STATISTICS — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 12  ·  Q–
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Class 9 Mathematics Exercise-12.1 NCERT Solutions Olympiad Board Exam
Chapter 12

STATISTICS

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

9 Questions
20–30 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):
Cause Female Fatality Rate (%)
Reproductive health conditions 31.8
Neuropsychiatric conditions 25.4
Injuries 12.4
Cardiovascular conditions 4.3
Respiratory conditions 4.1
Other causes 22.0
(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.
📘 Concept & Theory Concept Used ›

Statistics is the branch of Mathematics that deals with the collection, organisation, presentation, analysis and interpretation of data.

One of the most important methods of presenting data is through graphical representation. Graphs help us understand data quickly, compare different categories easily and identify trends at a glance.

For the given question, the data represent percentages of different causes responsible for illness and death among women aged 15–44 years worldwide. Since each category has a corresponding percentage value, a bar graph is an appropriate graphical representation.

📝 Theory Behind the Graph ›
  • Different causes are represented along the horizontal (x) axis.
  • Female fatality rates (%) are represented along the vertical (y) axis.
  • The height of each bar is proportional to its percentage value.
  • The tallest bar represents the category with the highest fatality rate.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Read the data carefully from the table.

  2. Choose a suitable scale for the vertical axis.

  3. Mark all causes on the horizontal axis.

  4. Draw bars corresponding to the given percentages.

  5. Compare the heights of the bars.

  6. Identify the category with the maximum percentage.

  7. Discuss possible reasons behind the major cause.

📊 Graph / Figure Graph / Figure ›
Female Fatality Rate (%) Among Women (15–44 Years) 0 10 20 30 Reproductive Neuropsychiatric Injuries Cardio Respiratory Others
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  14 steps
  1. given
  2. Cause Female Fatality Rate (%)
    Reproductive health conditions 31.8
    Neuropsychiatric conditions 25.4
    Injuries 12.4
    Cardiovascular conditions 4.3
    Respiratory conditions 4.1
    Other causes 22.0
  3. (i) Represent the information graphically
  4. Draw a bar graph using the following steps:
  5. Draw two mutually perpendicular axes.
  6. Take causes on the horizontal axis (x-axis).
  7. Take female fatality rate (%) on the vertical axis (y-axis).
  8. Choose a suitable scale, for example:\[\ \text{unit} = 5\%\]
  9. Draw bars of heights corresponding to the given percentages:
    • Reproductive health conditions = 31.8%
    • Neuropsychiatric conditions = 25.4%
    • Injuries = 12.4%
    • Cardiovascular conditions = 4.3%
    • Respiratory conditions = 4.1%
    • Other causes = 22.0%
  11. Thus, the required bar graph is obtained.
  12. (ii) Which condition is the major cause of women’s ill health and death worldwide?
  13. Comparing all percentages:\[31.8\% > 25.4\% > 22.0\% > 12.4\% > 4.3\% > 4.1\%\]
  14. The highest percentage is\[31.8\%\]
  15. corresponding to Reproductive Health Conditions.
  16. Therefore, Reproductive Health Conditions are the major cause of women’s ill health and death worldwide.
  17. (iii) Any two factors responsible for reproductive health conditions being the major cause
  18. Some important factors are:
    1. Lack of proper healthcare facilities

      In many regions, women do not receive adequate antenatal care, safe childbirth facilities, reproductive health services, immunisation and medical guidance.

    2. Poverty, malnutrition and social barriers

      Poor nutrition, early marriages, lack of education, gender discrimination and financial limitations increase health risks and make treatment difficult.

🎯 Exam Significance Exam Significance ›
  • Tests understanding of graphical representation of data.
  • Checks ability to interpret statistical information.
  • Develops data analysis and comparison skills.
  • Frequently asked as a short-answer or activity-based question.
  • Strengthens understanding of real-life applications of statistics.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Statistics helps organise and interpret numerical information.

  2. Bar graphs are useful for comparing different categories of data.

  3. The tallest bar represents the largest value.

  4. Graphical representation makes comparison easier than studying raw tables.

  5. Interpretation of graphs is as important as constructing them.
↑ Top
1 / 9  ·  11%
Q2 →
Q2
NUMERIC3 marks
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.
Section Number of girls per thousand boys
Scheduled Caste (SC) 940
Scheduled Tribe (ST) 970
Non SC/ST 920
Backward districts 950
Non-backward districts 920
Rural 930
Urban 910
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
📘 Concept & Theory Concept Used ›

Statistics helps us collect, organize, present and interpret data. One of the simplest and most effective methods of presenting categorical data is a bar graph.

A bar graph uses rectangular bars of equal width to represent numerical values. The height of each bar is proportional to the value it represents.

In this question, the data show the number of girls per thousand boys in different sections of Indian society. Since different categories are being compared, a bar graph is the most suitable representation.

📝 Theory Behind the Question ›

The number of girls per thousand boys is commonly known as the sex ratio. It is an important social indicator used to study gender balance in society.

A higher ratio indicates a better balance between girls and boys, while a lower ratio indicates gender disparity.

By studying the graph, we can:

  • Compare different sections of society.
  • Identify sections with the highest and lowest ratios.
  • Understand social trends related to gender equality.
  • Interpret data visually rather than through numerical tables alone.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Read the data carefully from the table.

  2. Choose suitable axes.

  3. Select an appropriate scale.

  4. Mark the different sections on the horizontal axis.

  5. Mark the number of girls per thousand boys on the vertical axis.

  6. Draw bars corresponding to the given values.

  7. Compare the heights of the bars.

  8. Draw conclusions from the graphical representation.

📊 Graph / Figure Graph / Figure ›
Number of Girls per Thousand Boys 900 910 920 930 940 950 970 SC ST Non SC/ST Backward Non-Backward Rural Urban
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  10 steps
  1. Given
  2. Section Number of Girls per Thousand Boys
    Scheduled Caste (SC) 940
    Scheduled Tribe (ST) 970
    Non SC/ST 920
    Backward Districts 950
    Non-Backward Districts 920
    Rural 930
    Urban 910
  3. Represent the information by a bar graph.
  4. Draw two perpendicular axes.
  5. Mark the different sections of society on the horizontal axis (x-axis).
  6. Mark the number of girls per thousand boys on the vertical axis (y-axis).
  7. Choose a suitable scale. For example, \[1 \text{ unit} = 10 \text{ girls}\]
  8. Draw bars corresponding to the values:
    • SC = 940
    • ST = 970
    • Non SC/ST = 920
    • Backward Districts = 950
    • Non-Backward Districts = 920
    • Rural = 930
    • Urban = 910
  9. The resulting bar graph represents the required information.
  10. Conclusions obtained from the graph
  11. By carefully observing the heights of the bars, the following conclusions can be drawn:
    1. The Scheduled Tribe (ST) category has the highest number of girls per thousand boys.

      2. \[ 970 \]

    2. The Urban category has the lowest number of girls per thousand boys.

      3. \[ 910 \]

    3. Backward districts have a better ratio than non-backward districts.

      4. \[ 950 > 920 \]

    4. Scheduled Castes have a better ratio than Non SC/ST groups.

      5. \[ 940 > 920 \]

    5. Rural areas have a better ratio than urban areas.

      6. \[ 930 > 910 \]

    6. None of the categories reaches the ideal ratio of \[ 1000 \] girls per thousand boys.
    7. The graph indicates that gender imbalance still exists across all sections of society.
  12. (ii) Classroom Discussion Points
    • Why do some sections show a higher girl-to-boy ratio than others?
    • What role do education and social awareness play in improving the sex ratio?
    • Why might urban areas have a lower ratio compared to rural areas?
    • How can government policies help achieve gender equality?
🎯 Exam Significance Exam Significance ›
  • Tests the ability to construct and interpret bar graphs.
  • Develops skills in comparing categorical data.
  • Checks understanding of statistical representation.
  • Frequently appears in competency-based and application-based questions.
  • Encourages interpretation of real-life social data.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Bar graphs are useful for comparing different categories of data.

  2. The highest sex ratio is observed in Scheduled Tribe (ST) communities.

  3. The lowest sex ratio is observed in Urban areas

  4. Statistical graphs help identify patterns and social trends quickly.

  5. Interpretation of data is an important part of statistics.
← Q1
2 / 9  ·  22%
Q3 →
Q3
NUMERIC3 marks
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:
Political Party A B C D E F
Seats Won 75 55 37 29 10 37
(i) Draw a bar graph to represent the polling results.
(ii)Which political party won the maximum number of seats?
📘 Concept & Theory Concept Used ›

In Statistics, data can be represented graphically to make comparison easy and meaningful. A bar graph is one of the simplest graphical tools used for comparing numerical values belonging to different categories.

In a bar graph:

  • Each category is represented by a rectangular bar.
  • All bars have equal width.
  • The height of a bar is proportional to the value it represents.
  • The tallest bar corresponds to the largest value.

In this question, different political parties form the categories and the number of seats won forms the numerical data. Therefore, a bar graph is the most suitable representation.

📝 Theory Behind the Question ›

Election results are often presented using tables, charts and graphs because they allow quick comparison among political parties.

A bar graph helps us:

  • Compare the performance of different political parties.
  • Identify the party with the maximum number of seats.
  • Interpret election results visually.
  • Understand trends in voting patterns.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Read the data from the table.

  2. Draw two perpendicular axes.

  3. Represent political parties on the horizontal axis.

  4. Represent the number of seats on the vertical axis.

  5. Select a suitable scale.

  6. Draw bars corresponding to the seats won by each party.

  7. Compare the heights of the bars

  8. Identify the political party with the maximum seats.

📊 Graph / Figure Graph / Figure ›
Polling Results of State Assembly Election 0 10 20 30 40 50 60 70 A B C D E F 75 55 37 29 10 37
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  10 steps
  1. Given
  2. Political Party A B C D E F
    Seats Won 75 55 37 29 10 37
  3. (i) Draw a bar graph to represent the polling results.
  4. Draw two perpendicular axes.
  5. Mark the political parties A, B, C, D, E and F on the horizontal axis (x-axis).
  6. Mark the number of seats on the vertical axis (y-axis).
  7. Choose a suitable scale.\[1 \text{ unit} = 10 \text{ seats}\]
  8. Draw bars corresponding to the following seat counts:
    • Party A = 75 seats
    • Party B = 55 seats
    • Party C = 37 seats
    • Party D = 29 seats
    • Party E = 10 seats
    • Party F = 37 seats
  9. The resulting bar graph represents the polling outcome of the state assembly election.
  10. (ii) Which political party won the maximum number of seats?
  11. Comparing the number of seats won:\[75,\;55,\;37,\;29,\;10,\;37\]
  12. The greatest value is:\[75\] which corresponds to Political Party A.
  13. Therefore, Political Party A won the maximum number of seats.
🎯 Exam Significance Exam Significance ›
  • Tests understanding of bar graph construction.
  • Checks the ability to interpret statistical data.
  • Develops graphical representation skills.
  • Frequently asked in competency-based and application-based questions.
  • Encourages analysis of real-life data such as election results.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  4 points

  1. A bar graph is used to compare values across different categories.

  2. The tallest bar represents the highest value.

  3. Political Party A secured the maximum seats.

  4. Graphs make election data easier to understand and compare.
← Q2
3 / 9  ·  33%
Q4 →
Q4
NUMERIC3 marks
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm) Number of leaves
118 - 126 3
127 - 135 5
136 - 144 9
145 - 153 12
154 - 162 5
163 - 171 4
172 - 180 2
  1. Draw a histogram to represent the given data.
  2. Is there any other suitable graphical representation for the same data?
  3. Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
📘 Concept & Theory Concept Used ›

When data are grouped into class intervals and represent a continuous variable, a histogram is the most suitable graphical representation.

A histogram consists of adjacent rectangles whose:

  • Base represents class intervals.
  • Height represents frequency.
  • There is no gap between consecutive rectangles because the data are continuous.

Before drawing a histogram, we must check whether the class intervals are continuous. If they are not continuous, we first convert them into continuous intervals using the correction factor.

📝 Theory Behind the Question ›

The lengths of leaves are measured to the nearest millimetre. Therefore, the classes:

\[ 118-126,\;127-135,\;136-144,\ldots \]

are not continuous because there is a gap between successive classes.

Gap between two consecutive classes:

\[ 127-126 = 1 \]

Therefore, the correction factor is:

\[ \frac{1}{2}=0.5 \]

Subtracting 0.5 from each lower limit and adding 0.5 to each upper limit converts the classes into continuous intervals.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Convert the class intervals into continuous intervals.

  2. Prepare the continuous frequency distribution table.

  3. Draw the histogram using class intervals and frequencies.

  4. Identify other possible graphical representations.

  5. Interpret the histogram correctly.

  6. Answer whether the exact length of the maximum number of leaves can be determined.

📊 Graph / Figure Graph / Figure ›
Histogram : Length of Leaves 0 2 4 6 8 10 12 117.5-126.5 126.5-135.5 135.5-144.5 144.5-153.5 153.5-162.5 162.5-171.5 171.5-180.5
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  20 steps
  1. (i) Draw a histogram to represent the given data.
  2. Since the lengths are measured correct to one millimetre, the given class intervals are not continuous.
  3. Gap between successive classes:\[127 - 126 = 1\]
  4. Correction factor:\[\frac{1}{2}=0.5\]
  5. Therefore:
    • Subtract 0.5 from each lower limit.
    • Add 0.5 to each upper limit.
  6. The continuous frequency distribution becomes:
    Continuous Length (mm) Frequency
    117.5 – 126.5 3
    126.5 – 135.5 5
    135.5 – 144.5 9
    144.5 – 153.5 12
    153.5 – 162.5 5
    162.5 – 171.5 4
    171.5 – 180.5 2
  7. Class width:\[126.5 - 117.5 = 9\]
  8. Since all class intervals have equal width, the heights of the rectangles are directly proportional to their frequencies.
  9. Draw adjacent rectangles corresponding to frequencies:\[3,\;5,\;9,\;12,\;5,\;4,\;2\]
  10. The resulting figure is the required histogram.
  11. (ii) Is there any other suitable graphical representation for the same data?
  12. Yes, The most suitable alternative graphical representation is a Frequency Polygon.
  13. A frequency polygon is obtained by:
    1. Finding the class marks (midpoints) of all intervals.
    2. Plotting the corresponding frequencies.
    3. Joining successive points by straight line segments.
  14. Frequency polygons are often used together with histograms to study the shape of a frequency distribution.
  15. herefore, besides a histogram, a frequency polygon is also a suitable graphical representation.
  16. (iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
  17. No, this conclusion is not correct.
  18. The highest frequency is: \(12\)
  19. corresponding to the class interval:\[145 - 153\] (or continuous interval \[ 144.5 - 153.5 \] ).
  20. This only tells us that 12 leaves have lengths lying somewhere within this interval.
  21. The data do not provide the exact lengths of individual leaves.
  22. Therefore, we cannot conclude that all or most of these leaves are exactly \[ 153\text{ mm} \] long.
  23. We can only conclude that the class interval \[ 145 - 153 \] contains the maximum number of leaves.
🎯 Exam Significance Exam Significance ›

Students frequently make the mistake of interpreting the class interval with the highest frequency as a single value. In grouped data, frequencies relate to an entire interval, not to any one value within that interval.

Why This Question Is Important for Board Examinations
  • Tests understanding of histograms.
  • Checks conversion of discontinuous classes into continuous classes.
  • Develops interpretation skills for grouped data.
  • Frequently appears as a long-answer competency question.
  • Strengthens understanding of frequency distributions.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points

  1. Histograms are used for continuous grouped data.

  2. Class intervals must be continuous before drawing a histogram.

  3. Correction factor: \[ 0.5 \]

  4. Frequency polygon is another suitable graphical representation.<

  5. Grouped data do not reveal exact observations within a class interval.

  6. The interval with highest frequency is not the same as the exact value with highest frequency.
← Q3
4 / 9  ·  44%
Q5 →
Q5
NUMERIC3 marks
The following table gives the lifetimes of 400 neon lamps:

Life time (in hours) Number of lamps
300 - 400 14
400 - 500 56
500 - 600 60
600 - 700 86
700 - 800 74
800 - 900 62
900 - 1000 48
  1. Represent the given information with the help of a histogram.
  2. How many lamps have a lifetime of more than 700 hours?
📘 Concept & Theory Concept Used ›

A histogram is a graphical representation of grouped continuous data. It consists of adjoining rectangles whose:

  • Base represents class intervals.
  • Height represents frequencies.
  • No gaps are left between adjacent rectangles.

Histograms help us visualize the distribution of data and identify the class intervals having higher or lower frequencies.

In this question, the lifetimes of neon lamps are grouped into continuous intervals. Therefore, a histogram is the most suitable graphical representation.

📝 Theory Behind the Question ›

The lifetime of a neon lamp is measured in hours and is a continuous variable. The data are already grouped into equal class intervals:

\[ 300-400,\;400-500,\;500-600,\;600-700,\;700-800,\;800-900,\;900-1000 \]

Since all class widths are equal:

\[ 400-300 = 100 \]

the frequencies can be plotted directly as the heights of the rectangles.

The histogram helps us identify the intervals where the maximum number of lamps occur and also enables us to answer questions regarding the number of lamps within specific lifetime ranges.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Verify whether the class intervals are continuous.

  2. Check whether the class widths are equal.

  3. Draw a histogram using the given frequencies.

  4. Identify the intervals corresponding to lifetimes greater than 700 hours.

  5. Add the frequencies of those intervals.

  6. Write the final answer.

📊 Graph / Figure Graph / Figure ›
Histogram : Lifetime of Neon Lamps 0 10 20 30 40 50 60 70 80 300-400 400-500 500-600 600-700 700-800 800-900 900-1000
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  17 steps
  1. Given Data
  2. Lifetime (in hours) Number of Lamps
    300 – 400 14
    400 – 500 56
    500 – 600 60
    600 – 700 86
    700 – 800 74
    800 – 900 62
    900 – 1000 48
  3. (i) Represent the given information with the help of a histogram.<
  4. Observe that all class intervals are continuous and have equal width.
  5. Class width:\[400 - 300 = 100\]
  6. Similarly,\[ 500 - 400 = 100\]
  7. and all remaining class widths are also equal to 100
  8. Therefore, frequencies can be taken directly as the heights of the rectangles.
  9. Draw two perpendicular axes.
  10. Take lifetime intervals on the horizontal axis and frequencies on the vertical axis.
  11. Draw adjacent rectangles having heights:
  12. \[14,\;56,\;60,\;86,\;74,\;62,\;48\]
  13. corresponding to the respective class intervals.
  14. The resulting figure is the required histogram.
  15. (ii) How many lamps have a lifetime of more than 700 hours?
  16. Lamps having lifetime greater than \[700 \text{ hours}\]
  17. belong to the following classes:
  18. Lifetime (hours) Number of Lamps
    700 – 800 74
    800 – 900 62
    900 – 1000 48
  19. Total number of such lamps:\[74 + 62 + 48=184\]
  20. Therefore, Number of lamps having lifetime more than 700 hours \[ = 184 \]
🎯 Exam Significance Exam Significance ›
  • Tests construction and interpretation of histograms.
  • Checks understanding of grouped frequency distributions.
  • Develops data interpretation skills.
  • Frequently appears in competency-based questions.
  • Strengthens graphical representation concepts.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  4 points

  1. Histograms are used for continuous grouped data.

  2. Equal class widths allow direct use of frequencies as heights.

  3. Histograms help visualize the distribution of observations effectively.

  4. Grouped data can be analyzed easily using graphical techniques.
← Q4
5 / 9  ·  56%
Q6 →
Q6
NUMERIC3 marks
The following table gives the distribution of students of two sections according to the marks obtained by them:

Section A Section B
Marks Frequency Marks Frequency
0-10 3 0-10 5
10-20 9 10-20 19
20-30 17 20-30 15
30-40 12 30-40 10
40-50 9 40-50 1

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.

📘 Concept & Theory Concept Used ›

A Frequency Polygon is a graphical representation of a frequency distribution obtained by plotting frequencies against the corresponding class marks (midpoints of class intervals) and joining the points by straight line segments.

Frequency polygons are especially useful when comparing two or more frequency distributions on the same graph.

In this question, the performance of students from Sections A and B is compared. Therefore, drawing two frequency polygons on the same coordinate system provides a clear visual comparison.

📝 Theory Behind the Question ›

To draw a frequency polygon, we first calculate the class mark of each class interval.

Class Mark =

\[ \frac{\text{Lower Limit + Upper Limit}}{2} \]

For example, for the class interval \[ 0-10 \]

Class Mark \[ = \frac{0+10}{2} = \frac{10}{2} = 5 \]

The class marks become the x-coordinates, while the frequencies become the y-coordinates.

To close the frequency polygon, one imaginary class interval is taken before the first class and another after the last class, both having frequency zero.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the class marks of all class intervals.

  2. Prepare separate coordinate points for Sections A and B.

  3. Plot the points on the same graph.

  4. Join the points by straight line segments.

  5. Close both polygons using zero-frequency end points.

  6. Compare the shapes of the two polygons.

  7. Draw conclusions about the performance of both sections.

📊 Graph / Figure Graph / Figure ›
Frequency Polygons of Sections A and B 0 5 10 15 20 5 15 25 35 45 Section A Section B
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  21 steps
  1. Calculation of Class Marks
  2. Marks Class Mark Frequency (Section A) Frequency (Section B)
    0 – 10 \[\frac{0+10}{2}=5\] 3 5
    10 – 20 \[\frac{10+20}{2}=15\] 9 19
    20 – 30 \[\frac{20+30}{2}=25\] 17 15
    30 – 40 \[\frac{30+40}{2}=35\] 12 10
    40 – 50 \[\frac{40+50}{2}=45\] 9 1
  3. Plot Frequency Polygon for Section A
  4. Plot the points:\[(5,3),\;(15,9),\;(25,17),\;(35,12),\;(45,9)\]
  5. To close the polygon, include:\[(-5,0)\;\text{and}\;(55,0)\]
  6. Join all points using straight line segments.
  7. Plot Frequency Polygon for Section B
  8. Plot the points:\[(5,5),\;(15,19),\;(25,15),\;(35,10),\;(45,1)\]
  9. Again include:\[(-5,0)\;\text{and}\;(55,0)\]
  10. Join the points by straight line segments using a different colour or line style.
  11. The resulting graph gives the required frequency polygons.
  12. Comparison of Performance of the Two Sections
  13. The frequency polygons show the distribution of marks in both sections.
  14. Section B has the highest frequency \[ 19 \] in the class interval \[ 10-20 \] indicating that many students scored relatively low marks.
  15. Section A has the highest frequency \[ 17 \] in the class interval \[ 20-30 \] which is a higher score range.
  16. In the higher mark intervals \[ 30-40 \] and \[ 40-50, \] Section A has more students than Section B.
  17. Section B has only \[ 1 \] student in the interval \[ 40-50. \]
  18. Section A has \[ 9 \] students in the interval \[ 40-50. \]
  19. Therefore, Section A performed better overall than Section B because a larger number of students obtained marks in the higher intervals.
  20. Verification Using Mean Marks (Advanced Observation)
  21. Estimated mean marks of Section A:\[\begin{aligned} \bar{x} &= \frac{(5\times3)+(15\times9)+(25\times17)+(35\times12)+(45\times9)} {50}\\ &= \frac{1400}{50}\\ &= 28 \end{aligned}\]
  22. Estimated mean marks of Section B: \[ \begin{aligned} \bar{x} &= \frac{(5\times5)+(15\times19)+(25\times15)+(35\times10)+(45\times1)} {50}\\ &= \frac{1080}{50}\\ &= 21.6 \end{aligned} \]
  23. Since\[ 28 > 21.6 \]
  24. this further confirms that Section A performed better.
🎯 Exam Significance Exam Significance ›
  • Tests construction of frequency polygons.
  • Checks understanding of class marks.
  • Develops graph interpretation skills.
  • Frequently appears as a competency-based question.
  • Encourages comparison of two data distributions.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Frequency polygons are drawn using class marks and frequencies.

  2. Class Mark Formula: \[ \frac{\text{Lower Limit + Upper Limit}}{2} \]

  3. Two frequency polygons can be drawn on the same graph for comparison.

  4. Overall performance of Section A is better than Section B.

  5. Frequency polygons provide a quick visual comparison of distributions.
← Q5
6 / 9  ·  67%
Q7 →
Q7
NUMERIC3 marks
The runs scored by two teams, A and B on the first 60 balls in a cricket match are given below

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 6 8
55-60 2 10
📘 Concept & Theory Concept Used ›

A Frequency Polygon is a graphical representation of grouped data obtained by plotting frequencies against class marks and joining the plotted points using straight line segments.

When two sets of data need to be compared, frequency polygons are particularly useful because both distributions can be represented on the same graph.

In this question, the runs scored by Teams A and B during different groups of six balls are compared. Therefore, drawing two frequency polygons on the same graph helps us compare the scoring patterns of the two teams.

📝 Theory Behind the Question ›

The intervals:

\[ 1-6,\;7-12,\;13-18,\ldots,55-60 \]

are not continuous because there is a gap between two successive classes.

Gap:

\[ 7-6=1 \]

Correction factor:

\[ \frac{1}{2}=0.5 \]

Therefore, continuous class intervals are obtained by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit.

After obtaining continuous classes, class marks are calculated using:

\[ \text{Class Mark} = \frac{\text{Lower Limit + Upper Limit}}{2} \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Convert the given class intervals into continuous intervals.

  2. Find the class marks of all intervals.

  3. Plot the frequencies (runs scored) for Team A.

  4. Plot the frequencies (runs scored) for Team B.

  5. Join the points to obtain two frequency polygons.

  6. Compare the performance of the two teams using the graph.

📊 Graph / Figure Graph / Figure ›
Runs Scored by Teams A and B in First 60 Balls 0 2 4 6 8 10 Team A Team B 3.5 15.5 27.5 39.5 51.5
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. Convert the Class Intervals into Continuous Form
  2. Since the gap between two consecutive intervals is 1
  3. the correction factor is 0.5
  4. Therefore, the continuous class intervals become:
    Number of Balls Team A Team B
    0.5 – 6.5 2 5
    6.5 – 12.5 1 6
    12.5 – 18.5 8 2
    18.5 – 24.5 9 10
    24.5 – 30.5 4 5
    30.5 – 36.5 5 6
    36.5 – 42.5 6 3
    42.5 – 48.5 10 4
    48.5 – 54.5 6 8
    54.5 – 60.5 2 10
  5. Calculate the Class Marks
  6. Continuous Class Interval Class Mark Team A Team B
    0.5 – 6.5 \[ \frac{0.5+6.5}{2}=3.5 \] 2 5
    6.5 – 12.5 \[ \frac{6.5+12.5}{2}=9.5 \] 1 6
    12.5 – 18.5 \[ \frac{12.5+18.5}{2}=15.5 \] 8 2
    18.5 – 24.5 \[ \frac{18.5+24.5}{2}=21.5 \] 9 10
    24.5 – 30.5 \[ \frac{24.5+30.5}{2}=27.5 \] 4 5
    30.5 – 36.5 \[ \frac{30.5+36.5}{2}=33.5 \] 5 6
    36.5 – 42.5 \[ \frac{36.5+42.5}{2}=39.5 \] 6 3
    42.5 – 48.5 \[ \frac{42.5+48.5}{2}=45.5 \] 10 4
    48.5 – 54.5 \[ \frac{48.5+54.5}{2}=51.5 \] 6 8
    54.5 – 60.5 \[ \frac{54.5+60.5}{2}=57.5 \] 2 10
  7. Draw the Frequency Polygons
  8. Plot the following points for Team A:\[(3.5,2),\;(9.5,1),\;(15.5,8),\;(21.5,9),\;(27.5,4),\]
  9. \[(33.5,5),\;(39.5,6),\;(45.5,10),\;(51.5,6),\;(57.5,2)\]
  10. Join the points using straight line segments.
  11. Similarly, plot the following points for Team B:\[(3.5,5),\;(9.5,6),\;(15.5,2),\;(21.5,10),\;(27.5,5),\]
  12. \[(33.5,6),\;(39.5,3),\;(45.5,4),\;(51.5,8),\;(57.5,10)\]
  13. Join these points using another coloured line to obtain the second frequency polygon.
  14. Comparison of the Two Teams
    1. Team A scored heavily in the middle part of the innings, especially in the intervals \[ 19-24 \] and \[ 43-48. \]
    2. Team B started strongly and also finished strongly.
    3. Team B scored \[ 10 \] runs during both intervals: \[ 19-24 \] and \[ 55-60. \]
    4. Team A achieved its highest score of \[ 10 \] runs during the interval \[ 43-48. \]
    5. Total runs scored by Team A:

      6. \[ 2+1+8+9+4+5+6+10+6+2 \]

      \[ =53 \]

    6. Total runs scored by Team B:

      7. \[ 5+6+2+10+5+6+3+4+8+10 \]

      \[ =59 \]

  16. Since

    \[ 59 > 53 \]

  17. Team B performed better overall in the first 60 balls.

🎯 Exam Significance Exam Significance ›
  • Tests understanding of frequency polygons.
  • Checks the conversion of discontinuous intervals into continuous intervals.
  • Develops graph plotting skills.
  • Enhances interpretation and comparison of data.
  • Frequently asked in competency-based and application-based questions.
← Q6
7 / 9  ·  78%
Q8 →
Q8
NUMERIC3 marks
A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years) Number of children
1-2 5
2-3 3
3-5 6
5-7 12
7-10 9
10-15 10
15-17 4
📘 Concept & Theory concept Used ›

A Histogram is a graphical representation of a frequency distribution using adjacent rectangles. For equal class intervals, the height of each rectangle is directly proportional to the frequency.

However, when class intervals are of unequal widths, we cannot directly use frequencies as heights because wider intervals would occupy more area and lead to incorrect interpretation.

In such cases, we use Frequency Density or Adjusted Frequency to determine the height of each rectangle.

📝 Theory Behind the Question ›

Observe the given age intervals:

\[ 1-2,\;2-3,\;3-5,\;5-7,\;7-10,\;10-15,\;15-17 \]

Their widths are:

\[ 1,\;1,\;2,\;2,\;3,\;5,\;2 \]

Since the class widths are not equal, a simple histogram cannot be drawn.

Therefore, we first calculate the adjusted heights using:

\[ \text{Adjusted Height} = \left( \frac{\text{Frequency}} {\text{Class Width}} \right) \times \text{Minimum Class Width} \]

Here, the minimum class width is:

\[ 1 \]

Hence,

\[ \text{Adjusted Height} = \frac{\text{Frequency}} {\text{Class Width}} \]

These adjusted heights are used as the heights of the rectangles in the histogram.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the width of every class interval.

  2. Identify the minimum class width.

  3. Calculate adjusted heights for each interval.

  4. Draw rectangles with actual widths and adjusted heights.

  5. Construct the histogram.

  6. Interpret the graphical representation.

📊 Graph / Figure Graph / Figure ›
Histogram : Children Playing in the Park 0 1 2 3 4 5 6 1-2 2-3 3-5 5-7 7-10 10-15 15-17
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  10 steps
  1. Calculate Class Widths
  2. Age Group (Years) Frequency (f) Class Width
    1 – 2 5 \[ 2-1=1 \]
    2 – 3 3 \[ 3-2=1 \]
    3 – 5 6 \[ 5-3=2 \]
    5 – 7 12 \[ 7-5=2 \]
    7 – 10 9 \[ 10-7=3 \]
    10 – 15 10 \[ 15-10=5 \]
    15 – 17 4 \[ 17-15=2 \]
  3. Determine Minimum Class Width
  4. The minimum class width is: 1
  5. Calculate Adjusted Heights
  6. Formula: \[ \text{Adjusted Height} = \left( \frac{f} {\text{Class Width}} \right) \times 1 \]
  7. Age Interval \(f\) Class Width Adjusted Height
    1 – 2 5 1 \[ \frac{5}{1}\times1=5 \]
    2 – 3 3 1 \[ \frac{3}{1}\times1=3 \]
    3 – 5 6 2 \[ \frac{6}{2}\times1=3 \]
    5 – 7 12 2 \[ \frac{12}{2}\times1=6 \]
    7 – 10 9 3 \[ \frac{9}{3}\times1=3 \]
    10 – 15 10 5 \[ \frac{10}{5}\times1=2 \]
    15 – 17 4 2 \[ \frac{4}{2}\times1=2 \]
  8. Important Correction: In the original working, the frequencies for the intervals \[ 7-10 \] and \[ 10-15 \] were interchanged while calculating heights. Using the given data correctly:

    For \[ 7-10, \]

    \[ \frac{9}{3}=3 \]

    For \[ 10-15, \]

    \[ \frac{10}{5}=2 \]

  9. Draw the Histogram
  10. Draw rectangles having:
    • Base equal to the actual class interval width.
    • Height equal to the adjusted height calculated above.
  11. Thus the heights of the rectangles will be:\[ 5,\;3,\;3,\;6,\;3,\;2,\;2\]
  12. respectively for the intervals:\[1-2,\;2-3,\;3-5,\;5-7,\;7-10,\;10-15,\;15-17\]
  13. The resulting figure is the required histogram.
  14. Interpretation of the Histogram
    • The age group \[ 5-7 \] years has the highest adjusted frequency density.
    • Most children in the park belong to the age group \[ 5-7 \] years.
    • Very young children \[ (1-3\text{ years}) \] and older children \[ (15-17\text{ years}) \] are comparatively fewer.
    • The histogram shows how children are distributed across different age groups.
🎯 Exam Significance Exam Significance ›
  • Tests drawing histograms with unequal class intervals.
  • Checks understanding of frequency density.
  • Develops interpretation of grouped data.
  • Frequently appears as a higher-order thinking question.
  • Strengthens understanding of graphical representation.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Unequal class widths require adjusted heights.

  2. Formula: \[ \text{Adjusted Height} = \left( \frac{f}{\text{Class Width}} \right) \times \text{Minimum Class Width} \]

  3. Minimum class width here is 1.

  4. The histogram area, not merely height, represents frequency.

  5. This concept forms the basis for advanced statistical graphs.
← Q7
8 / 9  ·  89%
Q9 →
Q9
NUMERIC3 marks
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:

Number of letters Number of surnames
1-4 6
4-6 30
6-8 44
8-12 16
12-20 4
  1. Draw a histogram to depict the given information.
  2. Write the class interval in which the maximum number of surnames lie
📘 Concept & Theory Concept Used ›

A Histogram is a graphical representation of grouped frequency data using adjacent rectangles. The area of each rectangle must be proportional to the frequency represented by that class.

When all class intervals have equal widths, the frequencies themselves are taken as the heights of the rectangles.

However, when class intervals are unequal, the heights must be adjusted using frequency density so that the area of each rectangle continues to represent the actual frequency.

📝 Theory Behind the Question ›

The given class intervals are:

\[ 1-4,\;4-6,\;6-8,\;8-12,\;12-20 \]

Their widths are:

\[ 3,\;2,\;2,\;4,\;8 \]

Since the class widths are unequal, we cannot directly use frequencies as the heights of the histogram bars.

Therefore, we use the adjusted height formula:

\[ \text{Adjusted Height} = \left( \frac{f} {\text{Class Width}} \right) \times \text{Minimum Class Width} \]

Here, the minimum class width is:

\[ 2 \]

Thus,

\[ \text{Adjusted Height} = \left( \frac{f} {\text{Class Width}} \right) \times 2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the width of each class interval.

  2. Identify the minimum class width.

  3. Calculate adjusted heights.

  4. Draw a histogram using actual class widths and adjusted heights.

  5. Determine the class interval having the maximum frequency.

  6. Interpret the graphical representation.

📊 Graph / Figure Graph / Figure ›
Histogram : Number of Letters in Surnames 0 5 10 15 20 25 30 35 40 1-4 4-6 6-8 8-12 12-20
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. Determine Class Widths
  2. Number of Letters Frequency Class Width
    1 – 4 6 \[ 4-1=3 \]
    4 – 6 30 \[ 6-4=2 \]
    6 – 8 44 \[ 8-6=2 \]
    8 – 12 16 \[ 12-8=4 \]
    12 – 20 4 \[ 20-12=8 \]
  3. Find the Minimum Class Width

  4. Minimum class width:

    \[ 2 \]

  5. Calculate Adjusted Heights

  6. Formula:

    \[ \text{Adjusted Height} = \left( \frac{f} {\text{Class Width}} \right) \times 2 \]

  7. Number of Letters \(f\) Class Width Adjusted Height
    1 – 4 6 3 \[ \frac{6}{3}\times2 =2\times2 =4 \]
    4 – 6 30 2 \[ \frac{30}{2}\times2 =15\times2 =30 \]
    6 – 8 44 2 \[ \frac{44}{2}\times2 =22\times2 =44 \]
    8 – 12 16 4 \[ \frac{16}{4}\times2 =4\times2 =8 \]
    12 – 20 4 8 \[ \frac{4}{8}\times2 =0.5\times2 =1 \]
  8. Draw a histogram to depict the given information.
  9. Since the class intervals are unequal, use the adjusted heights calculated above:\[4,\;30,\;44,\;8,\;1\]
  10. Draw rectangles whose:
    • Widths correspond to the actual class intervals.
    • Heights correspond to the adjusted heights.
  12. he resulting figure is the required histogram.
  13. (ii) Write the class interval in which the maximum number of surnames lie.
  14. Looking at the frequency table:
  15. Class Interval Frequency
    1 – 4 6
    4 – 6 30
    6 – 8 44
    8 – 12 16
    12 – 20 4
  16. The highest frequency is:
  17. Therefore, the maximum number of surnames lie in the class interval: \[6-8\]
  18. Interpretation of the Histogram
    • Most surnames contain between \[ 6 \] and \[ 8 \] letters.
    • Very few surnames contain between \[ 12 \] and \[ 20 \] letters.
    • The distribution is concentrated around medium-length surnames.
    • Extremely short and extremely long surnames occur less frequently.
🎯 Exam Significance Exam Significance ›

In questions involving unequal class intervals, students often identify the tallest rectangle in the histogram as having the highest frequency. Remember that after adjustment, the rectangle height represents frequency density, whereas the actual frequency must always be read from the frequency table.

Why This Question Is Important for Board Examinations
  • Tests histogram construction with unequal class intervals.
  • Checks understanding of adjusted frequencies.
  • Develops graphical interpretation skills.
  • Frequently appears in competency-based assessments.
  • Strengthens concepts of frequency density.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Unequal class intervals require adjusted heights.

  2. Formula: \[ \text{Adjusted Height} = \left( \frac{f} {\text{Class Width}} \right) \times \text{Minimum Class Width} \]

  3. Minimum class width: \[ 2 \]

  4. Adjusted heights: \[ 4,\;30,\;44,\;8,\;1 \]

  5. Highest frequency: \[ 44 \]

  6. Modal class: \[ 6-8 \]

  7. Histograms with unequal intervals rely on area rather than height alone.
← Q8
9 / 9  ·  100%
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NCERT Class 9 Maths Exercise 12.1 Solutions
NCERT Class 9 Maths Exercise 12.1 Solutions — Complete Notes & Solutions · academia-aeternum.com
Understanding and interpreting data is a key skill in today’s information-driven world. Chapter 12 of the NCERT Class 9 Mathematics textbook, “Statistics,” introduces students to fundamental concepts of data collection, organization, and representation. The chapter guides learners on how to classify data, construct frequency tables, create graphical representations such as bar graphs and histograms, and perform essential calculations like mean, median, and mode. This exercise solution set aims…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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