Class : 9 – Math (English) : Lesson 12. Statistics

EXPLANATION & SUMMARY

🔵 Detailed Explanation
📌 1. Introduction
🔴 Statistics is the branch of mathematics dealing with the collection, presentation, analysis, and interpretation of numerical data. In daily life—weather reports, exam scores, market trends—statistics helps us understand and compare data effectively.
💡 Concept: “Statistical data” refers to a set of numerical facts. The term comes from “status” (state), historically used for state administration data.

📌 2. Types of Data
🟢 (a) Primary Data ➡️ Collected by the investigator firsthand.
🟢 (b) Secondary Data ➡️ Obtained from published sources or others’ work.
✏️ Example: A teacher surveying her class = primary; using a newspaper’s poll result = secondary.

📌 3. Frequency and Frequency Distribution
🔴 Frequency (fᵢ) is the number of times a data value xᵢ occurs.
🔵 Frequency Distribution Table: Organizes data into classes and frequencies.
💡 Steps:
1️⃣ Choose class intervals.
2️⃣ Tally occurrences.
3️⃣ Count tallies to fill frequencies.

📌 4. Inclusive vs. Exclusive Classes
🟢 Exclusive form → Upper limit not included (e.g., 0–10, 10–20).
🟡 Inclusive form → Both limits included (e.g., 1–10, 11–20).

📌 5. Cumulative Frequency (CF)
🔴 Less than CF → Sum of frequencies up to an upper class boundary.
🔴 Greater than CF → Sum from lower boundary upward.
Formula:
CFₖ = Σ fᵢ (for i = 1 to k).

📌 6. Graphical Representation
🟢 Bar Graphs: Rectangles with equal spacing, heights ∝ frequencies.
🟡 Histograms: Adjacent rectangles (no gaps) for continuous data.
🔵 Frequency Polygons: Connect midpoints of histogram bars with line segments.
🔴 CF Curves (Ogives): Plot cumulative frequencies vs. class boundaries.
✏️ Tip: Use equal scales on axes for accuracy.

📌 7. Measures of Central Tendency
🟡 7.1 Mean (x̄)
Formula:
x̄ = Σ(xᵢ × fᵢ) / Σfᵢ
Steps:
1️⃣ Multiply each value xᵢ by its frequency fᵢ.
2️⃣ Find Σ(xᵢ × fᵢ) and Σfᵢ.
3️⃣ Compute x̄.
✔ Example:
Marks (xᵢ): 10, 20, 30, 40
fᵢ: 3, 5, 7, 5
Σ(xᵢ × fᵢ) = 10×3 + 20×5 + 30×7 + 40×5 = 30 + 100 + 210 + 200 = 540
Σfᵢ = 3 + 5 + 7 + 5 = 20
x̄ = 540 / 20 = 27.

🟡 7.2 Median (M)
Median = Middle value after arranging data.
For grouped data:
Locate class where CF ≥ N/2 (N = Σfᵢ).
Formula:
M = L + [(N/2 – CF_prev)/f] × h
where
L = lower boundary of median class
CF_prev = cumulative frequency before median class
f = frequency of median class
h = class width.

🟡 7.3 Mode (Z)
Mode = Value with highest frequency.
For grouped data:
Z = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
where
f₁ = frequency of modal class
f₀ = frequency of class before modal
f₂ = frequency of class after modal
h = class width.

📌 8. Steps to Draw Histogram and Polygon
1️⃣ Draw class intervals on x-axis, frequencies on y-axis.
2️⃣ Draw bars without gaps.
3️⃣ For polygon, connect midpoints of bars; optionally, close polygon by adding dummy classes with zero frequency.

📌 9. Real-Life Applications
🔴 Weather departments compare rainfall data.
🔵 Economists analyze market trends.
🟢 Education boards evaluate exam performance.
🟡 Sports analysts track scores.

📌 10. Important Tips
✔ Check class intervals for equal width.
✔ Use Σ (Sigma) symbol for summation.
✔ Always label axes with units.
✔ CF curves can estimate median/percentiles graphically.

📌 11. Worked Example — Mean
Class Intervals: 0–10, 10–20, 20–30, 30–40
fᵢ: 5, 7, 10, 8
Midpoints (xᵢ): 5, 15, 25, 35
xᵢ × fᵢ: 25, 105, 250, 280
Σ(xᵢ × fᵢ) = 660
Σfᵢ = 30
x̄ = 660 / 30 = 22.

📌 12. Worked Example — Median
Σfᵢ = 30 ⇒ N/2 = 15.
Locate median class (where CF ≥ 15).
Suppose 10–20 has CF_prev = 5 + 7 = 12, f = 10, L = 20, h = 10.
M = 20 + [(15 – 12)/10]×10 = 20 + 3 = 23.

📌 13. Worked Example — Mode
Modal class: 20–30, f₁ = 10, f₀ = 7, f₂ = 8, L = 20, h = 10.
Z = 20 + [(10 – 7)/(2×10 – 7 – 8)]×10
= 20 + (3/5)×10 = 20 + 6 = 26.

📌 14. Advantages and Limitations
🟢 Advantages: Summarizes data; aids decision-making.
🔴 Limitations: Sensitive to extreme values (mean); ignores distribution shape.

📌 15. Relation between Mean, Median, Mode (for moderately skewed data)
Z ≈ 3M – 2x̄

📌 16. Practice Points
• Always cross-check Σfᵢ.
• For ogives, maintain uniform scale.
• Prefer polygon when comparing two distributions.

🟢 Summary (~300 words)
Statistics in Class 9 introduces data handling fundamentals. Data can be primary or secondary. To make sense of data, we create frequency distribution tables—either inclusive or exclusive class intervals. Frequency is how often a value appears. Cumulative frequency (CF) tables and ogives help visualize distribution trends.
Graphical representations include bar graphs (separated bars for discrete data), histograms (adjacent bars for continuous data), frequency polygons (line graphs through midpoints), and cumulative frequency curves.
Measures of central tendency—mean (x̄), median (M), and mode (Z)—summarize data. The mean is Σ(xᵢfᵢ)/Σfᵢ. The median uses class boundaries and CF:
M = L + [(N/2 – CF_prev)/f]×h.
The mode identifies the most frequent value:
Z = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)]×h.
For moderately skewed data: Z ≈ 3M – 2x̄.
Real-life uses include weather forecasting, market analysis, academic assessment, and sports analytics. Drawing histograms and polygons requires accurate scales and consistent intervals. Limitations: the mean is influenced by outliers; median ignores distribution shape; mode may be unstable in small samples.
Statistics thus equips students with tools for summarizing and interpreting data effectively—skills crucial in science, economics, social studies, and daily decision-making.

📝 Quick Recap
🔵 • Data → Primary / Secondary.
🟢 • Frequency (fᵢ) = count of occurrences.
🟡 • Mean x̄ = Σ(xᵢfᵢ)/Σfᵢ.
🔴 • Median M = L + [(N/2 – CF_prev)/f]×h.
🔵 • Mode Z = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)]×h.
🟢 • Graphs → Bar graph, Histogram, Polygon, Ogive.
🟡 • Relation → Z ≈ 3M – 2x̄.
🔴 • Applications → Weather, Markets, Education, Sports.

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TEXT BOOK QUESTIONS

Exercise 12.1

🔵 Question 1
A survey conducted by an organization for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %):
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
📎 Data :
Reproductive health conditions — 31.8%
Neuropsychiatric conditions — 25.4%
Injuries — 12.4%
Cardiovascular conditions — 3.1%
Respiratory conditions — 4.1%
Other causes — 22.0%
🟢 Answer 1
✨ Step 1: Draw axes (x → Causes, y → Percentage).
✨ Step 2: Choose a scale (e.g., 1 big square = 5%).
✨ Step 3: Draw a bar graph (or pie chart) with heights equal to given percentages.
✨ Step 4: Title and label axes clearly; show legend if using pie chart.
✅ (ii) Major cause = Reproductive health conditions (31.8%).
💡 (iii) Two factors (illustrative): limited maternal/prenatal care; low awareness about reproductive health & family planning.


🔵 Question 2
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
📎 Data :
SC — 940
ST — 970
Non SC/ST — 920
Backward districts — 950
Non-backward districts — 920
Rural — 930
Urban — 910
🟢 Answer 2
✨ Step 1: Draw axes (x → Section, y → Girls per 1000 boys).
✨ Step 2: Choose a scale (e.g., 1 big square = 20).
✨ Step 3: Draw bars for all 7 categories.
✨ Step 4: Title and labels; read values neatly.
✅ (ii) Conclusions: ST highest (970); Urban lowest (910); Backward districts (950) > Non-backward (920); disparity visible in Non SC/ST and Urban groups.


🔵 Question 3
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
📎 Data :
Party A — 75
Party B — 55
Party C — 37
Party D — 29
Party E — 10
Party F — 37
🟢 Answer 3
✨ Step 1: Draw axes (x → Parties A–F, y → Seats).
✨ Step 2: Pick a scale (e.g., 1 big square = 5 seats).
✨ Step 3: Draw bars with heights as given.
✨ Step 4: Add title/labels.
✅ (ii) Party A won the maximum seats (75).


🔵 Question 4
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is given below:
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous.]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
📎 Data :
118–126 → 3
127–135 → 5
136–144 → 9
145–153 → 12
154–162 → 5
163–171 → 4
172–180 → 2
🟢 Answer 4
✨ Step 1: Make classes continuous by using 0.5 correction:
 117.5–126.5, 126.5–135.5, 135.5–144.5, 144.5–153.5, 153.5–162.5, 162.5–171.5, 171.5–180.5.
✨ Step 2: Draw axes (x → length in mm, y → frequency).
✨ Step 3: Draw a histogram with bar heights equal to frequencies (width = 9 mm each).
✨ Step 4: Title/labels; uniform scale on both axes.
✅ (ii) Frequency polygon is also suitable (join midpoints of histogram tops).
✅ (iii) No. Only the class 145–153 has the highest frequency (12). We cannot claim exactly 153 mm; any value in 145–153 mm is possible.


🔵 Question 5
The following table gives the life times of 400 neon lamps:
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
📎 Data :
300–400 → 14
400–500 → 56
500–600 → 60
600–700 → 86
700–800 → 74
800–900 → 62
900–1000 → 48
🟢 Answer 5
✨ Step 1: Draw axes (x → life time in hours, y → number of lamps).
✨ Step 2: Note equal class width = 100; draw histogram with heights as frequencies.
✨ Step 3: Title and labels.
✅ (ii) Lamps with life time > 700 h = 74 + 62 + 48 = 184.


🔵 Question 6
The following table gives the distribution of students of two sections according to the marks obtained by them. Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
📎 Data:
Section A — 0–10: 3, 10–20: 9, 20–30: 17, 30–40: 12, 40–50: 9
Section B — 0–10: 5, 10–20: 19, 20–30: 15, 30–40: 10, 40–50: 1
🟢 Answer 6
✨ Step 1: Compute midpoints (class width 10): 5, 15, 25, 35, 45.
✨ Step 2: Plot points (midpoint, frequency) for Section A and join with straight lines (add zero-frequency dummy classes at both ends if you want a closed polygon).
✨ Step 3: Repeat for Section B on the same axes.
✨ Step 4: Compare heights/spread of the two polygons.
✅ Comparison: Section B has more students in 10–20; Section A has higher counts in 20–40, indicating slightly better central performance for Section A overall.


🔵 Question 7
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below. Represent the data of both the teams on the same graph by frequency polygons. [Hint: First make the class intervals continuous.]
📎 Data :
Number of balls → Team A, Team B
1–6 → 2, 5
7–12 → 1, 6
13–18 → 8, 2
19–24 → 9, 10
25–30 → 4, 5
31–36 → 5, 6
37–42 → 6, 3
43–48 → 10, 4
49–54 → 5, 8
55–60 → 2, 10
🟢 Answer 7
✨ Step 1: Make classes continuous: 0.5–6.5, 6.5–12.5, …, 54.5–60.5 (width = 6).
✨ Step 2: Take midpoints: 3.5, 9.5, 15.5, 21.5, 27.5, 33.5, 39.5, 45.5, 51.5, 57.5.
✨ Step 3: Plot (midpoint, frequency) for Team A, join to get polygon.
✨ Step 4: Plot similarly for Team B on same axes.
✅ Observation: Team A peaks around 19–24; Team B finishes stronger at 55–60.


🔵 Question 8
A random survey of the number of children of various age groups playing in a park was found as follows. Draw a histogram to represent the data above.
📎 Data :
1–2 → 5
2–3 → 3
3–5 → 6
5–7 → 12
7–10 → 9
10–15 → 10
15–17 → 4
🟢 Answer 8
✨ Step 1: Since class widths vary, use frequency density for bar height.
 • Widths: 1, 1, 2, 2, 3, 5, 2.
 • Densities fᵢ/width: 5/1=5, 3/1=3, 6/2=3, 12/2=6, 9/3=3, 10/5=2, 4/2=2.
✨ Step 2: Draw histogram using heights = densities and widths = class widths.
✨ Step 3: Title and labels (x → Age in years, y → Frequency density).
✅ Interpretation: Highest bar is 5–7 (density 6), indicating most concentration there.


🔵 Question 9
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
📎 Data :
1–4 → 6
4–6 → 30
6–8 → 44
8–12 → 16
12–20 → 4
🟢 Answer 9
✨ Step 1: Note unequal widths; use frequency density for heights.
 • Widths: 3, 2, 2, 4, 8.
 • Densities: 6/3=2, 30/2=15, 44/2=22, 16/4=4, 4/8=0.5.
✨ Step 2: Draw histogram with heights = densities and widths = given class widths.
✨ Step 3: Title and labels (x → Number of letters, y → Frequency density).
✅ (ii) Maximum number of surnames are in 6–8 letters class (44 surnames).

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OTHER IMPORTANT QUESTIONS FOR EXAMS

🅐 Section A — 6 × 1 mark
🔵 Question 1 (MCQ)
Which of the following is primary data?
A) Marks list from a newspaper report
B) Temperatures taken by you every hour today
C) Census table downloaded from a website
D) Rainfall normals published by IMD
Answer: B
🟢 Question 2 (MCQ)
In a frequency distribution, the sum Σfᵢ equals:
A) Number of classes
B) Sum of midpoints
C) Total number of observations
D) Median
Answer: C
🟡 Question 3 (MCQ)
For continuous classes 10–20, 20–30, 30–40, the class width (h) is:
A) 9
B) 10
C) 20
D) 30
Answer: B
🔴 Question 4 (MCQ)
The mode of a distribution is:
A) Arithmetic mean of all values
B) Middle term of ordered data
C) Most frequently occurring value
D) Average of mean and median
Answer: C
🔵 Question 5 (MCQ)
A bar graph is most appropriate for:
A) Continuous grouped data
B) Discrete category data
C) Cumulative frequency data only
D) Bivariate scatter data
Answer: B
🟢 Question 6 (1 mark)
Write the formula of mean for grouped data using midpoints xᵢ and frequencies fᵢ.
Answer: x̄ = Σ(xᵢ fᵢ) / Σfᵢ

🅑 Section B — 6 × 2 marks
🟡 Question 7 (2 marks)
The marks in Mathematics of 8 students are: 6, 8, 9, 7, 8, 10, 9, 8. Find the mode.
Answer:
Step 1: Tally frequencies → 6:1, 7:1, 8:3, 9:2, 10:1.
Step 2: Most frequent value = 8.
Final: Mode = 8
🔴 Question 8 (2 marks)
For classes 0–10, 10–20, 20–30 with frequencies 5, 9, 6, compute Σfᵢ and Σ(xᵢ fᵢ).
Answer:
Step 1: Midpoints xᵢ = 5, 15, 25.
Step 2: xᵢ fᵢ = 5×5=25, 15×9=135, 25×6=150.
Step 3: Σfᵢ = 5+9+6 = 20.
Step 4: Σ(xᵢ fᵢ) = 25+135+150 = 310.
Final: Σfᵢ = 20, Σ(xᵢ fᵢ) = 310
🔵 Question 9 (2 marks, MCQ)
Grouped data have unequal class widths. To draw a histogram, bar heights should be proportional to:
A) Raw frequencies fᵢ
B) Class widths only
C) Frequency density fᵢ / width
D) Midpoints xᵢ
Answer: C
🟢 Question 10 (2 marks)
Given a less-than cumulative frequency table with N = 50. The median is located at which cumulative frequency position?
Answer:
Step 1: Median position for grouped data uses N/2.
Step 2: N/2 = 50/2 = 25.
Final: Locate CF ≥ 25 to find the median class.
🟡 Question 11 (2 marks)
Convert the following to continuous classes: 70–79, 80–89, 90–99.
Answer:
Step 1: Each class is of width 10 with integer limits.
Step 2: Use 0.5 correction on boundaries.
Final: 69.5–79.5, 79.5–89.5, 89.5–99.5
🔴 Question 12 (2 marks)
For the grouped data below, find the mean.
Classes: 10–20, 20–30, 30–40; Frequencies fᵢ: 2, 3, 5.
Answer:
Step 1: Midpoints xᵢ = 15, 25, 35.
Step 2: xᵢ fᵢ = 15×2=30, 25×3=75, 35×5=175.
Step 3: Σ(xᵢ fᵢ) = 30+75+175 = 280.
Step 4: Σfᵢ = 2+3+5 = 10.
Step 5: x̄ = Σ(xᵢ fᵢ) / Σfᵢ = 280 / 10 = 28.
Final: x̄ = 28

🅒 Section C — Q13 – Q22 (3 marks each)
🔵 Question 13
Marks of 30 students:
0–10:3, 10–20:5, 20–30:7, 30–40:6, 40–50:5, 50–60:4.
Find the mean.
🟢 Answer 13
Step 1: Midpoints xᵢ = 5, 15, 25, 35, 45, 55.
Step 2: xᵢ fᵢ = 5×3=15, 15×5=75, 25×7=175, 35×6=210, 45×5=225, 55×4=220.
Step 3: Σ(xᵢ fᵢ) = 15+75+175+210+225+220 = 920.
Step 4: Σfᵢ = 3+5+7+6+5+4 = 30.
Step 5: x̄ = 920 / 30 = 30.67.
✅ Final: Mean ≈ 30.7.

🔴 Question 14 (OR)
(a) Using Q13’s data, estimate the median.
OR
(b) Draw a less-than cumulative frequency curve.
🟢 Answer 14 (a)
Step 1: CFs = 3, 8, 15, 21, 26, 30.
Step 2: N = 30 ⇒ N/2 = 15.
Step 3: Median class = 20–30 (CF ≥ 15). L = 20, CF_prev = 8, f = 7, h = 10.
Step 4: M = 20 + ((15 – 8) / 7)×10 = 20 + (7/7)×10 = 30.
✅ Final: Median = 30.

🔵 Question 15
Daily wages of 100 workers:
0–20:6, 20–40:20, 40–60:24, 60–80:30, 80–100:20.
Find the modal wage.
🟢 Answer 15
Step 1: Modal class = 60–80 (f₁=30).
Step 2: f₀=24, f₂=20, L=60, h=20.
Step 3: Z = 60 + ((30 – 24) / (2×30 – 24 – 20))×20.
Step 4: Z = 60 + (6 / 16)×20 = 60 + 7.5.
✅ Final: Mode ≈ 67.5.

🟡 Question 16
The following table shows the ages (in years) of 50 people:
10–20:5, 20–30:8, 30–40:12, 40–50:15, 50–60:10.
Find the median age.
🟢 Answer 16
Step 1: CFs = 5, 13, 25, 40, 50.
Step 2: N = 50 ⇒ N/2 = 25.
Step 3: Median class = 30–40 (CF ≥ 25). L = 30, CF_prev = 13, f = 12, h = 10.
Step 4: M = 30 + ((25 – 13) / 12)×10 = 30 + (12/12)×10 = 40.
✅ Final: Median age = 40 years.

🔴 Question 17
Convert to continuous classes and draw a histogram:
10–19:5, 20–29:7, 30–39:12, 40–49:6.
🟢 Answer 17
Step 1: Continuous classes: 9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5.
Step 2: Draw axes (x=classes, y=frequency).
Step 3: Draw bars with heights=5,7,12,6.
✅ Histogram complete.

🔵 Question 18
Lifetimes of bulbs:
300–400:14, 400–500:56, 500–600:60, 600–700:86, 700–800:74, 800–900:62, 900–1000:48.
Estimate the mean life.
🟢 Answer 18
Step 1: Midpoints xᵢ=350,450,550,650,750,850,950.
Step 2: xᵢ fᵢ=4900,25200,33000,55900,55500,52700,45600.
Step 3: Σ(xᵢ fᵢ)=272800.
Step 4: Σfᵢ=14+56+60+86+74+62+48=400.
Step 5: x̄=272800 / 400=682.
✅ Final: Mean ≈ 682 hours.

🟡 Question 19 (OR)
(a) Find N/2 and locate the median class for Q18 data.
OR
(b) Draw a less-than ogive.
🟢 Answer 19 (a)
Step 1: N=400 ⇒ N/2=200.
Step 2: CFs:14,70,130,216,290,352,400.
Step 3: Median class=600–700 (CF ≥ 200).
✅ Median class located.

🔵 Question 20
A histogram of daily temperatures shows the tallest bar for 30–35 °C. State the modal class.
🟢 Answer 20
✅ Modal class = 30–35 °C.

🔴 Question 21
Find Σfᵢ and Σ(xᵢ² fᵢ) for:
Classes 0–10:2, 10–20:3, 20–30:5.
🟢 Answer 21
Step 1: Midpoints xᵢ=5,15,25.
Step 2: xᵢ² fᵢ=5²×2=50, 15²×3=675, 25²×5=3125.
Step 3: Σ(xᵢ² fᵢ)=50+675+3125=3850.
Step 4: Σfᵢ=2+3+5=10.
✅ Final: Σfᵢ=10, Σ(xᵢ² fᵢ)=3850.

🟡 Question 22
Less-than CF table of 60 children’s weights (kg):
<20:4, <30:12, <40:20, <50:32, <60:48, <70:60.
Find the median weight.
🟢 Answer 22
Step 1: N=60 ⇒ N/2=30.
Step 2: Median class = 40–50 (CF ≥ 30).
Step 3: L=40, CF_prev=20, f=12, h=10.
Step 4: M=40+((30–20)/12)×10=40+(10/12)×10=48.33.
✅ Final: Median ≈ 48.3 kg.

🅓 Section D — Q23–Q30 (4 marks each)
🔵 Question 23
The weekly wages of 120 workers are:
0–10:4, 10–20:16, 20–30:20, 30–40:24, 40–50:28, 50–60:18, 60–70:10.
Find the mean wage using the step-deviation method.
🟢 Answer 23
Step 1: Midpoints xᵢ=5,15,25,35,45,55,65.
Step 2: Assume A=35, class width h=10.
Step 3: uᵢ=(xᵢ–A)/h → –3,–2,–1,0,1,2,3.
Step 4: Compute fᵢuᵢ=4×–3=–12,16×–2=–32,20×–1=–20,24×0=0,28×1=28,18×2=36,10×3=30.
Step 5: Σfᵢ=120, Σ(fᵢuᵢ)=–12–32–20+0+28+36+30=30.
Step 6: x̄=A+(Σ(fᵢuᵢ)/Σfᵢ)×h=35+(30/120)×10=35+2.5=37.5.
✅ Final: Mean wage=37.5.

🔴 Question 24 (OR)
(a) For Q23 data, construct a less-than ogive and estimate the median.
OR
(b) Draw a greater-than ogive and verify the two curves intersect near the median.
🟢 Answer 24 (a)
Step 1: Compute CF:4,20,40,64,92,110,120.
Step 2: N=120 ⇒ N/2=60.
Step 3: Locate CF≥60 ⇒ class 30–40. L=30, CF_prev=40, f=24, h=10.
Step 4: M=30+((60–40)/24)×10=30+(20/24)×10=30+8.33=38.33.
✅ Final: Median ≈38.3.

🟡 Question 25
A company recorded the number of defective items in 50 samples:
0–5:5, 5–10:9, 10–15:12, 15–20:10, 20–25:8, 25–30:6.
Find the mode.
🟢 Answer 25
Step 1: Modal class=10–15 (f₁=12).
Step 2: f₀=9, f₂=10, L=10, h=5.
Step 3: Z=10+((12–9)/(2×12–9–10))×5=10+(3/(24–19))×5=10+(3/5)×5=10+3=13.
✅ Final: Mode=13 defective items.

🔴 Question 26
Compute the median weight from:
Classes:40–50:5,50–60:8,60–70:12,70–80:20,80–90:10,90–100:5.
🟢 Answer 26
Step 1: CF=5,13,25,45,55,60.
Step 2: N=60 ⇒ N/2=30.
Step 3: Median class=70–80. L=70, CF_prev=25, f=20, h=10.
Step 4: M=70+((30–25)/20)×10=70+(5/20)×10=70+2.5=72.5.
✅ Final: Median=72.5 kg.

🔵 Question 27 (OR)
(a) The following are class intervals of the daily study hours of students:
0–2:6,2–4:10,4–6:14,6–8:12,8–10:8. Find the mean using the assumed mean method.
OR
(b) Draw a histogram and a frequency polygon for the same data.
🟢 Answer 27 (a)
Step 1: Midpoints=1,3,5,7,9. Assume A=5,h=2.
Step 2: uᵢ=(xᵢ–A)/h=–2,–1,0,1,2.
Step 3: fᵢuᵢ=6×–2=–12,10×–1=–10,14×0=0,12×1=12,8×2=16.
Step 4: Σfᵢ=50, Σ(fᵢuᵢ)=–12–10+0+12+16=6.
Step 5: x̄=A+(Σ(fᵢuᵢ)/Σfᵢ)×h=5+(6/50)×2=5+0.24=5.24.
✅ Final: Mean≈5.24 h.

🟡 Question 28
The lengths of 100 leaves (cm) are grouped:
2–4:5,4–6:8,6–8:10,8–10:20,10–12:25,12–14:15,14–16:10,16–18:7.
Find the median length.
🟢 Answer 28
Step 1: CF=5,13,23,43,68,83,93,100.
Step 2: N=100 ⇒ N/2=50.
Step 3: Median class=10–12 (CF≥50). L=10,CF_prev=43,f=25,h=2.
Step 4: M=10+((50–43)/25)×2=10+(7/25)×2=10+0.56=10.56.
✅ Final: Median length≈10.56 cm.

🔴 Question 29
The marks of 80 students are:
0–10:5,10–20:8,20–30:10,30–40:15,40–50:16,50–60:12,60–70:8,70–80:6.
Estimate the mode.
🟢 Answer 29
Step 1: Modal class=40–50 (f₁=16).
Step 2: f₀=15,f₂=12,L=40,h=10.
Step 3: Z=40+((16–15)/(2×16–15–12))×10=40+(1/(32–27))×10=40+(1/5)×10=40+2=42.
✅ Final: Mode=42 marks.

🟡 Question 30
The table shows daily incomes of workers:
100–120:8,120–140:10,140–160:20,160–180:22,180–200:18,200–220:12,220–240:10.
Compute the mean income using the direct method.
🟢 Answer 30
Step 1: Midpoints=110,130,150,170,190,210,230.
Step 2: xᵢfᵢ=880,1300,3000,3740,3420,2520,2300.
Step 3: Σfᵢ=8+10+20+22+18+12+10=100.
Step 4: Σ(xᵢfᵢ)=880+1300+3000+3740+3420+2520+2300=17160.
Step 5: x̄=17160/100=171.6.
✅ Final: Mean income=171.6.

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