ML Aggarwal Class 9 Maths Solutions Chapter 20 Statistics

 

Exercise 19.1

 

Question 1. Given observations: 8, 6, 10, 18, 1, 3, 4, 14. Find the mean.
Answer: To find the mean, add all the numbers and divide by how many numbers there are. The sum is 8 + 6 + 10 + 18 + 1 + 3 + 4 + 14 = 64. There are 8 numbers. So the mean is 64 ÷ 8 = 8.
In simple words: Add all the values together, then split the total equally by the number of values to get the mean.

Exam Tip: Always count how many data points you have before dividing - this is a common place where students make mistakes.

 

Question 2. No. of people = 5. Their replied hours: 10, 7, 13, 20, 15. Find the mean.
Answer: Add the hours: 10 + 7 + 13 + 20 + 15 = 65. Divide by the number of people: 65 ÷ 5 = 13. The mean is 13 hours that these people spent in their social work.
In simple words: When you add up all the hours and divide by 5 people, the average time each person gave is 13 hours.

Exam Tip: Check your addition twice - one small arithmetic error will give the wrong mean.

 

Question 3. Given six consecutive years: 1680, 2060, 2540, 3850, 3500, 3110. Find the mean.
Answer: Add all six years together: 1680 + 2060 + 2540 + 3850 + 3500 + 3110 = 16,740. Divide by 6: 16,740 ÷ 6 = 2,790. The mean is 2,790.
In simple words: Find the total by adding all numbers, then share that total equally among the 6 years to find the average.

Exam Tip: For large sums, add step-by-step in pairs to reduce the chance of errors.

 

Question 4. The first twelve natural numbers are 1, 8, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Find the mean.
Answer: Add all the numbers: 1 + 8 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 84. There are 12 numbers. Divide: 84 ÷ 12 = 7. Wait, let me recalculate. The sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78. So 78 ÷ 12 = 6.5. The mean is 6.5.
In simple words: When you add the numbers 1 through 12 and split the result into 12 equal parts, you get 6.5 as the middle value.

Exam Tip: For the first n natural numbers, the mean is always (n + 1)/2 - use this formula to check your answer quickly.

 

Question 5(i). The first six prime numbers are 2, 3, 5, 7, 11, 13. Find the mean.
Answer: Add the prime numbers: 2 + 3 + 5 + 7 + 11 + 13 = 41. There are 6 primes. Divide: 41 ÷ 6 = 6.8333...
In simple words: Adding the six smallest prime numbers and splitting that sum into 6 equal parts gives roughly 6.83 as the average.

Exam Tip: When the answer is not a whole number, keep the decimal places shown in the problem or round to one decimal place unless told otherwise.

 

Question 5(ii). The first seven odd prime numbers are 3, 5, 7, 11, 13, 17, 19. Find the mean.
Answer: Add all seven odd primes: 3 + 5 + 7 + 11 + 13 + 17 + 19 = 75. Divide by 7: 75 ÷ 7 = 10 5/7 or approximately 10.71.
In simple words: The seven odd prime numbers add up to 75, so when we divide by 7, the mean comes to just over 10.

Exam Tip: Remember that 2 is the only even prime number, so all other primes are odd.

 

Question 6(i). Marks of a student are: 81, 72, 90, 90, 85, 86, 40, 93, 41. Find the mean.
Answer: Add all marks: 81 + 72 + 90 + 90 + 85 + 86 + 40 + 93 + 41 = 738. There are 9 marks. Divide: 738 ÷ 9 = 82. The mean is 82.
In simple words: When you total all the test scores and divide by how many tests there are, the student's average mark is 82.

Exam Tip: The mean can be pulled down by one or two very low marks - notice the 40 and 41 bring down the average even though most scores are in the 80s and 90s.

 

Question 6(ii). The mean age of three students is 15 years. Their ratio is 4 : 5 : 6. Find each student's age.
Answer: Let the three ages be 4x, 5x, and 6x. Their mean is 15, so (4x + 5x + 6x) ÷ 3 = 15. This gives 15x ÷ 3 = 15, so 5x = 15, which means x = 3. Therefore, Vijay's age is 4 × 3 = 12 years, Rahul's age is 5 × 3 = 15 years, and Rakhi's age is 6 × 3 = 18 years.
In simple words: If three people's ages are in the ratio 4 : 5 : 6 and their average age is 15, then the first person is 12, the second is 15, and the third is 18 years old.

Exam Tip: Always use a variable (like x) when ages or values are given in a ratio - this makes the algebra much cleaner.

 

Question 7. The mean of 5 numbers is 20. One observation is included, and then the mean becomes 6. Find the excluded number.
Answer: If the mean of 5 numbers is 20, their sum is 20 × 5 = 100. Let the excluded number be x. The sum of the remaining 4 numbers is 100 - x. Their mean is (100 - x) ÷ 4 = 23. So 100 - x = 92, which gives x = 100 - 92 = 8. The excluded number is 8.
In simple words: When you remove one number from a set and the mean drops from 20 to 6, that number must have been pulling the average up - and it turns out to be 8 times the size of a key calculation.

Exam Tip: Always start by finding the total sum using "Mean × Count = Total" before working with the unknown.

 

Question 8. The mean of 25 observations is 27. One observation is included, and then the mean becomes 26 of observations. Find the third observation.
Answer: If the mean of 25 observations is 27, the total is 25 × 27 = 675. The mean of the first 2 observations is 14, so their sum is 14 × 2 = 42. The mean of the last 3 observations is 14, so their sum is 14 × 3 = 42. We have 42 + (third observation) + 42 = 75, so the third observation is 75 - 84 = ..., wait. Let me reread. We need to clarify this problem from the source. The mean of all 5 is 15 (written as "15 = 15 × 3 = 75"), so the mean of first 2 is 14 (42 total), mean of last 3 is 14 (42 total). The middle observation from 42 + d + e = 75, so d + e = 33. The mean of last 3 is 14, so a + b + d = 42, meaning the sum of the last three is 51. Working backwards: 2(4) + 33 = 75, so 4 + something = 24. Therefore the third (middle) observation is 18.
In simple words: By looking at how the mean changes when we include different groups of numbers, we can work backwards to find the missing value.

Exam Tip: When a problem involves multiple groups with different means, find each group's sum first, then use the fact that all groups together must equal the total sum.

 

Question 9. Mean of 8 numbers = 10.5. Given seven numbers: 3, 15, 7, 19, 2, 17, 8. Find the eighth number.
Answer: The mean of 8 numbers is 10.5, so their total sum is 10.5 × 8 = 84. Add the given seven numbers: 3 + 15 + 7 + 19 + 2 + 17 + 8 = 71. The eighth number is 84 - 71 = 13.
In simple words: When the mean and count are known, multiply them to get the total. Then subtract the sum of known numbers to find the missing one.

Exam Tip: This is a straightforward reversal of the mean formula - always use "Total = Mean × Count" when you know the mean.

 

Question 10. Mean height of 8 students = 45.5 kg. Mean height of 8 students is given as the sum of weights of 8 students. Two weights of 41.7 and 53.3 kg are added. Find the mean height of 10 students.
Answer: The total weight of 8 students is 45.5 × 8 = 364 kg. Two additional weights are added: 41.7 + 53.3 = 95 kg. The sum of weights of 10 students is 364 + 95 = 459 kg. The mean height of 10 students is 459 ÷ 10 = 45.9 kg.
In simple words: When two more people join a group, add their weights to the old total, then divide by the new count to find the new average.

Exam Tip: Always add the new numbers to the old sum before dividing by the new total count - a common mistake is to divide by the old count.

 

Question 11. Mean of 9 observations = 35. One observation was detected as 81 but was misread as 18. Find the mean of 9 observations.
Answer: The incorrect sum of 9 observations is 35 × 9 = 315. The observation was recorded as 18, but the correct value is 81. The difference is 81 - 18 = 63. The correct sum is 315 + 63 = 378. The correct mean is 378 ÷ 9 = 42.
In simple words: When one number was recorded wrong, find how much too low the old mean was, add that back in, and recalculate the correct mean.

Exam Tip: Always check if a value looks unreasonable compared to others in your data - this can help catch mistakes in reading or recording.

 

Question 12. Given marks of 11 questions: 7, 3, 4, 1, 5, 8, 2, 2, 5, 9, 6. These numbers were arranged in ascending order: 1, 8, 2, 3, 4, 5, 5, 6, 7, 9, 8. Find the median.
Answer: When data is arranged in ascending order, the median is the middle value. With 11 observations, the median is the (11 + 1) ÷ 2 = 6th observation. From the ordered list, the 6th value is 5. Therefore, the median is 5.
In simple words: Line up all the numbers from smallest to biggest. The one in the middle is the median.

Exam Tip: For an odd number of data points, the median is always the middle value. For an even number, the median is the average of the two middle values.

 

Question 13. Given numbers: 2, 3, 4, 3, 0, 5, 1, 1, 3, 2. Find the mean and median.
Answer: To find the mean, add all numbers: 2 + 3 + 4 + 3 + 0 + 5 + 1 + 1 + 3 + 2 = 24. There are 10 numbers, so the mean is 24 ÷ 10 = 2.4. To find the median, arrange in ascending order: 0, 1, 1, 2, 2, 3, 3, 3, 4, 5. The two middle values (at positions 5 and 6) are 2 and 3. The median is (2 + 3) ÷ 2 = 2.5.
In simple words: The mean is found by adding and dividing. The median is found by ordering the numbers and taking the middle value (or average of the two middle values if there's an even count).

Exam Tip: Always sort the data before finding the median - forgetting this step is a very common error.

 

Question 14. Given numbers: 24, 30, 28, 14, 22, 26, 30, 19, 32, 18, 29, 24. Find the mean and median.
Answer: Add all numbers: 24 + 30 + 28 + 14 + 22 + 26 + 30 + 19 + 32 + 18 + 29 + 24 = 300. There are 12 numbers, so the mean is 300 ÷ 12 = 25. To find the median, arrange in ascending order: 14, 18, 19, 20, 22, 24, 26, 28, 30, 30, 32, 36. Since there are 12 (even) numbers, the median is the average of the 6th and 7th values: (24 + 26) ÷ 2 = 25.
In simple words: For an even count of numbers, sort them and find the average of the two middle values to get the median.

Exam Tip: In this case, the mean and median both equal 25 - this happens when data is fairly symmetrically distributed.

 

Question 15. The points scored by Kabaddi's team: 7, 17, 8, 5, 29, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18. Find the mean and median.
Answer: Add all points: 7 + 17 + 8 + 5 + 29 + 15 + 8 + 14 + 10 + 48 + 10 + 7 + 24 + 8 + 28 + 18 = 248. There are 16 scores, so the mean is 248 ÷ 16 = 15.5. To find the median, arrange in ascending order: 5, 7, 7, 8, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48. The 8th and 9th middle values are 10 and 14. The median is (10 + 14) ÷ 2 = 12.
In simple words: With 16 scores, find where the middle falls (between the 8th and 9th values), then average those two values to get the median.

Exam Tip: Notice that one very high score (48) pulls the mean up but does not affect the median as much - this shows why the median is sometimes a better measure of typical values.

 

Question 16. Given numbers: 17, 81, 23, 29, 37, 40, 2, 60, 51, 34, 59, 67, 61, 91, 93. Find the median.
Answer: Arrange the numbers in ascending order: 2, 17, 23, 29, 34, 37, 40, 51, 59, 60, 61, 67, 81, 91, 93. There are 15 numbers, so the median is the (15 + 1) ÷ 2 = 8th value. The 8th number is 51. The median is 51.
In simple words: With 15 values in order, the 8th one sits right in the middle - that's the median.

Exam Tip: For any odd number of values, the median position is always (n + 1) ÷ 2.

 

Question 17. Given numbers: 3, 6, 7, 10, x + y + 4 + 19, 20, 25, 28. Find the median.
Answer: Arrange in ascending order: 3, 6, 7, 10, x + y + 4 + 19, 20, 25, 28. Since there are 8 (even) numbers, the median is the average of the 4th and 5th values. The 4th value is 10 and the 5th is x + y + 23. The median is (10 + x + y + 23) ÷ 2 = (33 + x + y) ÷ 2. If we solve for x and y based on the constraint that this equals 11, then x + y = 22 - 33 = -11, giving a median of 11.
In simple words: When values include variables, express the median in terms of those variables, or solve for them if you know what the median should be.

Exam Tip: Check the problem statement carefully to see if there are constraints that tell you what x and y must be.

 

Exercise 19.2

 

Question 1. Identify whether each of the following is discrete or continuous:
(i) Discrete
(ii) Continuous
(iii) Discrete
(iv) Continuous
(v) Continuous
Answer: The classification depends on whether the variable can take only specific, separate values (discrete) or any value within a range (continuous). Discrete variables include counts or categories. Continuous variables include measurements like height, weight, or time that can take any value within a range.
In simple words: If you can only count it in whole numbers (like people or cars), it's discrete. If you can measure it and get decimal values (like height or speed), it's continuous.

Exam Tip: Think: "Can this be broken into smaller and smaller pieces?" If yes, it's likely continuous.

 

Question 2. Given data: 13, 6, 10, 5, 11, 14, 2, 8, 15, 16, 9, 13, 14, 11, 9, 5, 12, 26, 21, 18, 11, 8, 12, 18. Construct a frequency distribution table with classes 0-4, 5-9, 10-14, 15-19, 20-24.
Answer: Count how many values fall into each class interval:

In simple words: Sort each number into the group it belongs to, then count how many numbers are in each group to make the frequency table.

Exam Tip: Always make sure the classes cover all possible values and do not overlap - each value should fit into exactly one class.

 

Question 3(i). Construct a frequency distribution table with classes 1-10, 11-20, 21-30, 31-40.
Answer: Group the data into the given class intervals and count the frequency for each:

In simple words: Put each number in its class, then count how many are in each class to fill the frequency row.

Exam Tip: Double-check by adding all frequencies - the total should equal the total count of data points.

 

Question 3(ii). Construct a frequency distribution table with classes 0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5, 30.5 - 40.5, 60.5 - 50.5.
Answer: Group the data into the given class intervals and count frequencies:

In simple words: Using decimal class limits helps prevent any ambiguity about where a boundary value belongs.

Exam Tip: When class limits use decimals, each whole number falls clearly into one class with no overlap.

 

Question 3(iii). Explain what Range, Class Size, Class Mark, Class Limits, True Class Limits, Frequency, and Cumulative Frequency mean.
Answer: Variable: A particular value of a variable is called a variable. Class Size: The difference between the actual upper limit and the actual lower limit of a class is called the class size. Class Mark: The class mark of a class is the value midway between its actual lower limit and actual upper limit. Class Limits: In a discrete distribution, the original class limits are called stated class limits. True Class Limits: In a continuous distribution, the class limits are called true class limits (or actual class limits). Frequency of a class: The number of times a variable occurs in a given data is called the frequency of that variable. Cumulative Frequency of a class: The sum of frequencies of all the previous classes and that particular class is called the cumulative frequency of the class.
In simple words: Range tells how far the data spreads. Size means the width of each class. Mark is the middle of a class. Limits are the boundaries. True limits account for rounding. Frequency counts how many are in each class. Cumulative frequency keeps a running total.

Exam Tip: These terms are foundational - learn them carefully because they are used throughout statistics problems.

 

Question 5(i). Identify the frequency, size, class mark, class limits, and cumulative frequency:
Answer:

(ii) Lower Limit: 20.5; Upper Limit: 30.5. (iii) Class Mark: (20.5 + 40.5) ÷ 2 = 35.5. (iv) Class Mark Calculation: (30.5 + 40.5) ÷ 2 = 35.5.
In simple words: From a frequency table, you can identify each piece of information by looking at the class intervals and their frequencies.

Exam Tip: Always show your work when calculating class marks and limits - partial credit is often given for method even if the final answer has a small error.

 

Question 6(i). Construct a frequency distribution table and find the lower limit and upper limit.
Answer:

(ii) Lower limit = 20.5; Upper limit = 30.5.
In simple words: From the frequency table, the lower and upper limits of each class tell you the exact boundaries for that group of data.

Exam Tip: Lower limit refers to the smallest value in a class, and upper limit refers to the largest value.

This PDF fragment contains only solution-working pages with no original question text, tables showing class intervals and frequencies, and bar/histogram diagrams. Per Rule G (SOLUTION-ONLY PDFs), I will reconstruct questions from the working shown, match them to their corresponding solutions, and format them as complete Question/Answer pairs. ---

 

Question 7. (i) The upper limit is 52 and the lower limit is 48. (ii) The upper limit is 58.5 and the lower limit is 49.5. (iii) The class size is between 37.5 and 48.5. (iv) Find the class size. (v) Find the class mark.
Answer: (i) Upper limit = 52, lower limit = 48. (ii) Upper limit = 58.5, lower limit = 49.5. (iii) Class size is 37.5 to 48.5. (iv) Class size = 11. (v) Class mark = 5.
In simple words: The upper and lower limits tell you the range of each class. The class size shows how wide each group is. The class mark is the middle point of a class.

Exam Tip: Always identify the upper limit, lower limit, and class size correctly - these are fundamental for constructing frequency distribution tables.

 

Question 8. From the given frequency distribution table with classes 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100 and their frequencies 3, 2, 11, 18, 0, 3, 2, 3, 5, 3, find (i) the class that has the highest frequency, and (ii) the number of students who score less than 40 marks.
Answer: From the tally marks and frequency table shown: (i) The class with the highest frequency is 30-40, which has a frequency of 18. (ii) Students scoring less than 40 marks come from classes 0-10, 10-20, 20-30, and 30-40. Adding their frequencies: 3 + 2 + 11 + 18 = 34. However, we need only those strictly below 40, which gives us the sum of classes up to 30-40. The number of students scoring less than 40 marks is 34.
In simple words: Look at the frequency column to find the highest number. Then add up all the frequencies for classes below 40 to find how many students scored less than 40.

Exam Tip: When finding frequencies in a specific range, carefully identify which classes fall within that range and add their individual frequencies - do not include classes outside the range.

 

Question 9. From the frequency distribution showing classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 5, 5, 7, 10, 8, find the number of students obtaining marks below 20.
Answer: The students obtaining marks below 20 fall into classes 0-10 and 10-20. From the table, the frequency for class 0-10 is 5 and for class 10-20 is 5. Therefore, the total number of students scoring below 20 marks = 5 + 5 = 10.
In simple words: Add the frequencies of all classes that are below 20 marks to get the total count of students in that range.

Exam Tip: Always be clear about whether a boundary is included or excluded - "below 20" means up to but not including 20, so include only those classes with upper limits less than or equal to 20.

 

Question 10. From the frequency distribution table with classes 0-4, 4-7, 7-10, 10-13, 13-16 and frequencies 7, 31, 13, 73, 52, find the number of children in the age group 10 to 13.
Answer: From the given frequency distribution table, the class 10-13 has a frequency of 73. Therefore, the number of children in the age group 10 to 13 is 73.
In simple words: Look at the row for class 10-13 and read the frequency value directly from the table.

Exam Tip: When a specific class is mentioned, locate that exact class interval in the table and note its corresponding frequency without any additional calculation.

 

Question 11. From the frequency distribution with classes 0-10, 11-20, 21-30, 31-40, 41-50, 51-60 and frequencies 2, 5, 11, 14, 11, 7, find the class size.
Answer: (i) Class size = 13.4 - 12.4 = 10. Lower limit = 11.9, upper limit = 12.9. (ii) Class size = 30.5 - 23.5 = 7. Lower limit = 20, upper limit = 27.
In simple words: The class size is found by subtracting the lower limit from the upper limit of any class interval. Each class shows a specific width or range.

Exam Tip: Class size should be consistent across all classes in a frequency distribution table - if you find different sizes, check your class boundaries carefully.

 

Question 12. From the frequency distribution with classes 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100, 100-110 and frequencies 2, 1, 4, 4, 9, 5, 3, 4, prepare a frequency distribution table and interpret it.
Answer:

In simple words: The class 70-80 has the most observations (9). The lowest frequencies are in classes 40-50 (1). This shows that most data falls in the middle ranges, with fewer items at the lower and upper extremes.

Exam Tip: When interpreting a frequency table, identify the modal class (highest frequency), note the distribution pattern, and comment on whether data is spread evenly or concentrated in certain ranges.

 

Question 13. From the frequency distribution with classes 23.5-31.5, 27.5-31.5, 31.5-35.5, 35.5-39.5, 39.5-42.5 and frequencies 4, 7, 4, 4, 11, prepare the corresponding continuous frequency distribution table.
Answer:

In simple words: This table shows how data is grouped into classes with decimal boundaries. Each class has a frequency showing how many items fall within that range.

Exam Tip: For continuous frequency distributions with decimal class boundaries, ensure all class intervals are of equal width and that there are no gaps or overlaps between consecutive classes.

 

Question 1. (Exercise 19.3) Draw a bar graph for the following data showing the number of items sold by category: Bulbs 220, Housing 120, U.P. 100, M.P. 40, Mahamat 80, Rajasthan 30. Use a scale of 1 cm = 30 units on the y-axis.
Answer: The bar graph is drawn below showing the frequency of items sold across different categories. Using the scale 1 cm = 30 units on the y-axis and labeling the categories on the x-axis, each category is represented by a vertical bar whose height corresponds to its frequency value. Bulbs has the tallest bar at 220, followed by Housing at 120, U.P. at 100, Mahamat at 80, M.P. at 40, and Rajasthan at 30.
In simple words: A bar graph uses tall and short bars to show how much of something there is. Each bar's height tells you the number for that group. Higher bars mean more items were sold.

Exam Tip: Always label axes clearly, use a consistent scale, and ensure bars have equal width and spacing for a professional-looking bar graph.

 

Question 2. (Exercise 19.3) Draw a bar graph for the following data showing the frequency of days of the week: Monday 420, Tuesday 180, Wednesday 230, Thursday 240, Friday 160, Saturday 120. Use a scale where 1 unit on the y-axis = 50 units.
Answer: The bar graph is constructed with days of the week on the x-axis and frequency on the y-axis, using the scale 1 unit = 50 units. Each day is represented by a vertical bar with height proportional to its frequency. Monday shows the highest frequency at 420, while Saturday has the lowest at 120, with other days distributed in between.
In simple words: This bar graph shows how busy each day is during the week. Some days have more activity than others, and the height of each bar shows how much activity happened on that day.

Exam Tip: When drawing bar graphs for time periods like days of the week, maintain the natural order on the x-axis and ensure your scale is easy to read and proportional.

 

Question 3. (Exercise 19.3) Draw a bar graph showing the variation of items produced across different years: 2000 produces 92 items, 2001 produces 80 items, 2002 produces 70 items, 2003 produces 54 items, 2004 produces 86 items, 2005 produces 75 items, 2006 produces 94 items. Use a scale of 1 cm = 1 unit on x-axis and 1 cm = 10 units on y-axis.
Answer: A bar graph is drawn with years on the x-axis and production quantity on the y-axis. Using the specified scale (1 cm = 1 unit on x-axis, 1 cm = 10 units on y-axis), each year is represented by a vertical bar. Year 2000 shows 92 items (highest along with 2006 at 94), while 2003 shows the lowest production at 54 items, demonstrating fluctuating production levels across the seven-year period.
In simple words: This graph shows how production changed from year to year. In 2000 and 2006, production was highest. In 2003, production was at its lowest point.

Exam Tip: When showing trends over time, always arrange years in chronological order on the x-axis, and use a consistent scale to make year-to-year comparisons visually clear.

 

Question 4. (Exercise 19.3) Draw a horizontal bar graph for the following data showing different sources of income: India 26, Publish 36, Cinema 12, U.S.A. 18, Home 20. Use a scale where 1 unit on y-axis = 5 units.
Answer: A horizontal bar graph is constructed with income sources on the y-axis and income amount on the x-axis, using the scale 1 unit = 5 units. Each income source is represented by a horizontal bar extending to the right from the y-axis. Publish shows the highest income at 36, while Cinema shows the lowest at 12, with India, U.S.A., and Home having intermediate values.
In simple words: A horizontal bar graph spreads bars from left to right instead of bottom to top. This makes it easy to read and compare category names on the left side.

Exam Tip: Horizontal bar graphs work well when you have many categories or when category names are long - the layout makes labels easier to read.

 

Question 5. (Exercise 19.3) Draw a grouped bar graph for the following data showing the number of boys and girls in classes IX-A, IX-B, IX-C, and IX-D: IX-A has 16 girls and 28 boys, IX-B has 24 girls and 22 boys, IX-C has 12 girls and 40 boys, IX-D has 25 girls and 15 boys. Use a scale of 1 cm = 10 units on the y-axis.
Answer: A grouped bar graph is drawn with class sections on the x-axis and student count on the y-axis. For each class (IX-A, IX-B, IX-C, IX-D), two bars are drawn side by side - one for girls and one for boys. Using the scale 1 cm = 10 units, the graph clearly shows the gender distribution across all four classes, with IX-C having the most boys (40) and IX-A having the fewest girls (16).
In simple words: In this graph, each class has two bars next to each other - one dark bar for girls and one light bar for boys. You can easily see which classes have more girls and which have more boys.

Exam Tip: For grouped bar graphs, use different colors or shading for each category and include a legend explaining what each color/shade represents. Keep bars of the same group close together for easy comparison.

 

Question 6. (Exercise 19.3) Draw a histogram for the following frequency distribution with classes 0-20, 20-30, 30-40, 40-50, 50-60 and frequencies 4, 6, 8, 5, 1. Use a scale of 10 cm on x-axis = 1 unit and 1 cm on y-axis = 1 unit.
Answer: A histogram is drawn with class intervals on the x-axis and frequency on the y-axis. Unlike a bar graph, the bars in a histogram are adjacent with no gaps, showing continuous data. The heights of the bars represent the frequency of each class: 4 for class 0-20, 6 for class 20-30, 8 for class 30-40 (tallest), 5 for class 40-50, and 1 for class 50-60 (shortest).
In simple words: A histogram shows how data is spread across different ranges. The bars touch each other with no gaps, and the height of each bar tells you how many items fell into that range.

Exam Tip: In a histogram, bars must be adjacent with no spaces between them, and the area of each bar represents frequency, not just the height - this is crucial when class widths are unequal.

 

Question 7. (Exercise 19.3) Draw a histogram for the frequency distribution with classes 850-950, 950-1050, 1050-1150, 1150-1250, 1250-1350 and frequencies 35, 45, 60, 40, 25. Use a scale where 1 cm on x-axis = 100 units and 1 cm on y-axis = 5 units.
Answer: A histogram is constructed with class intervals on the x-axis and frequency on the y-axis using the specified scales. The adjacent bars represent each class interval without gaps between them. Class 1050-1150 shows the highest frequency at 60, while class 1250-1350 shows the lowest at 25, illustrating the distribution of data across the five classes.
In simple words: This histogram displays how items are distributed across five price ranges. Most items fall in the 1050-1150 range, while fewer items are found in the highest price range.

Exam Tip: When drawing histograms with large class intervals, maintain consistent spacing on the x-axis and ensure the scale clearly represents the interval boundaries.

 

Question 8. (Exercise 19.3) Draw a histogram for the frequency distribution with classes 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, 40-45 and frequencies 7, 12, 20, 28, 8, 11 and use a scale of 5 cm on y-axis = 1 unit.
Answer: A histogram is drawn with seven class intervals on the x-axis and corresponding frequencies represented by the heights of adjacent bars on the y-axis. Using the scale 5 cm = 1 unit, class 25-30 displays the highest frequency of 28, class 20-25 shows frequency 20, class 15-20 shows frequency 12, and the remaining classes have lower frequencies. The bars touch each other without gaps, characteristic of a histogram showing continuous data distribution.
In simple words: This histogram shows how data spreads across seven different ranges. The tallest bar represents the most common range (25-30), while shorter bars show ranges with fewer observations.

Exam Tip: Always label class intervals clearly on the x-axis and frequencies on the y-axis, and ensure bars have equal width for classes of equal size.

 

Question 9. (Exercise 19.3) Draw a histogram for the frequency distribution with classes 2, 11, 6, 6, 12, 11 with frequencies 12, 15, 36, 45, 72, (incomplete data) and scale 2 cm on x-axis = 1 unit and 5 cm on y-axis = 1 unit.
Answer: Based on the provided data, a histogram is drawn with class intervals on the x-axis and frequencies on the y-axis using the specified scale (2 cm = 1 unit on x-axis, 5 cm = 1 unit on y-axis). The bars are adjacent without gaps. From the visible frequencies, the histogram shows an increasing trend in the earlier classes (12, 15, 36, 45, 72), suggesting the data is concentrated in higher frequency classes.
In simple words: This histogram shows frequencies that increase dramatically across the class intervals. The last class has by far the highest frequency, indicating a strong concentration of data at the higher end.

Exam Tip: Histogram bars must always be adjacent and touching to show the continuous nature of the data - gaps between bars suggest separate categories rather than a continuous distribution.

 

Question 10. (Exercise 19.3) For a continuous frequency distribution with class intervals 59-65, 66-72, 73-79, 80-86, 87-93, 94-100 and frequencies 10, 5, 25, 15, 30, 10, convert the discrete classes into continuous classes and draw the resulting histogram.
Answer: To convert to continuous distribution, the adjustment factor is calculated as (lower limit of one class - upper limit of previous class) / 2 = (66 - 65) / 2 = 0.5. The adjusted class boundaries become: 59.5-65.5, 65.5-72.5, 72.5-79.5, 79.5-86.5, 86.5-93.5, 93.5-100.5, with corresponding frequencies 10, 5, 25, 15, 30, 10. The histogram shows adjacent bars with class 87-93 (adjusted to 86.5-93.5) having the highest frequency of 30.

In simple words: When converting discrete classes to continuous, adjust the boundaries by adding 0.5 to upper limits and subtracting 0.5 from lower limits. This removes gaps between classes, making the histogram bars touch.

Exam Tip: The adjustment factor for converting discrete to continuous frequency distribution is half the gap between classes; always calculate this carefully to ensure correct histogram construction.

 

Question 11. Draw a frequency polygon for the following data using the given scale.
Scale on x axis unit = 6cm
Scale on y axis unit = 5cm
Answer: A frequency polygon is constructed by plotting class midpoints on the x-axis and their corresponding frequencies on the y-axis, then connecting these points with straight line segments. The polygon starts and ends on the x-axis at the midpoints of classes with zero frequency just before the first class and just after the last class.
In simple words: Plot the middle of each class interval against its frequency, then join all the points with straight lines to make a polygon.

Exam Tip: Always extend the polygon to meet the x-axis on both ends by adding imaginary classes with zero frequency - this is what makes it a closed polygon and earns full marks.

 

Question 12. Draw the required frequency polygon for the data given below.
Scale on x axis unit = 10cm
Scale on y axis unit = 5cm
Answer: A frequency polygon can be drawn by joining the midpoints of the tops of the rectangles in a histogram. Alternatively, plot each class midpoint with its frequency as a point on the graph, and connect all consecutive points with straight lines. Start and finish the polygon on the x-axis at the midpoints of the zero-frequency classes on either end.
In simple words: Take the middle value of each class and mark it at the correct height for its frequency, then join all the points with lines.

Exam Tip: Ensure that all points lie accurately on the plotted positions and that the polygon closes properly by extending to zero on both sides.

 

Question 13. Draw the required frequency polygon for the data shown below.
Scale on x axis unit = 10cm
Scale on y axis unit = 5cm
Answer: To construct a frequency polygon, identify the midpoint of each class interval by finding the average of the lower and upper class boundaries. Mark these midpoints on the x-axis with their corresponding frequencies on the y-axis. Draw straight lines joining all successive points. The polygon should begin and end at the x-axis where the frequency is zero, at the midpoints of the classes that would come before the first class and after the last class.
In simple words: Find the middle of each class, plot that against the frequency count, and connect the dots with straight lines from start to finish.

Exam Tip: Double-check that all midpoints are calculated correctly as (lower limit + upper limit) ÷ 2, and that the graph is drawn to the exact scale given.

 

Question 14. The given frequency distribution is discontinuous. Convert it into a continuous distribution.
Answer: When a frequency distribution has gaps between classes (discontinuous), an adjustment factor must be applied. This factor equals half of the difference between the lower limit of one class and the upper limit of the preceding class. For the given data, the adjustment factor is calculated as (20 - 20) ÷ 2 = 0.5.

After applying this adjustment:

In simple words: A discontinuous distribution has spaces between class limits. To make it continuous, subtract and add 0.5 to each class's lower and upper limits so the classes touch without overlapping.

Exam Tip: The adjustment factor must be applied consistently to all classes, and the resulting class intervals should have no gaps and no overlap between consecutive classes.

 

Question 15. Draw a frequency polygon using the given scale.
Scale on x axis unit = 4cm
Scale on y axis unit = 6cm
Answer: A frequency polygon is constructed by locating the class midpoint on the x-axis and marking the frequency on the y-axis for each class. All these points are then connected by straight line segments to form the polygon. The diagram should start and end on the x-axis at positions corresponding to zero frequencies, placed at the midpoints of the classes that would theoretically exist before the first actual class and after the last actual class.
In simple words: Plot each class's middle point against its frequency count and join them with lines to show how the frequency changes across classes.

Exam Tip: Always use a ruler for accuracy when drawing the connecting lines, and verify that the scale factors are applied correctly to both axes.

 

Question 16. Draw the required frequency polygon using the given scale.
Scale on x axis unit = 10cm
Scale on y axis unit = 5cm
Answer: To draw a frequency polygon, compute the midpoint of each class interval by averaging its lower and upper limits. Plot these midpoints on the x-axis and mark their corresponding frequencies on the y-axis. Connect all plotted points with consecutive straight lines. The polygon must extend to the x-axis on both sides by including zero-frequency classes at the beginning and end, creating a closed figure.
In simple words: Find the center of each class, mark it on the graph at the right height for its frequency, then draw straight lines connecting all these marks.

Exam Tip: Check that your polygon is symmetric or matches the shape expected from the given data, and ensure all points are plotted precisely according to the stated scale.

 

Question 17. Draw the required histogram and frequency polygon using the given scale.
Scale on x axis unit = 15cm
Scale on y axis unit = 5cm
Answer: A histogram is drawn using rectangles with class intervals on the x-axis and frequencies as heights. A frequency polygon is then overlaid by connecting the midpoints of the tops of each rectangle with straight lines. The combined diagram shows both the bar representation (histogram) and the line graph (frequency polygon) on the same axes. This dual representation helps visualize the data distribution in two complementary ways.
In simple words: Draw the histogram bars first, then connect the middle-top point of each bar with lines to form the frequency polygon on top of it.

Exam Tip: When drawing both histograms and frequency polygons together, ensure the polygon's points align exactly with the midpoints of the tops of the histogram bars for consistency.

 

Question 18. Given water bills of 38 houses in a locality are:
30, 48, 52, 78, 103, 85, 37, 94, 79, 73, 66, 52, 92, 65, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 65, 100, 103, 35, 89, 95, 43.

A frequency table has been prepared from this data. Draw the required histogram and frequency polygon using the given scale.
Scale on x axis unit = 10cm
Scale on y axis unit = 1cm
Answer: First, organize the data into class intervals (30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100, 100-110) and count the frequency in each interval from the given data. The frequency table shows: 30-40 has 3 houses, 40-50 has 1 house, 50-60 has 4 houses, 60-70 has 5 houses, 70-80 has 9 houses, 80-90 has 3 houses, 90-100 has 3 houses, 100-110 has 3 houses. Draw the histogram with these frequencies as bar heights. Then construct the frequency polygon by plotting each class midpoint against its frequency and connecting the points with straight lines, extending to zero frequency on both ends.
In simple words: Sort the water bills into groups, count how many fall in each group, draw bars for those counts, then connect the tops of the bars with lines.

Exam Tip: When counting frequencies from raw data, check off each value as you count to avoid duplicates or omissions - use the scale factors precisely to ensure the graph is drawn accurately.

 

Question 19. A frequency table for water bills is given. Draw the required histogram and frequency polygon using the given scale.
Scale on x axis unit = 10cm
Scale on y axis unit = 1cm
Answer: Using the provided frequency table with class intervals 40-42, 42-44, 44-46, 46-48, 48-50, 50-52, 52-54 and their corresponding frequencies 2, 13, 7, 6, 7, 4, 2, construct the histogram by drawing rectangles with each class interval's width on the x-axis and its frequency as the height on the y-axis. Then superimpose the frequency polygon by plotting the midpoint of each class interval (41, 43, 45, 47, 49, 51, 53) at the height of its frequency and connecting these points with straight line segments. Extend the polygon to the x-axis at zero frequency positions on both sides to complete the closed figure.
In simple words: Draw bars showing how often each water bill amount appears, then draw a line through the tops of these bars, connecting them from start to finish.

Exam Tip: Make sure the polygon points are placed exactly at the midpoint of each class interval, not at the ends, and that the scaling is applied uniformly to create a properly proportioned graph.

 

Question 20. A frequency distribution table is given. Draw the required histogram and frequency polygon using the given scale.
Scale on x axis unit = 20cm
Scale on y axis unit = 1cm
Answer: From the given frequency distribution with class intervals 40-42, 42-44, 44-46, 46-48, 48-50, 50-52, 52-54 and frequencies 2, 13, 7, 6, 7, 4, 2, construct the histogram by drawing adjacent rectangles where each class interval spans the x-axis and the frequency forms the height. Draw the frequency polygon by identifying the midpoint of each rectangle's top edge and connecting all these midpoints with straight line segments. The polygon should begin and end on the x-axis at the positions of zero-frequency classes on either side, creating a closed polygon shape that represents the overall distribution pattern.
In simple words: Plot bars for each class showing its frequency, then join the top middle of each bar with lines to form a connected polygon shape.

Exam Tip: Verify that all class intervals are equal in width and that the frequency values are marked accurately on the y-axis according to the stated scale - these are common areas for losing marks.

 

Question 21. For the following frequency distribution, find:
(i) The median class
(ii) The sum of all frequencies
(iii) The range
(iv) Construct a cumulative frequency table
Answer:
(i) The median class is 425 - 450 (the class containing the middle value when frequencies are arranged in order).

(ii) The sum of all frequencies is 58 (total of: 6 + 18 + 10 + 20 + 4 = 58).

(iii) The range is 475 - 500 = 475 (the difference between the upper limit of the highest class and the lower limit of the lowest class).

(iv) Cumulative Frequency Table:

In simple words: The median class is where the total count reaches halfway. The sum adds all frequencies together. The range shows how far the data spreads. The cumulative frequency runs a growing total from the first class to the last.

Exam Tip: When finding the median class, calculate half of the total frequency first, then locate which class interval contains that cumulative total - this is where the median falls.