≈ 42.1 years (41.5–42.5 accepted). [1 Mark] (b) At age 52, cf ≈ 91 ⇒ Above 52 = 100 − 91 = 9 policy holders . [1 Mark]

Q: The following distribution gives the daily wages of 60 workers of a factory: Daily income in ₹ 200–300 300–400 400–500 500–600 600–700 700–800 Number of workers ($f$) 6 10 14 16 10 4 Use graph paper to answer this question. Take 2 cm = ₹100 along one axis and 2 cm = 2 workers along the other axis. D

Daily Wages (₹) Workers 200 300 400 500 600 700 800 Mode = ₹525 Figure: Histogram with modal class 500–600 and intersecting diagonals giving Mode = ₹525 Step 1 Highest frequency bar corresponds to class 500–600 (frequency 16). Draw diagonals connecting upper corners of the modal bar to the adjacent bars. 1.5 Marks Step 2 Drop perpendicular from intersection point to horizontal axis. Reading from the graph: $\mathbf{\text{Mode} = ₹525}$ (acceptable range ₹520 to ₹530). Formula Check: $\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 500 + \frac{16 - 14}{32 - 14 - 10} \times 100 = 500 + \frac{2}{8} \times 100 = 500 + 25 = ₹525$. 1.5 Marks

Q: Use graph paper to answer this question. During a medical checkup of 60 students in a school, weights were recorded as follows: Weight (in kg) 28–30 30–32 32–34 34–36 36–38 38–40 40–42 42–44 Number of Students 2 4 10 13 15 9 5 2 Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the othe

Weight (kg) cf Median ≈ 36.1 Q3 ≈ 38.2 28 32 36 40 44 30 45 60 Figure: Cumulative frequency ogive showing Median reading at $N/2=30$ and $Q_3$ at $3N/4=45$ Cumulative Frequency Table Points to plot (Upper Boundary, $cf$): $(28, 0), (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58), (44, 60)$. Total number of students $N = 60$. 2 Marks (a) Median $\frac{N}{2} = \frac{60}{2} = 30^\text{th}$ student. Locate $cf = 30$ on vertical axis, draw horizontal line to curve and drop perpendicular to weight axis. Median $\approx 36.1\text{ kg}$ (acceptable range $36.0\text{ to } 36.3\text{ kg}$). 1.5 Marks (b) Upper Quartile $Q_3$ $\frac{3N}{4} = \frac{3 \times 60}{4} = 45^\text{th}$ student. From $cf = 45$, drop perpendicular from curve to weight axis. Upper Quartile $Q_3 \approx 38.2\text{ kg}$ (acceptable range $38.0\text{ to } 38.4\text{ kg}$). 1.5 Marks (c) Students Weighing Above 37 kg From weight $= 37\text{ kg}$ on horizontal axis, draw vertical line to ogive curve. Corresponding cumulative frequency reading $\approx 36.5$ (approx 36 or 37 students weigh $\le 37\text{ kg}$). Number of students weighing above $37\text{ kg} = 60 - 36.5 \approx \mathbf{23\text{ to } 24\text

Q: Draw a Histogram for the given data, using a graph paper: Weekly Wages (in ₹): 3000–4000 (4), 4000–5000 (9), 5000–6000 (18), 6000–7000 (6), 7000–8000 (7), 8000–9000 (2), 9000–10000 (4). Estimate the mode from the graph.

Weekly Wages (₹) No. of People 3k 4k 5k 6k 7k 8k 9k 10k Mode ≈ ₹5,400 Figure: Histogram of weekly wages with intersecting diagonals for mode estimation Step 1 Plot class intervals along the horizontal axis (taking a kink between 0 and 3000) and number of people along the vertical axis ($1\text{ cm} = 2\text{ people}$). Construct rectangles for each class interval. [2 Marks] Step 2 The highest bar corresponds to the modal class 5000 – 6000 (frequency = 18). Join the top-left vertex to top-left of following bar and top-right vertex to top-right of preceding bar. [1 Mark] Step 3 Draw a vertical line from the intersection point to the horizontal axis. Reading on the x-axis gives: Mode ≈ ₹5,400 (acceptable range: ₹5,350 to ₹5,450). [1 Mark]

Q: 40 students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below: Distance in m: 12–13 (3), 13–14 (9), 14–15 (12), 15–16 (9), 16–17 (4), 17–18 (2), 18–19 (1). Use a graph paper to draw an ogive for the above distribution. Use a scale of $2\text{ cm} = 1\text{ m

Distance thrown in m Cumulative Frequency (N=40) Median ≈ 14.7 m Q3 ≈ 15.7 m 12 13 14 15 16 17 18 19 0 10 20 30 40 Figure: Cumulative frequency curve (Ogive) showing median, upper quartile, and $16.5\text{ m}$ threshold Cumulative Table (Upper Limit, cf): $(12, 0), (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39), (19, 40)$. Smooth ogive plotted on graph paper with given scale. [3 Marks] (i) Median $= \frac{N}{2} = \frac{40}{2} = 20\text{th term}$. Horizontal line from $y = 20$ meets curve at $x \approx \mathbf{14.7\text{ m}}$ (acceptable: $14.6 - 14.8$). [1 Mark] (ii) Upper Quartile $Q_3 = \frac{3N}{4} = \frac{3 \times 40}{4} = 30\text{th term}$. Horizontal line from $y = 30$ meets curve at $x \approx \mathbf{15.7\text{ m}}$ (acceptable: $15.6 - 15.8$). [1 Mark] (iii) Above 16.5 m Vertical line from $x = 16.5$ meets curve at $cf \approx 35$. Number of students covering $> 16.5\text{ m} = 40 - 35 = \mathbf{5\text{ students}}$ (acceptable: $4 - 6$). [1 Mark]

Question 1

The histogram drawn on the graph represents the number of students of different heights. (a) State the modal class. (b) Find the mode of the distribution graphically from the histogram.

Accepted Answer

Part (a) Modal Class The bar with the highest frequency (f = 14) is for height 130–140 cm.
Hence, Modal Class = 130–140 cm. [1 Mark]

Part (b) Mode from Graph Join the top-left vertex of the modal bar to the top-left vertex of the right adjacent bar, and top-right of modal to top-right of left adjacent bar.
From their point of intersection, draw a vertical line down to the x-axis.
The reading on the x-axis gives Mode = 137 cm (± 0.5 cm). [3 Marks]

Question 2

Using a graph paper, draw an ogive for the following distribution showing records of weight of 100 students (Scale: 2 cm = 5 kg on x-axis, 2 cm = 10 students on y-axis):
Weight: 35–40 (f=4), 40–45 (f=6), 45–50 (f=10), 50–55 (f=24), 55–60 (f=26), 60–65 (f=17), 65–70 (f=8), 70–75 (f=5).
Use your ogive to estimate the:
(a) median weight of the students.
(b) percentage of students whose weight is 60 kg or more.

Accepted Answer

Weight (kg) Cumulative Frequency (N=100) 50 (Median) 56.5 kg 70 60 kg 35 45 70 Ogive (Cumulative Frequency) curve with Median ≈ 56.5 kg and ≥60 kg estimate = 30% Cumulative Frequencies (40, 4), (45, 10), (50, 20), (55, 44), (60, 70), (65, 87), (70, 95), (75, 100). [1.5 Marks] Ogive Plot Plotted points joined smoothly with a free-hand S-curve starting from (35, 0). [1.5 Marks] Part (a) Median Total N = 100. Median position = N/2 = 50th student. From y = 50, project horizontally to the curve and down to the x-axis. Median Weight ≈ 56.5 kg (± 0.5 kg) . [1.5 Marks] Part (b) ≥ 60 kg From x = 60 kg, draw vertical line to meet curve at y = 70. Number of students weighing 60 kg or more = 100 − 70 = 30 students. Percentage = (30 / 100) × 100 = 30% . [1.5 Marks]

Question 3

Draw a Histogram for the distribution (Marks: 30–40, 40–50, 50–60, 60–70, 70–80; Students: 10, 26, 40, 54, 34) and find Mode. (Scale: 2 cm = 10 Marks, 2 cm = 10 students).

Accepted Answer

1. Tallest bar is modal class 60–70 with frequency 54. [1.5 Marks]
2. Connecting top corners diagonally to adjacent bars gives mode at x ≈ 65 marks (calculated: 60 + (14/34)×10 = 64.1 marks). [1.5 Marks]

Question 4

Using the salary ogive (points plotted at 2k, 4k, 6k, 8k, 10k, 12k with cf 10, 20, 35, 55, 70, 75 up to N=80), find:
(a) total number of workers.
(b) median class.
(c) lower-quartile class.
(d) workers earning ≥ ₹6,000 but < ₹10,000.

Accepted Answer

(a) Total number of workers = 80 workers . [1 Mark] (b) Median = N/2 = 40th worker ⇒ lies in ₹6,000 – ₹8,000 class . [1 Mark] (c) Lower quartile Q1 = N/4 = 20th worker ⇒ ₹2,000 – ₹4,000 class . [1 Mark] (d) Number of workers = cf(10,000) − cf(6,000) = 70 − 35 = 35 workers . [1 Mark]

Question 5

Histogram of plant heights: 30–40 (4), 40–50 (2), 50–60 (8), 60–70 (12), 70–80 (6), 80–90 (3), 90–100 (4).
(a) Frequency table.
(b) State the modal class.
(c) Identify mode of distribution.
(d) Number of plants with height 80 cm to 90 cm.

Accepted Answer

(a) Table completed. [2 Marks] (b) Modal class = 60 − 70 cm . [1 Mark] (c) From intersection of diagonals: Mode = 64 cm . [1 Mark] (d) Reading bar 80–90 cm: 3 plants . [1 Mark]

Question 6

Ages of 100 policy holders: 20–25 (2), 25–30 (4), 30–35 (12), 35–40 (20), 40–45 (28), 45–50 (22), 50–55 (8), 55–60 (4). Draw ogive to find: (a) Median age (b) Policy holders above 52 years.

Accepted Answer

Plot at upper limits: (20,0), (25,2), (30,6), (35,18), (40,38), (45,66), (50,88), (55,96), (60,100). [2 Marks]
(a) N/2 = 50 ⇒ Median age ≈ 42.1 years (41.5–42.5 accepted). [1 Mark]
(b) At age 52, cf ≈ 91 ⇒ Above 52 = 100 − 91 = 9 policy holders. [1 Mark]

Question 7

The following distribution gives the daily wages of 60 workers of a factory: Daily income in ₹ 200–300 300–400 400–500 500–600 600–700 700–800 Number of workers ($f$) 6 10 14 16 10 4 Use graph paper to answer this question. Take 2 cm = ₹100 along one axis and 2 cm = 2 workers along the other axis. D

Accepted Answer

Daily Wages (₹) Workers 200 300 400 500 600 700 800 Mode = ₹525 Figure: Histogram with modal class 500–600 and intersecting diagonals giving Mode = ₹525 Step 1 Highest frequency bar corresponds to class 500–600 (frequency 16). Draw diagonals connecting upper corners of the modal bar to the adjacent bars. 1.5 Marks Step 2 Drop perpendicular from intersection point to horizontal axis. Reading from the graph: $\mathbf{	ext{Mode} = ₹525}$ (acceptable range ₹520 to ₹530). Formula Check: $	ext{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} 	imes h = 500 + \frac{16 - 14}{32 - 14 - 10} 	imes 100 = 500 + \frac{2}{8} 	imes 100 = 500 + 25 = ₹525$. 1.5 Marks

Question 8

Use graph paper to answer this question.
During a medical checkup of 60 students in a school, weights were recorded as follows:

Weight (in kg)	28–30	30–32	32–34	34–36	36–38	38–40	40–42	42–44
Number of Students	2	4	10	13	15	9	5	2

Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the other axis, draw an ogive. Use your graph to find the:

median
upper quartile
number of students whose weight is above 37 kg.

Accepted Answer

Weight (kg) cf Median ≈ 36.1 Q3 ≈ 38.2 28 32 36 40 44 30 45 60 Figure: Cumulative frequency ogive showing Median reading at $N/2=30$ and $Q_3$ at $3N/4=45$ Cumulative Frequency Table Points to plot (Upper Boundary, $cf$): $(28, 0), (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58), (44, 60)$. Total number of students $N = 60$. 2 Marks (a) Median $\frac{N}{2} = \frac{60}{2} = 30^	ext{th}$ student. Locate $cf = 30$ on vertical axis, draw horizontal line to curve and drop perpendicular to weight axis. Median $\approx 36.1	ext{ kg}$ (acceptable range $36.0	ext{ to } 36.3	ext{ kg}$). 1.5 Marks (b) Upper Quartile $Q_3$ $\frac{3N}{4} = \frac{3 	imes 60}{4} = 45^	ext{th}$ student. From $cf = 45$, drop perpendicular from curve to weight axis. Upper Quartile $Q_3 \approx 38.2	ext{ kg}$ (acceptable range $38.0	ext{ to } 38.4	ext{ kg}$). 1.5 Marks (c) Students Weighing Above 37 kg From weight $= 37	ext{ kg}$ on horizontal axis, draw vertical line to ogive curve. Corresponding cumulative frequency reading $\approx 36.5$ (approx 36 or 37 students weigh $\le 37	ext{ kg}$). Number of students weighing above $37	ext{ kg} = 60 - 36.5 \approx \mathbf{23	ext{ to } 24	ext

Question 9

Draw a Histogram for the given data, using a graph paper:
Weekly Wages (in ₹): 3000–4000 (4), 4000–5000 (9), 5000–6000 (18), 6000–7000 (6), 7000–8000 (7), 8000–9000 (2), 9000–10000 (4).
Estimate the mode from the graph.

Accepted Answer

Weekly Wages (₹) No. of People 3k 4k 5k 6k 7k 8k 9k 10k Mode ≈ ₹5,400 Figure: Histogram of weekly wages with intersecting diagonals for mode estimation Step 1 Plot class intervals along the horizontal axis (taking a kink between 0 and 3000) and number of people along the vertical axis ($1	ext{ cm} = 2	ext{ people}$). Construct rectangles for each class interval. [2 Marks] Step 2 The highest bar corresponds to the modal class 5000 – 6000 (frequency = 18). Join the top-left vertex to top-left of following bar and top-right vertex to top-right of preceding bar. [1 Mark] Step 3 Draw a vertical line from the intersection point to the horizontal axis. Reading on the x-axis gives: Mode ≈ ₹5,400 (acceptable range: ₹5,350 to ₹5,450). [1 Mark]

Question 10

40 students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below: Distance in m: 12–13 (3), 13–14 (9), 14–15 (12), 15–16 (9), 16–17 (4), 17–18 (2), 18–19 (1). Use a graph paper to draw an ogive for the above distribution. Use a scale of $2	ext{ cm} = 1	ext{ m

Accepted Answer

Distance thrown in m Cumulative Frequency (N=40) Median ≈ 14.7 m Q3 ≈ 15.7 m 12 13 14 15 16 17 18 19 0 10 20 30 40 Figure: Cumulative frequency curve (Ogive) showing median, upper quartile, and $16.5	ext{ m}$ threshold Cumulative Table (Upper Limit, cf): $(12, 0), (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39), (19, 40)$. Smooth ogive plotted on graph paper with given scale. [3 Marks] (i) Median $= \frac{N}{2} = \frac{40}{2} = 20	ext{th term}$. Horizontal line from $y = 20$ meets curve at $x \approx \mathbf{14.7	ext{ m}}$ (acceptable: $14.6 - 14.8$). [1 Mark] (ii) Upper Quartile $Q_3 = \frac{3N}{4} = \frac{3 	imes 40}{4} = 30	ext{th term}$. Horizontal line from $y = 30$ meets curve at $x \approx \mathbf{15.7	ext{ m}}$ (acceptable: $15.6 - 15.8$). [1 Mark] (iii) Above 16.5 m Vertical line from $x = 16.5$ meets curve at $cf \approx 35$. Number of students covering $> 16.5	ext{ m} = 40 - 35 = \mathbf{5	ext{ students}}$ (acceptable: $4 - 6$). [1 Mark]

Question 11

Use graph paper for this question. The marks obtained by 120 students in an English test are given below: Marks 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100 No. of students 5 9 16 22 26 18 11 6 4 3 Draw the ogive and hence, estimate: the median marks. the number of students who did no

Accepted Answer

Marks Cumulative Frequency (Students) 0 10 20 30 40 50 60 70 80 90 100 20 40 60 80 90 120 Median ≈ 43 Failed = 78 Q₃ ≈ 57 Vector Diagram: Cumulative frequency curve (Ogive) with graphical projections for Median, Fail count, and Q₃ Step 1 Cumulative Frequency Table ($N = 120$): Upper Limits & Cumulative Frequencies: $(10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120)$. Plot points on graph paper and join with a smooth, continuous freehand S-curve. [2 Marks] Step 2 (i) Determine Median Marks: $	ext{Median position} = \frac{N}{2} = \frac{120}{2} = 60	ext{th term}$. Draw a horizontal line from $y = 60$ on the vertical axis to meet the curve, then drop a vertical perpendicular to the x-axis. $\mathbf{	ext{Median Marks} \approx 43	ext{ marks}}$ (accepted range: $42 - 44$). [1.5 Marks] Step 3 (ii) Number of students who did not pass ($< 50	ext{ marks}$): Pass mark $= 50$. At $x = 50$ on the horizontal axis, draw a vertical line up to meet the ogive. From that intersection point, project horizontally to the y-axis, which corresponds to $y = 78$. $\mathbf{	ext{Number of students who failed} = 78	ext{ students}}$. [1.5 Marks] Step 4

Question 12

Use graph paper for this question; draw the histogram and estimate mode for the following data. Use $2	ext{ cm} = 10	ext{ units}$ along the x-axis, and $2	ext{ cm} = 2	ext{ units}$ along the y-axis. Class 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 10 12 18 9 11

Accepted Answer

Class Frequency 30 40 50 Mode = 54 60 70 80 18 12 10 Vector Diagram: Histogram and intersection method determining Mode = 54 Step 1 Identify Modal Class: Highest frequency is $18$, corresponding to modal class $50 - 60$. [1 Mark] Step 2 Draw histogram and criss-cross lines: Join top corners of modal bar to opposite top corners of adjacent bars to locate intersection point. [1 Mark] Step 3 Drop perpendicular from intersection to x-axis: $\mathbf{	ext{Estimated Mode} = 54}$ (acceptable range: $53 - 55$). [1 Mark]

Question 13

Use graph sheet for this question. Take $2	ext{ cm} = 10	ext{ students}$ along one axis and $2	ext{ cm} = 5	ext{ cm}$ of height along the other axis. The following table shows the height of 80 students in a class: Height (in cms) 150 – 155 155 – 160 160 – 165 165 – 170 170 – 175 175 – 180 No. of

Accepted Answer

Height (cm) Cumulative Frequency Median ≈ 163.5 cm Q₁ ≈ 159 cm cf = 71 (>173 cm: 80 - 71 = 9) 150 155 160 165 170 175 180 40 20 Vector Diagram: Height Ogive for 80 students with projections for Median, Q₁, and students > 173 cm Step 1 Prepare Cumulative Frequency Table ($N = 80$): $(155, 8), (160, 23), (165, 47), (170, 64), (175, 75), (180, 80)$. Plot cumulative frequencies against upper limits on graph paper and join with smooth S-curve. [2 Marks] Step 2 (a) Median height ($N/2 = 40	ext{th student}$): Horizontal line from $y = 40$ to ogive, then drop perpendicular to x-axis $\implies \mathbf{	ext{Median} \approx 163.5	ext{ cm}}$ (acceptable range: $163 - 164	ext{ cm}$). [1 Mark] Step 3 (b) Lower-quartile height $Q_1$ ($N/4 = 20	ext{th student}$): From $y = 20$ on y-axis to curve, drop perpendicular $\implies \mathbf{Q_1 \approx 159	ext{ cm}}$ (acceptable range: $158.5 - 159.5	ext{ cm}$). [1 Mark] Step 4 (c) Number of students taller than $173	ext{ cm}$: From $x = 173$ on x-axis, vertical line to curve meets at $y \approx 71$. Number of students with height $> 173	ext{ cm} = 80 - 71 = \mathbf{9	ext{ students}}$. [1 Mark]

Question 14

Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data:

Runs scored	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000
No. of batsmen	4	18	9	6	7	2	4

Accepted Answer

5 10 15 20 3000 4000 5000 6000 7000 8000 9000 10000 Runs Scored → No. of Batsmen → Mode ≈ 4600 Step 1 Modal class identification: The class with the highest frequency ($18$) is $4000 - 5000$. [1 Mark] Step 2 Drawing the Histogram: Plot class intervals on the x-axis and frequencies on the y-axis, drawing rectangles of corresponding heights. [2 Marks] Step 3 Graphical Mode Estimation: Inside the modal rectangle ($4000 - 5000$), connect the upper left corner to the top left of the adjacent right bar, and the upper right corner to the top right of the adjacent left bar. Drop a perpendicular from their intersection to the x-axis. $$	ext{Estimated Mode} = \mathbf{4600	ext{ runs}} \quad (	ext{acceptable range: } 4550 - 4650)$$ [1 Mark] Answer: Estimated Mode $\approx 4600$ runs.

Question 15

Use Graph paper for this question. A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded: Height in cm 135 – 140 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 165 – 170 No. of boys 4 8 20 14 7 6 1 Taking 2cm = height of 10 c

Accepted Answer

10 20 30 40 50 60 135 140 145 150 155 160 165 170 Median ≈ 149.5 Q1 ≈ 146 cf = 50 Height (in cm) → Cumulative Frequency → Cumulative Frequency Table • 135 – 140: $f = 4, cf = 4$ → Plot $(140, 4)$ • 140 – 145: $f = 8, cf = 12$ → Plot $(145, 12)$ • 145 – 150: $f = 20, cf = 32$ → Plot $(150, 32)$ • 150 – 155: $f = 14, cf = 46$ → Plot $(155, 46)$ • 155 – 160: $f = 7, cf = 53$ → Plot $(160, 53)$ • 160 – 165: $f = 6, cf = 59$ → Plot $(165, 59)$ • 165 – 170: $f = 1, cf = 60$ → Plot $(170, 60)$ Base point: $(135, 0)$. [2 Marks for table & ogive] (i) Median Total $N = 60 \implies \frac{N}{2} = 30^{	ext{th}}$ term. From $y = 30$ on y-axis, draw horizontal line to meet curve, then drop vertically to x-axis: $$	ext{Median} \approx \mathbf{149.5	ext{ cm}} \quad (	ext{accept } 149 - 150	ext{ cm})$$ [1.5 Marks] (ii) Lower Quartile Q₁ $Q_1 = \frac{N}{4} = \frac{60}{4} = 15^{	ext{th}}$ term. From $y = 15$ on y-axis, drop vertically to x-axis: $$Q_1 \approx \mathbf{146	ext{ cm}} \quad (	ext{accept } 145.5 - 146.5	ext{ cm})$$ [1.5 Marks] (iii) Tall Boys (> 158 cm) At height $x = 158	ext{ cm}$, the cumulative frequency from the graph is $50$. Number of boys with height $> 158	ext{ cm} = 60

Question 16

The daily wages of 80 workers in a project are given below:

Wages (in ₹)	400-450	450-500	500-550	550-600	600-650	650-700	700-750
No. of workers	2	6	12	18	24	13	5

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = ₹50 on x-axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate:
(i) the median wage of the workers.
(ii) the lower quartile wage of workers.
(iii) the number of workers who earn more than ₹625 daily.

Accepted Answer

20 40 60 80 400 450 500 550 600 650 700 750 Median ≈ ₹605 Q1 = ₹550 cf = 50 Daily Wages (in ₹) → Cumulative Frequency → Cumulative Frequency Table • 400-450: $cf = 2$ → $(450, 2)$ • 450-500: $cf = 8$ → $(500, 8)$ • 500-550: $cf = 20$ → $(550, 20)$ • 550-600: $cf = 38$ → $(600, 38)$ • 600-650: $cf = 62$ → $(650, 62)$ • 650-700: $cf = 75$ → $(700, 75)$ • 700-750: $cf = 80$ → $(750, 80)$ Base point: $(400, 0)$. [2 Marks] (i) Median Wage $\frac{N}{2} = 40^{	ext{th}}$ worker. From $y = 40$ on graph, drop to x-axis: $$	ext{Median} \approx \mathbf{₹605} \quad (	ext{acceptable: } 600 - 610)$$ [1.5 Marks] (ii) Lower Quartile Wage $\frac{N}{4} = 20^{	ext{th}}$ worker. From $y = 20$ on graph: $$Q_1 = \mathbf{₹550}$$ [1.5 Marks] (iii) Workers Earning > ₹625 At $x = 625$, the cumulative frequency is $50$. $$	ext{Number of workers} = 80 - 50 = \mathbf{30	ext{ workers}}$$ [1 Mark] Answers: (i) Median $\approx ₹605$, (ii) Lower quartile $= ₹550$, (iii) 30 workers.

23. Ogives and Histograms Previous Year Questions

CISCE Examiner Observation & Common Deduction Pitfall

Archive Expansion Slot • 2014–2025 PYQs

Daily income in ₹	200–300	300–400	400–500	500–600	600–700	700–800
Number of workers ($f$)	6	10	14	16	10	4

Marks	0–10	10–20	20–30	30–40	40–50	50–60	60–70	70–80	80–90	90–100
No. of students	5	9	16	22	26	18	11	6	4	3

Class	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Frequency	10	12	18	9	11

Height (in cms)	150 – 155	155 – 160	160 – 165	165 – 170	170 – 175	175 – 180
No. of students	8	15	24	17	11	5

Height in cm	135 – 140	140 – 145	145 – 150	150 – 155	155 – 160	160 – 165	165 – 170
No. of boys	4	8	20	14	7	6	1