CG Board Class 9 Mathematics Chapter 16 Solutions: Data Handling and Analysis (आँकड़ों का प्रबंधन एवं विश्लेषण) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 16 Solutions: Data Handling and Analysis (आँकड़ों का प्रबंधन एवं विश्लेषण) in Bilingual Format (English & Hindi Medium).

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Question

InText Questions Try This 1 Page 293 Question…


📝 Question:

In-Text Questions Try This – 1 (Page 293) Question 1: Write any five conclusions that can be drawn from the bar diagram

💡 Solution:

Based on Table 3 and the corresponding Bar Diagram (Blood Group Data of 30 Students)

  1. The blood group B+ has the highest frequency (9 students).
  2. The blood group O- has the lowest frequency (1 student).
  3. The total number of students with Rh+ blood factor is 5 + 9 + 3 + 3 = 20 students.
  4. The total number of students with Rh- blood factor is 3 + 4 + 2 + 1 = 10 students.
  5. The number of students having blood group A (A+ and A-) is 5 + 3 = 8 students.

Question 2

Similarly also represent the data of group 2…


📝 Question:

Similarly also represent the data of group -2 as a bar diagram. -2

💡 Solution:

Data of Group 2 (Means of Transport to School for 30 Students)

Question 3

Make the attendance table for the students of…


📝 Question:

Make the attendance table for the students of your class in the month of January and answer the following: (i) On which day was the attendance maximum? (ii) When was the attendance minimum? (iii) Write some more conclusions

💡 Solution:

(Sample Class Data for January, Total Strength = 40)

(i) Maximum attendance occurred on Wednesday, January 15 (39 students present).

(ii) Minimum attendance occurred on Saturday, January 25 (22 students present due to extreme winter cold).

(iii) Additional conclusions

  • Average daily attendance during January was 34 students.
  • Attendance on Mondays and Fridays was generally higher than on Saturdays.

Question 4

From hockey cricket kabbadi football and volleyball which…


📝 Question:

From hockey, cricket, kabbadi, football and volleyball, which is the sport your class fellows like the most? Collect data for this and make a frequency table to find the answer to the following: (i) Which is the most popular sport? (ii) Which sport is liked by less children? , , , ?

💡 Solution:

Sample Collected Data (Total 40 Students)

  • Hockey: 5 students (Tally mark: ||||)
  • Cricket: 18 students (Tally mark: |||| |||| |||| |||)
  • Kabaddi: 8 students (Tally mark: |||| |||)
  • Football: 6 students (Tally mark: |||| |)
  • Volleyball: 3 students (Tally mark: |||)

Total = 40 students.

(i) The most popular sport is Cricket (liked by 18 students).

(ii) The sport liked by the least children is Volleyball (liked by only 3 students).

Question

InText Questions on Table 5 Page 295 Question…


📝 Question:

In-Text Questions on Table 5 (Page 295) Question: Look at the frequency table given below. The heights of a group of people is recorded in inclusive classes: Table 5: Class Interval (Height in Centimetre) | Frequency 141-150 | 9 151-160 | 11 161-170 | 15 171-180 | 10 Total | 45 Now discuss the following questions with your friends: 1. How many classes are there in this frequency table? 2. Which group has 180 as its upper limit? 3. What are the lower and upper limits of the class 151 to 160? 4. Which class has the highest frequency? 5. What is the frequency of the first class? 6. What is the meaning of the statement that the class 171 to 180 has a frequency 10? 7. Are these classes inclusive? Yes or no, give reason for your choice? : 5: | 141-150 | 9 151-160 | 11 161-170 | 15 171-180 | 10 | 45 : 1. ? 2. 180 ? 3. 151 160 ? 4. ? 5. ? 6. 171 180 10 ? 7. ? , ?

💡 Solution:

  1. There are 4 classes in this frequency table (141-150, 151-160, 161-170, 171-180).
  2. The class interval 171-180 has 180 as its upper limit.
  3. For the class 151 to 160: Lower limit = 151, Upper limit = 160.
  4. The class 161-170 has the highest frequency (frequency = 15).
  5. The frequency of the first class (141-150) is 9.
  6. It means that there are 10 people whose heights are between 171 cm and 180 cm (including both 171 cm and 180 cm).
  7. Yes, these classes are inclusive because the upper limit of one class is not equal to the lower limit of the next class (e.g., upper limit 150 of 1st class ≠ lower limit 151 of 2nd class), and both upper and lower limits are included in the same class.

Question

Change the inclusive classes of Table 5 to…


📝 Question:

Change the inclusive classes of Table 5 to exclusive classes and add two people with heights 150.5 cm and 160.5 cm to the data. 5 150.5 cm 160.5 cm

💡 Solution:

Adjustment Factor Calculation

Difference between lower limit of second class (151) and upper limit of first class (150) = 151 – 150 = 1.

Adjustment factor = 1 / 2 = 0.5.

Subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Exclusive Classes Table before adding new data

  • 140.5 – 150.5 (Frequency = 9)
  • 150.5 – 160.5 (Frequency = 11)
  • 160.5 – 170.5 (Frequency = 15)
  • 170.5 – 180.5 (Frequency = 10)

Adding the two persons in exclusive classes

  • Person with height 150.5 cm goes into class 150.5 – 160.5 (as lower limit value is included in exclusive class). Frequency increases from 11 to 12.
  • Person with height 160.5 cm goes into class 160.5 – 170.5. Frequency increases from 15 to 16.

Updated Exclusive Frequency Table

  • 140.5 – 150.5 : 9
  • 150.5 – 160.5 : 12
  • 160.5 – 170.5 : 16
  • 170.5 – 180.5 : 10

Total Frequency = 9 + 12 + 16 + 10 = 47.

Question

The areas of the frequency polygon and of…


📝 Question:

The areas of the frequency polygon and of the histogram are the same, Why? (Hint: Use the property of congruency of triangles) , ?

💡 Solution:

Explanation

  1. When a frequency polygon is constructed over a histogram, the line segments of the polygon cut off triangular regions from the rectangles of the histogram.
  2. For every class rectangle, the triangular area included inside the polygon but outside the rectangle is congruent to the triangular area included inside the rectangle but excluded from the polygon.
  3. By SAS or ASA congruency criteria, these corresponding small right-angled triangles are congruent and thus equal in area.
  4. Hence, the total area added outside the histogram rectangles equals the total area removed from inside the histogram rectangles.
  5. Therefore, the total area enclosed by the frequency polygon is strictly equal to the total area of the histogram.

SECTION / EXERCISE

Exercise 16.1

Question 1

Explain the following Class interval size of a…


📝 Question:

Explain the following: Class interval, size of a class, class mark, class frequency, class limits : , , , ,

💡 Solution:

  1. Class interval: When a large set of numerical data is divided into small suitable groups, each group is called a class interval (e.g., 10-20, 20-30).
  2. Size of a class (Class width): The difference between the upper class limit and lower class limit of a class (or difference between two consecutive class marks) is called the class size. Formula: Class Size = Upper Limit – Lower Limit.
  3. Class mark: The middle point of a class interval is called its class mark. Formula: Class Mark = (Lower Limit + Upper Limit) / 2.
  4. Class frequency: The number of observations that fall within a particular class interval is called the class frequency.
  5. Class limits: The two extreme values that specify a class interval are called class limits. The smaller value is the Lower Class Limit and the larger value is the Upper Class Limit.

Question 2

Give the difference between inclusive and exclusive classes


📝 Question:

Give the difference between inclusive and exclusive classes

💡 Solution:

  1. Inclusive Classes:
  • The upper limit of one class is not equal to the lower limit of the next class (e.g., 1-10, 11-20).
  • Both lower and upper limits are included in the same class.
  • Suitable for discrete data (like whole numbers).
  1. Exclusive Classes:
  • The upper limit of one class is equal to the lower limit of the next class (e.g., 0-10, 10-20).
  • The lower limit is included in the class, but the upper limit is excluded (it goes into the next class).
  • Suitable for continuous data (like height, weight, temperature).

Question 3

The weather department has given the following as…


📝 Question:

The weather department has given the following as the data of maximum temperatures in August for Delhi. Make a frequency table of this data. 32.5, 33.3, 33.8, 31.0, 28.6, 33.9, 33.3, 32.4, 30.4, 32.6, 34.7, 34.9, 31.6, 35.2, 33.3, 33.3, 36.4, 36.6, 37.0, 34.5, 32.5, 31.4, 34.4, 33.6, 37.3, 37.5, 36.9, 37.0, 36.3, 36.9, 36.9 : 32.5, 33.3, 33.8, 31.0, 28.6, 33.9, 33.3, 32.4, 30.4, 32.6, 34.7, 34.9, 31.6, 35.2, 33.3, 33.3, 36.4, 36.6, 37.0, 34.5, 32.5, 31.4, 34.4, 33.6, 37.3, 37.5, 36.9, 37.0, 36.3, 36.9, 36.9

💡 Solution:

Minimum temperature = 28.6 °C, Maximum temperature = 37.5 °C.

Taking class intervals of 1 °C starting from 28.5 – 29.5

Frequency Distribution Table

  • 28.5 – 29.5 : Tally mark = | , Frequency = 1 (Data: 28.6)
  • 29.5 – 30.5 : Tally mark = | , Frequency = 1 (Data: 30.4)
  • 30.5 – 31.5 : Tally mark = || , Frequency = 2 (Data: 31.0, 31.4)
  • 31.5 – 32.5 : Tally mark = || , Frequency = 2 (Data: 31.6, 32.4)
  • 32.5 – 33.5 : Tally mark = |||| || , Frequency = 7 (Data: 32.5, 32.5, 32.6, 33.3, 33.3, 33.3, 33.3)
  • 33.5 – 34.5 : Tally mark = |||| , Frequency = 4 (Data: 33.6, 33.8, 33.9, 34.4)
  • 34.5 – 35.5 : Tally mark = |||| , Frequency = 4 (Data: 34.5, 34.7, 34.9, 35.2)
  • 35.5 – 36.5 : Tally mark = || , Frequency = 2 (Data: 36.3, 36.4)
  • 36.5 – 37.5 : Tally mark = |||| || , Frequency = 7 (Data: 36.6, 36.9, 36.9, 36.9, 37.0, 37.0, 37.3)
  • 37.5 – 38.5 : Tally mark = | , Frequency = 1 (Data: 37.5)

Total Frequency = 31 days.

Question 4

The following are the distances of the work…


📝 Question:

The following are the distances of the work places of 40 teachers from their homes: 7, 9, 5, 3, 7, 8, 10, 20, 3, 5, 11, 25, 15, 12, 7, 13, 18, 12, 11, 3, 12, 6, 12, 14, 7, 2, 9, 15, 6, 15, 17, 2, 16, 32, 19, 10, 12, 17, 18, 11 Make a frequency distribution table of class size 5 for this data. 40 : 7, 9, 5, 3, 7, 8, 10, 20, 3, 5, 11, 25, 15, 12, 7, 13, 18, 12, 11, 3, 12, 6, 12, 14, 7, 2, 9, 15, 6, 15, 17, 2, 16, 32, 19, 10, 12, 17, 18, 11 5

💡 Solution:

Minimum distance = 2 km, Maximum distance = 32 km.

Class intervals of size 5: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35.

Frequency Distribution Table

  • 0 – 5 : Tally mark = |||| , Frequency = 5 (Values: 2, 2, 3, 3, 3)
  • 5 – 10 : Tally mark = |||| |||| | , Frequency = 11 (Values: 5, 5, 6, 6, 7, 7, 7, 7, 8, 9, 9)
  • 10 – 15 : Tally mark = |||| |||| || , Frequency = 12 (Values: 10, 10, 11, 11, 11, 12, 12, 12, 12, 12, 13, 14)
  • 15 – 20 : Tally mark = |||| |||| , Frequency = 9 (Values: 15, 15, 15, 16, 17, 17, 18, 18, 19)
  • 20 – 25 : Tally mark = | , Frequency = 1 (Value: 20)
  • 25 – 30 : Tally mark = | , Frequency = 1 (Value: 25)
  • 30 – 35 : Tally mark = | , Frequency = 1 (Value: 32)

Total Frequency = 5 + 11 + 12 + 9 + 1 + 1 + 1 = 40.

Question 5

The value of Pi up to 50 places…


📝 Question:

The value of Pi up to 50 places of decimal is given below: 3.14159265358979323846264338327950288419716939937510 (i) Make a frequency distribution table of the digits from 0 to 9 occurring in this expansion. (ii) Which is the number that occurs the least? (iii) Which is the number that occurs the most? What can we conclude from this? 50 π : 3.14159265358979323846264338327950288419716939937510

💡 Solution:

Counting the 50 decimal digits after the decimal point

14159265358979323846264338327950288419716939937510

(i) Frequency Table for Digits 0 to 9

  • Digit 0: Tally mark = || , Frequency = 2
  • Digit 1: Tally mark = |||| , Frequency = 5
  • Digit 2: Tally mark = |||| , Frequency = 5
  • Digit 3: Tally mark = |||| ||| , Frequency = 8
  • Digit 4: Tally mark = |||| , Frequency = 4
  • Digit 5: Tally mark = |||| , Frequency = 5
  • Digit 6: Tally mark = |||| , Frequency = 4
  • Digit 7: Tally mark = |||| , Frequency = 4
  • Digit 8: Tally mark = |||| , Frequency = 5
  • Digit 9: Tally mark = |||| ||| , Frequency = 8

Total = 50 digits.

(ii) The digit that occurs the least is 0 (occurs 2 times).

(iii) The digits that occur the most are 3 and 9 (each occurs 8 times).

Conclusion: The digits in the decimal expansion of π appear randomly without any recurring pattern, which reflects the irrational nature of π.

Question 6

The per hectare production of rice in 40…


📝 Question:

The per hectare production of rice in 40 fields of a village in quintals is given below, make a frequency distribution for this data. 31, 20, 25, 18, 28, 20, 18, 26, 15, 12, 25, 16, 30, 20, 22, 24, 45, 28, 30, 16, 30, 40, 20, 30, 20, 30, 28, 47, 40, 35, 28, 45, 20, 35, 32, 18, 20, 26, 23, 16 Write the conclusions that you draw from this frequency distribution table. 40 , : 31, 20, 25, 18, 28, 20, 18, 26, 15, 12, 25, 16, 30, 20, 22, 24, 45, 28, 30, 16, 30, 40, 20, 30, 20, 30, 28, 47, 40, 35, 28, 45, 20, 35, 32, 18, 20, 26, 23, 16

💡 Solution:

Minimum yield = 12 quintals, Maximum yield = 47 quintals.

Class intervals of size 5 (Exclusive): 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50.

Frequency Distribution Table

  • 10 – 15 : Frequency = 1 (Value: 12)
  • 15 – 20 : Frequency = 7 (Values: 15, 16, 16, 16, 18, 18, 18)
  • 20 – 25 : Frequency = 10 (Values: 20, 20, 20, 20, 20, 20, 20, 22, 23, 24)
  • 25 – 30 : Frequency = 8 (Values: 25, 25, 26, 26, 28, 28, 28, 28)
  • 30 – 35 : Frequency = 7 (Values: 30, 30, 30, 30, 30, 31, 32)
  • 35 – 40 : Frequency = 2 (Values: 35, 35)
  • 40 – 45 : Frequency = 2 (Values: 40, 40)
  • 45 – 50 : Frequency = 3 (Values: 45, 45, 47)

Total Frequency = 1 + 7 + 10 + 8 + 7 + 2 + 2 + 3 = 40.

Conclusions

  1. Maximum number of fields (10 fields) have a yield between 20 and 25 quintals.
  2. Most fields (32 out of 40, or 80%) have yield in the range of 15 to 35 quintals.
  3. Only 5 fields (12.5%) produce 40 quintals or more per hectare.

Question 7

40 children were asked about the number of…


📝 Question:

40 children were asked about the number of hours they watched TV in the previous week, their responses were: 1, 5, 6, 2, 7, 4, 10, 12, 5, 8, 10, 12, 36, 22, 6, 15, 3, 1, 2, 4, 21, 16, 17, 13, 14, 2, 7, 9, 23, 26, 31, 33, 5, 35, 25, 26, 29, 30, 9, 31 (i) Make a frequency distribution table of class size 5. (ii) What is the lower limit of the first class? (iii) Give the limits of the fourth class (iv) What is the class mark of the 7th class (v) How many children watched television for 20 or more hours in the week? 40 , : 1, 5, 6, 2, 7, 4, 10, 12, 5, 8, 10, 12, 36, 22, 6, 15, 3, 1, 2, 4, 21, 16, 17, 13, 14, 2, 7, 9, 23, 26, 31, 33, 5, 35, 25, 26, 29, 30, 9, 31

💡 Solution:

(i) Frequency Distribution Table (Class size = 5)

  • 0 – 5 : Tally mark = |||| ||| , Frequency = 8 (Values: 1, 1, 2, 2, 2, 3, 4, 4)
  • 5 – 10 : Tally mark = |||| |||| , Frequency = 10 (Values: 5, 5, 5, 6, 6, 7, 7, 8, 9, 9)
  • 10 – 15 : Tally mark = |||| | , Frequency = 6 (Values: 10, 10, 12, 12, 13, 14)
  • 15 – 20 : Tally mark = ||| , Frequency = 3 (Values: 15, 16, 17)
  • 20 – 25 : Tally mark = ||| , Frequency = 3 (Values: 21, 22, 23)
  • 25 – 30 : Tally mark = |||| , Frequency = 4 (Values: 25, 26, 26, 29)
  • 30 – 35 : Tally mark = |||| , Frequency = 4 (Values: 30, 31, 31, 33)
  • 35 – 40 : Tally mark = || , Frequency = 2 (Values: 35, 36)

Total = 40.

(ii) Lower limit of the first class (0-5) = 0.

(iii) Limits of the 4th class (15-20): Lower limit = 15, Upper limit = 20.

(iv) 7th class is 30-35. Class mark = (30 + 35) / 2 = 65 / 2 = 32.5.

(v) Children watching TV for 20 or more hours = Frequencies of classes (20-25) + (25-30) + (30-35) + (35-40) = 3 + 4 + 4 + 2 = 13 children.

SECTION / EXERCISE

Exercise 16.2

Question 1

Choose the correct alternative from the following i…


📝 Question:

Choose the correct alternative from the following: (i) In an inclusive class: a) Both the limits are included in different classes b) Both the limits are included in the same class c) It is not decided in which class the limits are included d) None of the above (ii) Method to make the cumulative frequency table is: a) Less than b) More than c) Both a and b d) None of the above (iii) Using the histogram we can find out: a) Mode b) Median c) Both a and b d) None of the above (iv) Using the cumulative frequency curve we can find out: a) Mode b) Median c) Both a and b d) None of the above (v) The width of the rectangle in a histogram depends on: a) Class interval of the class b) Class mark c) On the class frequency d) On all of the above : a) – b) c) d) a) b) c) a b d) a) b) c) a b d) a) b) c) a b d) a) b) c) d)

💡 Solution:

(i) Correct option: b) Both the limits are included in the same class.

(ii) Correct option: c) Both a and b.

(iii) Correct option: c) Both a and b (Mode can be directly determined from peak height, and Median via combined graph/area).

(iv) Correct option: b) Median.

(v) Correct option: a) Class interval of the class.

Question 2

The time taken in second by 25 students…


📝 Question:

The time taken (in second) by 25 students in doing a question is as follows: 16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20 (i) Make frequency table of this data using a class interval of 10 second. (ii) Make a histogram depicting this frequency distribution. (iii) Use the histogram to make a frequency polygon. 25 : 16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20

💡 Solution:

(i) Frequency Table (Class Interval = 10 s)

  • 15 – 25 : Frequency = 3 (Values: 16, 20, 20)
  • 25 – 35 : Frequency = 5 (Values: 26, 27, 28, 30, 33)
  • 35 – 45 : Frequency = 5 (Values: 37, 38, 40, 42, 43)
  • 45 – 55 : Frequency = 8 (Values: 46, 46, 46, 48, 49, 50, 52, 53)
  • 55 – 65 : Frequency = 4 (Values: 58, 59, 60, 64)

Total Frequency = 3 + 5 + 5 + 8 + 4 = 25 students.

(ii) & (iii) Graphical Construction Steps

  1. Plot X-axis with time intervals (15 to 65 s) and Y-axis with frequency (0 to 8).
  2. Draw adjacent vertical bars (rectangles) over intervals with heights 3, 5, 5, 8, and 4 to form the Histogram.
  3. Mark mid-points of the top edge of each rectangle: B(20, 3), C(30, 5), D(40, 5), E(50, 8), F(60, 4).
  4. Mark mid-points on X-axis of preceding class 5-15 at A(10, 0) and succeeding class 65-75 at G(70, 0).
  5. Connect points A-B-C-D-E-F-G with straight line segments to form the Frequency Polygon ABCDEFG.

Question 3

The length of 30 leaves of a plant…


📝 Question:

The length of 30 leaves of a plant is given in milimetre below: Length of leaves (in mm.): 111-120 | 121-130 | 131-140 | 141-150 | 151-160 | 161-170 Number of leaves: 3 | 5 | 7 | 9 | 4 | 2 (i) Make a histogram to depict the frequency distribution. (Hint: Make the class intervals continuous) (ii) Make a frequency polygon using the direct method. (iii) In which class the number of leaves is the largest? 30 : : 111-120 | 121-130 | 131-140 | 141-150 | 151-160 | 161-170 : 3 | 5 | 7 | 9 | 4 | 2

💡 Solution:

(i) Converting Inclusive Classes to Continuous Exclusive Classes

Adjustment factor = (121 – 120) / 2 = 0.5.

Continuous Class Intervals and Class Marks

  • 110.5 – 120.5 : Frequency = 3, Class mark = 115.5
  • 120.5 – 130.5 : Frequency = 5, Class mark = 125.5
  • 130.5 – 140.5 : Frequency = 7, Class mark = 135.5
  • 140.5 – 150.5 : Frequency = 9, Class mark = 145.5
  • 150.5 – 160.5 : Frequency = 4, Class mark = 155.5
  • 160.5 – 170.5 : Frequency = 2, Class mark = 165.5

Histogram Construction

Plot continuous class boundaries on X-axis and number of leaves on Y-axis. Construct rectangles of heights 3, 5, 7, 9, 4, 2.

(ii) Direct Method for Frequency Polygon

  1. Coordinates for points: P1(115.5, 3), P2(125.5, 5), P3(135.5, 7), P4(145.5, 9), P5(155.5, 4), P6(165.5, 2).
  2. Take preceding class mark 105.5 at P0(105.5, 0) and succeeding class mark 175.5 at P7(175.5, 0) on X-axis.
  3. Join P0-P1-P2-P3-P4-P5-P6-P7 with straight line segments.

(iii) The largest number of leaves is in the class 141-150 (or continuous class 140.5-150.5 mm) with frequency = 9.

Question 4

The number of runs made by two teams…


📝 Question:

The number of runs made by two teams A and B in a 10 over (60 ball) match is given below: Number of Balls | Team A | Team B 1-6 | 3 | 6 7-12 | 2 | 6 13-18 | 7 | 1 19-24 | 8 | 10 25-30 | 3 | 6 31-36 | 8 | 5 37-42 | 5 | 3 43-48 | 11 | 4 49-54 | 6 | 7 55-60 | 2 | 11 Use the above data make a frequency polygon on one graph paper. (Hint: First make the class intervals continuous) 10 A B : | A | B 1-6 | 3 | 6 7-12 | 2 | 6 13-18 | 7 | 1 19-24 | 8 | 10 25-30 | 3 | 6 31-36 | 8 | 5 37-42 | 5 | 3 43-48 | 11 | 4 49-54 | 6 | 7 55-60 | 2 | 11

💡 Solution:

  1. Convert to Continuous Class Intervals & Calculate Class Marks

Adjustment factor = (7 – 6) / 2 = 0.5.

  • 0.5 – 6.5 : Class mark = 3.5 | Team A = 3 | Team B = 6
  • 6.5 – 12.5 : Class mark = 9.5 | Team A = 2 | Team B = 6
  • 12.5 – 18.5 : Class mark = 15.5 | Team A = 7 | Team B = 1
  • 18.5 – 24.5 : Class mark = 21.5 | Team A = 8 | Team B = 10
  • 24.5 – 30.5 : Class mark = 27.5 | Team A = 3 | Team B = 6
  • 30.5 – 36.5 : Class mark = 33.5 | Team A = 8 | Team B = 5
  • 36.5 – 42.5 : Class mark = 39.5 | Team A = 5 | Team B = 3
  • 42.5 – 48.5 : Class mark = 45.5 | Team A = 11 | Team B = 4
  • 48.5 – 54.5 : Class mark = 51.5 | Team A = 6 | Team B = 7
  • 54.5 – 60.5 : Class mark = 57.5 | Team A = 2 | Team B = 11
  1. Terminal Class Marks on X-axis

Preceding class mark = -2.5 (at 0 runs/frequency)

Succeeding class mark = 63.5 (at 0 runs/frequency)

  1. Graph Construction
  • Plot Class Marks on X-axis and Runs Scored on Y-axis.
  • Draw Team A polygon using solid line joining (-2.5, 0), (3.5, 3), (9.5, 2), (15.5, 7), (21.5, 8), (27.5, 3), (33.5, 8), (39.5, 5), (45.5, 11), (51.5, 6), (57.5, 2), and (63.5, 0).
  • Draw Team B polygon using dashed line joining (-2.5, 0), (3.5, 6), (9.5, 6), (15.5, 1), (21.5, 10), (27.5, 6), (33.5, 5), (39.5, 3), (45.5, 4), (51.5, 7), (57.5, 11), and (63.5, 0).

Question 5

The frequency distribution of marks obtained by 100…


📝 Question:

The frequency distribution of marks obtained by 100 students is as follows: Marks obtained: 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total Number of students: 7 | 10 | 23 | 51 | 6 | 3 | 100 Make a cumulative frequency curve using the above data. 100 : : 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | : 7 | 10 | 23 | 51 | 6 | 3 | 100

💡 Solution:

“Less than” Cumulative Frequency Table

  • Marks less than 10 : Cumulative Frequency = 7
  • Marks less than 20 : Cumulative Frequency = 7 + 10 = 17
  • Marks less than 30 : Cumulative Frequency = 17 + 23 = 40
  • Marks less than 40 : Cumulative Frequency = 40 + 51 = 91
  • Marks less than 50 : Cumulative Frequency = 91 + 6 = 97
  • Marks less than 60 : Cumulative Frequency = 97 + 3 = 100

Construction of Ogive Curve

  1. Plot upper class limits (10, 20, 30, 40, 50, 60) on X-axis and Cumulative Frequencies on Y-axis.
  2. Plot coordinates: (0, 0), (10, 7), (20, 17), (30, 40), (40, 91), (50, 97), (60, 100).
  3. Connect all points sequentially with a smooth freehand curve to obtain the “Less than” Ogive.

Question 6

Make the frequency table and the cumulative frequency…


📝 Question:

Make the frequency table and the cumulative frequency curve for the following data: Marks Obtained | Number of Students less than 10 | 3 less than 20 | 8 less than 30 | 12 less than 40 | 19 less than 50 | 31 less than 60 | 42 less than 70 | 60 : | 10 | 3 20 | 8 30 | 12 40 | 19 50 | 31 60 | 42 70 | 60

💡 Solution:

Frequency Table Derivation from Cumulative Frequencies

  • 0 – 10 : Frequency = 3
  • 10 – 20 : Frequency = 8 – 3 = 5
  • 20 – 30 : Frequency = 12 – 8 = 4
  • 30 – 40 : Frequency = 19 – 12 = 7
  • 40 – 50 : Frequency = 31 – 19 = 12
  • 50 – 60 : Frequency = 42 – 31 = 11
  • 60 – 70 : Frequency = 60 – 42 = 18

Total Frequency = 3 + 5 + 4 + 7 + 12 + 11 + 18 = 60 students.

Cumulative Frequency Curve (Ogive) Construction

Plot points (0, 0), (10, 3), (20, 8), (30, 12), (40, 19), (50, 31), (60, 42), (70, 60) with upper limits on X-axis and cumulative frequency on Y-axis. Join with a smooth curve.

Question 7

The marks obtained by two groups of students…


📝 Question:

The marks obtained by two groups of students in a test are given below: Class Interval | Group A | Group B 50-52 | 4 | 8 47-49 | 10 | 9 44-46 | 15 | 10 41-43 | 18 | 14 38-40 | 20 | 12 35-37 | 12 | 17 32-34 | 13 | 22 Total | 92 | 92 Make frequency polygons for each of these groups on one graph paper. : | A | B 50-52 | 4 | 8 47-49 | 10 | 9 44-46 | 15 | 10 41-43 | 18 | 14 38-40 | 20 | 12 35-37 | 12 | 17 32-34 | 13 | 22 | 92 | 92

💡 Solution:

  1. Arrange Class Intervals in Ascending Order and Convert to Continuous Exclusive Classes

Adjustment factor = (35 – 34) / 2 = 0.5.

Class Intervals, Class Marks & Frequencies

  • 31.5 – 34.5 : Class mark = 33 | Group A = 13 | Group B = 22
  • 34.5 – 37.5 : Class mark = 36 | Group A = 12 | Group B = 17
  • 37.5 – 40.5 : Class mark = 39 | Group A = 20 | Group B = 12
  • 40.5 – 43.5 : Class mark = 42 | Group A = 18 | Group B = 14
  • 43.5 – 46.5 : Class mark = 45 | Group A = 15 | Group B = 10
  • 46.5 – 49.5 : Class mark = 48 | Group A = 10 | Group B = 9
  • 49.5 – 52.5 : Class mark = 51 | Group A = 4 | Group B = 8
  1. Terminal Class Marks

Preceding class mark = 30 (Frequency = 0)

Succeeding class mark = 54 (Frequency = 0)

  1. Plotting Frequencies on Graph
  • X-axis: Class marks (30, 33, 36, 39, 42, 45, 48, 51, 54)
  • Y-axis: Number of students / Frequencies (0 to 22)
  • Group A Polygon: Join points (30, 0), (33, 13), (36, 12), (39, 20), (42, 18), (45, 15), (48, 10), (51, 4), and (54, 0) with a solid line.
  • Group B Polygon: Join points (30, 0), (33, 22), (36, 17), (39, 12), (42, 14), (45, 10), (48, 9), (51, 8), and (54, 0) with a dashed line.

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