RS Aggarwal Class 9 Mathematics Solutions Chapter 14 Statistics

 

Question 1. What is statistics?
Answer: Statistics is a branch of science that deals with gathering, presenting, analyzing, and drawing conclusions from numerical data.
In simple words: Statistics helps us collect numbers, organize them, and understand what they tell us about the world around us.

Exam Tip: Always remember the four key functions: collection, presentation, analysis, and interpretation. These are the foundation of any statistics definition.

 

Question 2. What are the fundamental characteristics of statistics?
Answer:
(i) It works exclusively with information that can be expressed as numbers.
(ii) Qualities that seem non-numerical - such as lack of education, mental ability, or economic hardship - can be changed into numbers and studied through statistics.
(iii) The conclusions drawn from statistical analysis are never completely precise or certain; they always involve some degree of uncertainty.
In simple words: Statistics needs numbers, it can turn qualities into numbers, and its answers are never 100 percent certain.

Exam Tip: The three characteristics form the foundation of statistical analysis. Be ready to explain why qualitative data must be converted to numerical form and why statistical inferences carry an element of uncertainty.

 

Question 3. Distinguish between primary data and secondary data.
Answer: Primary data refers to information collected directly by the researcher through their own systematic investigation. Since the researcher gathers this data themselves, it tends to be highly accurate and trustworthy, and they can ensure it matches their research goals precisely.

Secondary data, on the other hand, is information that another person or organization has already gathered. Since it was not collected by the current researcher, it may lack the same level of reliability. Additionally, because it was gathered for a different purpose, it might not be fully suited to the current investigation.
In simple words: Primary data is what you collect yourself - it is fresh and reliable. Secondary data is what others have collected - it may not fit your needs perfectly and could be less trustworthy.

Exam Tip: Use real-world examples to illustrate the difference: a government census is primary data for census officials but secondary data for a researcher using it later. Always highlight accuracy and relevance as key distinguishing factors.

 

Question 4. Define the following terms: (i) Variate (ii) Class Interval (iii) Class-Size (iv) Class-mark (v) Class limit (vi) True class limits (vii) Frequency of a class (viii) Cumulative frequency of a class
Answer:
(i) Variate: Any measurable property that can take on different numerical values is called a variate. For example, height, weight, or age are all variates because they differ from person to person.

(ii) Class Interval: A class interval is a range or division into which numerical data is grouped. All values that fall within that range belong to that single class interval.

(iii) Class-Size: The class-size is the difference between the true upper boundary and the true lower boundary of any class interval. It tells us the width of the interval.

(iv) Class-mark: The class-mark is the midpoint of a class interval - that is, the average of the upper limit and lower limit. It is calculated as: Class mark = \( \frac{\text{upper limit} + \text{lower limit}}{2} \)

(v) Class limit: Class limits are the two boundary values that define a class interval. The boundary on the left side is known as the lower limit, while the boundary on the right side is known as the upper limit.

(vi) True class limits: In exclusive form classification, the stated upper and lower limits are the actual true limits. In inclusive form classification, the true lower limit is found by subtracting 0.5 from the lower limit, and the true upper limit is found by adding 0.5 to the upper limit.

(vii) Frequency of a class: The frequency of a class is the count of how many observations or data points fall within that particular class interval.

(viii) Cumulative frequency of a class: The cumulative frequency of a class is the total count of all observations in that class plus all observations in every class before it.
In simple words: A variate is something that changes from person to person. A class interval bundles data into groups. Class-size is how wide the group is. Class-mark is the middle point. Class limits are the boundaries. True limits adjust for inclusive form. Frequency counts how many are in each group. Cumulative frequency adds up all the counts from the beginning to that group.

Exam Tip: Always provide both the definition and a numerical example for each term. Pay special attention to the difference between class limits and true class limits - this is commonly tested in exams.

 

Question 5. The minimum observation is 0 and the maximum observation is 6. Form the frequency distribution table with classes of equal size covering the given data.
Answer: Since the minimum value is 0 and the maximum value is 6, we set up class intervals of equal width covering this range. Using intervals of width 2, the classes are: (0-2), (2-4), (4-6), and (6-8).

The frequency distribution table will have columns for Class Interval, Tally Marks, and Frequency, with each row representing one class and showing the count of observations within that range.
In simple words: Divide the range from 0 to 6 into equal-width classes. Each class groups a set of numbers, and we count how many observations fall in each class.

Exam Tip: Always ensure class intervals are of equal size and cover the entire range from minimum to maximum. The choice of class width affects the clarity of your frequency distribution.

 

Question 6. The minimum observation is 1 and the maximum observation is 24. Form the frequency distribution table with classes of equal size covering the given data.
Answer: Since the minimum observation is 1 and the maximum observation is 24, we form classes of equal width to cover this entire range. Using intervals of width 5, the class intervals are: (0-5), (5-10), (10-15), (15-20), and (20-25).

The frequency distribution table displays the class intervals in the first column, tally marks in the second column, and the frequency (count) in the third column, with each row representing one class group.
In simple words: Create equal-width classes starting from 0 and going up to 25 to cover all observations from 1 to 24. Count how many observations fall into each class and record the totals.

Exam Tip: Ensure that all observations fit within the class intervals without overlap. A clear, organized table makes it easy to see the distribution pattern of your data.

 

Question 7. The minimum observation is 6 and the maximum observation is 23. Form the frequency distribution table with classes of equal size covering the given data.
Answer: Since the minimum observation is 6 and the maximum observation is 23, the range spans 17 units. Using intervals of width 3, the class intervals are: (6-9), (9-12), (12-15), (15-18), (18-21), and (21-24).

The frequency distribution table presents the class intervals, tally marks, and frequencies for each class, allowing us to see how observations are distributed across the different ranges.
In simple words: Organize the data from 6 to 23 into equal-width classes. Count the number of observations in each class and present them in a clear table format.

Exam Tip: When selecting class width, aim for classes that are neither too narrow (which creates many classes) nor too wide (which loses detail). A good balance improves data interpretation.

 

Question 8. The minimum observation is 210 and the maximum observation is 320. Form the frequency distribution table with classes of equal size covering the given data.
Answer: The minimum observation is 210 and the maximum observation is 320, giving a range of 110. Using intervals of width 20, the class intervals are: (210-230), (230-250), (250-270), (270-290), (290-310), and (310-330).

The frequency distribution table lists these class intervals and records the frequency (number of observations) for each class through a systematic count of all data points.
In simple words: Divide the range from 210 to 320 into equal-width classes. For each class, count how many observations fall within its boundaries and display the results in a table.

Exam Tip: Large data ranges often benefit from larger class widths. Always verify that your class intervals cover the complete range and that frequencies sum to the total number of observations.

 

Question 9. The minimum observation is 30 and the maximum observation is 110. Form the frequency distribution table with cumulative frequencies for the given data.
Answer: The minimum observation is 30 and the maximum observation is 110, giving a range of 80. Using intervals of width 10, the class intervals are: (30-40), (40-50), (50-60), (60-70), (70-80), (80-90), (90-100), (100-110), and (110-120).

The frequency distribution table shows each class interval with its corresponding frequency. The cumulative frequency column is built by adding each class frequency to all the frequencies of the classes before it. This creates a running total that shows how many observations are at or below the upper boundary of each class.
In simple words: Organize data from 30 to 110 into equal-width classes. For each class, record the count. Then add each count to all previous counts to get the cumulative frequency, which shows the total up to each point.

Exam Tip: The cumulative frequency of the last class should equal the total number of observations - use this as a check on your work. Cumulative frequencies are crucial for finding quartiles and medians.

 

Question 10. The minimum observation is 804 and the maximum observation is 898. Form the frequency distribution table for the given data.
Answer: The minimum observation is 804 and the maximum observation is 898, giving a range of 94. Using intervals of width 10, the class intervals are: (800-810), (810-820), (820-830), (830-840), (840-850), (850-860), (860-870), (870-880), (880-890), and (890-900).

The frequency distribution table lists each class interval and records the count of observations falling within each interval based on a careful tally of all data points.
In simple words: Divide 804 to 898 into equal-width classes of 10. Count how many observations belong to each class and present the results in a table.

Exam Tip: When working with large numbers, using round class boundaries (like multiples of 10) simplifies the construction and reading of frequency tables.

 

Question 11. The minimum observation is 52 and the maximum observation is 130. Form the frequency distribution table for the given data.
Answer: The minimum observation is 52 and the maximum observation is 130, creating a range of 78. Using intervals of width 10, the class intervals are: (50-60), (60-70), (70-80), (80-90), (90-100), (100-110), (110-120), (120-130), and (130-140).

The frequency distribution table presents each class interval with the corresponding frequency, which represents the number of observations counted in each class range through a systematic grouping of the data.
In simple words: Group the observations from 52 to 130 into classes of width 10. Count how many values belong to each class and organize them in a clear table.

Exam Tip: Always choose starting and ending class boundaries that comfortably contain your minimum and maximum values. This makes the frequency distribution easier to read and more professional.

 

Question 12. Complete the frequency distribution table by filling in the missing cumulative frequency values.

Answer: To find the cumulative frequency for each class, we add the current class frequency to the cumulative frequency of the previous class. Starting with 90 for the first class, we add 50 to get 140 for the second class, then add 60 to get 200, then add 80 to get 280, then add 50 to get 330, and finally add 30 to get 360. The last cumulative frequency (360) matches the total number of patients, confirming our calculations are correct.
In simple words: Start with the first frequency as the first cumulative value. Then keep adding the next frequency to the running total. Each cumulative frequency is the sum of all frequencies from the beginning up to that class.

Exam Tip: Always verify that your final cumulative frequency equals the total frequency count. This check ensures you have not made an arithmetic error in building the cumulative frequency column.

 

Question 13. Convert the cumulative frequency distribution into a regular frequency distribution.

Answer: To convert cumulative frequency to regular frequency, we subtract each cumulative value from the next higher cumulative value. For the class 0-10, the frequency is 5 (since this is the first class). For 10-20, the frequency is 12 - 5 = 7. For 20-30, it is 32 - 12 = 20. For 30-40, it is 40 - 32 = 8. The regular frequency for each class represents the count of students who scored within that specific range.
In simple words: To get the regular frequency from cumulative frequency, subtract the previous cumulative total from the current one. The difference gives you how many observations are in that particular class.

Exam Tip: Always remember that the first frequency equals the first cumulative frequency (there is nothing before it to subtract from). This conversion is essential when you need to analyze data in standard frequency distribution format.

 

Question 14. Convert the cumulative frequency distribution into a regular frequency distribution.

Answer: To convert cumulative frequency to regular frequency, subtract the previous cumulative value from the current cumulative value. For the 0-10 class, the frequency is 17. For 10-20, it is 22 - 17 = 5. For 20-30, it is 29 - 22 = 7. For 30-40, it is 37 - 29 = 8. For 40-50, it is 50 - 37 = 13. For 50-60, it is 60 - 50 = 10. Each result shows the actual count of students in that marks range.
In simple words: Find how many observations fall in each class by taking the difference between consecutive cumulative frequencies. This difference tells you exactly how many students scored in each range.

Exam Tip: As a verification step, add all the regular frequencies together - they should sum to the final cumulative frequency (60 in this case). This confirms that your conversion is accurate.

 

Question 15. Convert the "more than" cumulative frequency distribution into a regular frequency distribution.

Answer: A "more than" cumulative frequency table lists observations that exceed each boundary. To convert it to regular frequency, subtract consecutive values working downward. The frequency for 50-60 is 16 - 0 = 16. For 40-50, it is 40 - 16 = 24. For 30-40, it is 75 - 40 = 35. For 20-30, it is 87 - 75 = 12. For 10-20, it is 92 - 87 = 5. For 0-10, it is 100 - 92 = 8. Each result represents the actual count of students in that marks interval.
In simple words: When you have a "more than" table, find the difference between consecutive cumulative values. Each difference gives you how many observations fall within that specific class range.

Exam Tip: Always work systematically from top to bottom when converting "more than" cumulative frequency. The total of all regular frequencies should equal the largest cumulative frequency value (100 in this case).

 

Exercise 14B

 

Question 1. Draw a bar chart for the following data showing different types of games and number of students.
Answer: Along the x-axis, place the various types of games. Along the y-axis, place the number of students. Using the scale 1 small square = 3 units on the y-axis, construct rectangular bars of equal width with uniform spacing between them. Each bar's height corresponds to the number of students who play that particular game. This visual representation makes it easy to compare participation across different sports at a glance.
In simple words: Mark games on the horizontal axis and student count on the vertical axis. Draw bars of different heights to show how many students play each game. Equal spacing and equal widths make the chart clear and professional.

Exam Tip: Always include a scale on the y-axis and label both axes clearly. The bar chart becomes much more readable when you maintain uniform bar width and consistent spacing.

 

Question 2. Draw a bar chart for the following data showing times of day and corresponding temperatures.
Answer: Mark the times of day along the x-axis and the temperatures along the y-axis. Using a scale of 1 small square = 5 units on the y-axis, draw rectangular bars for each time point. Each bar's height represents the temperature recorded at that particular time. All bars should have equal width with uniform spacing maintained between consecutive bars to ensure a professional appearance.
In simple words: Place different times across the bottom. Show temperatures going up the side. For each time, draw a bar that is as tall as the temperature at that time. Keep all bars the same width and equally spaced.

Exam Tip: Temperature data often shows a trend - ensure your bar chart makes this trend visible. A clear scale helps readers understand temperature variations throughout the day.

 

Question 3. Draw a bar chart for the following data showing modes of transport and their velocities.
Answer: Position the different modes of transport along the x-axis and their corresponding velocities along the y-axis. Using a scale of 1 small square = 10 units on the y-axis, construct rectangular bars of equal width with consistent spacing. Each bar's height directly represents the velocity of that particular mode of transport, making speed comparisons immediate and visual.
In simple words: Put different types of vehicles on the bottom. Show velocity going up the side. Draw a bar for each vehicle as tall as its speed. Use equal bar widths and spacing throughout.

Exam Tip: Transportation data typically shows a clear hierarchy of speeds. Your bar chart should make these differences obvious - faster modes should have noticeably taller bars.

 

Question 4. Draw a bar chart for the following data showing different types of sports and number of students participating.
Answer: Along the x-axis, list the various sports. Along the y-axis, mark the number of students. Apply a scale of 1 small square = 10 units on the y-axis. Create rectangular bars for each sport, all of the same width, with equal spacing between consecutive bars. The height of each bar shows how many students participate in that particular sport.
In simple words: Show different sports on the horizontal line. Show student numbers on the vertical line. For each sport, draw a bar as high as the number of students playing it. Keep bars equally wide and evenly spaced.

Exam Tip: Sports participation data helps identify popular activities. Make sure your scale allows all bars to fit comfortably and that differences in participation are clearly visible.

 

Question 5. Draw a bar chart for the following data showing academic years and number of students enrolled.
Answer: Position the academic years along the x-axis and the number of students along the y-axis. Use a scale of 1 large division = 200 units on the y-axis. Draw rectangular bars of equal width with uniform spacing between them. Each bar's height reflects the student enrollment for that specific academic year, revealing enrollment trends over time.
In simple words: List years on the horizontal axis. List student numbers on the vertical axis. For each year, draw a bar showing how many students enrolled. Use consistent bar widths and spacing to make the trend easy to see.

Exam Tip: Enrollment data often shows growth or decline trends. A well-constructed bar chart makes these trends immediately apparent to readers.

 

Question 6. Draw a bar chart for the following data showing years and number of scooters sold.
Answer: Mark the years along the x-axis and the number of scooters along the y-axis. Use a scale of 1 large division = 5000 units on the y-axis. Create bars of equal width with consistent spacing between them for each year. The height of each bar shows the number of scooters sold in that particular year, making year-to-year comparisons straightforward.
In simple words: Put years on the bottom. Put scooter numbers on the side. For each year, draw a bar showing scooter sales. Keep all bars the same width with equal spacing between them.

Exam Tip: When dealing with large numbers like scooter sales, choose a scale that prevents bars from becoming too tall or too short. The 5000-unit scale ensures visibility and clarity.

 

Question 7. Draw a bar chart for the following data showing countries and their birth rates per thousand population.
Answer: Position countries along the x-axis and birth rates along the y-axis. Apply a scale of 1 large division = 5 units on the y-axis. Draw rectangular bars of equal width with uniform spacing for each country. Each bar's height represents that nation's birth rate per thousand, enabling easy comparison of demographic indicators across different countries.
In simple words: Show countries across the bottom. Show birth rates up the side. For each country, draw a bar matching its birth rate. Maintain equal bar width and spacing throughout the chart.

Exam Tip: Demographic data like birth rates varies significantly between developed and developing nations. Your bar chart should make these differences clear and visually impactful.

 

Question 8. Draw a bar chart for the following data showing years and interest earned in thousand crore rupees.
Answer: Mark years along the x-axis and interest (in Thousand Crore Rupees) along the y-axis. Use a scale of 1 large division = 20 units on the y-axis. Construct rectangular bars of equal width with consistent spacing between them for each year. Each bar's height represents the interest earned that year, making financial trends visible and comparable across the time period shown.
In simple words: Place years on the horizontal axis. Place interest amounts on the vertical axis. For each year, draw a bar showing interest earned. Use equal bar widths and consistent spacing for a professional look.

Exam Tip: Financial data often exhibits growth patterns. A clear bar chart helps stakeholders understand revenue and income trends at a glance.

 

Question 9. Draw a bar chart for the following data showing cities and their distances from Delhi in kilometers.
Answer: Along the x-axis, list the cities. Along the y-axis, mark distances from Delhi. Use a scale of 1 large division = 200 units on the y-axis. Create bars of equal width with uniform spacing for each city. Each bar's height shows the distance of that city from Delhi, making geographic distances immediately comparable.
In simple words: Show city names on the bottom. Show distances on the side. For each city, draw a bar as high as its distance from Delhi. Keep all bars equally wide with even spacing.

Exam Tip: Geographic data helps visualize relative locations. A bar chart makes it easy to see which cities are far and which are near to the reference point.

 

Question 10. Draw a bar chart for the following data showing countries and their life expectancy in years.
Answer: Position countries along the x-axis and life expectancy along the y-axis. Apply a scale of 1 large division = 10 units on the y-axis. Draw rectangular bars of equal width with consistent spacing for each country. Each bar's height reflects that nation's average life expectancy, allowing direct comparison of this important health indicator across populations.
In simple words: Put countries across the bottom. Put life expectancy years up the side. For each country, draw a bar showing its life expectancy. Use equal bar widths and spacing throughout.

Exam Tip: Life expectancy is a key indicator of health and development. Variations between countries are often significant - ensure your scale captures these differences visibly.

 

Question 11. Draw a bar chart for the following data showing number of weeks and rate per 10 grams in rupees.
Answer: Mark the week numbers along the x-axis and the rate per 10 grams (in Rs.) along the y-axis. Use a scale of 1 large division = 1000 units on the y-axis. Construct rectangular bars of equal width with uniform spacing between them for each week. Each bar's height shows the price per 10 grams during that week, making price trends over time immediately visible.
In simple words: Show week numbers on the horizontal line. Show prices per 10 grams on the vertical line. For each week, draw a bar showing the price. Keep bars equally wide and evenly spaced to reveal price trends.

Exam Tip: Price data often shows trends and seasonal variations. A clear bar chart helps identify when prices are highest and lowest, valuable for purchasing decisions.

 

Question 12. Draw a bar chart for the following data showing modes of transport and number of students using each.
Answer: List the modes of transport along the x-axis and the number of students along the y-axis. Use a scale of 1 large division = 100 units on the y-axis. Create rectangular bars of equal width with consistent spacing for each transport mode. Each bar's height represents the count of students who use that particular mode, revealing transportation preferences among the student population.
In simple words: Put transport types on the bottom. Put student numbers on the side. For each type, draw a bar showing how many students use it. Maintain equal bar width and spacing throughout.

Exam Tip: Transportation usage patterns help plan infrastructure and services. The bar chart should clearly show which transport modes are most and least popular among students.

 

Question 13. Analyze the bar graph showing marks obtained by a student in various subjects and answer the following: (i) What does the bar graph show? (ii) In which subject is the student very good? (iii) In which subject is the student poor? (iv) What is the average marks obtained?
Answer:
(i) The bar graph displays the marks earned by a student across different school subjects. Each bar's height indicates the score achieved in that particular subject.

(ii) By examining the bar heights, identify the tallest bar - this subject shows where the student performed best. The student demonstrates strong understanding and competence in this subject.

(iii) The shortest bar reveals the subject where the student struggles most. This subject requires additional focus and effort from the student to improve performance.

(iv) Average marks are found by adding all subject marks together and dividing by the number of subjects. This single value summarizes the student's overall academic performance across all subjects.
In simple words: The bar graph shows how a student did in each subject. The highest bar shows their best subject. The lowest bar shows their weakest subject. Average marks tell us the student's overall performance when all subjects are considered together.

Exam Tip: Always calculate average correctly by dividing the total by the number of items. Identify highest and lowest bars carefully - these comparisons form the basis of data interpretation questions.

 

Exercise 14C

 

Question 1. Draw a histogram for the following frequency distribution of daily wages.

Answer: A histogram visually represents a frequency distribution using adjacent rectangular bars with no gaps between them. The class intervals (daily wages in Rs.) are marked along the x-axis, while frequencies (number of workers) are marked along the y-axis. Since each class interval has equal width (40 Rs.), rectangles of equal width are drawn with heights corresponding to the frequencies. This is an exclusive form frequency distribution where the upper limit of one class becomes the lower limit of the next class. A kink or break is shown near the origin on the x-axis to indicate that the scale does not begin at zero but at 140, ensuring the histogram is drawn accurately to scale starting from that point.
In simple words: A histogram shows frequency distribution using bars that touch each other with no gaps. The base of each bar covers one class interval, and its height shows how many workers earn wages in that range. The break near the start tells us the scale doesn't begin at zero.

Exam Tip: Always ensure bars in a histogram are touching (no gaps), have equal width, and that the y-axis scale is clearly marked. The break symbol is essential when the x-axis doesn't start at zero.

 

Question 2. Draw a histogram for the following frequency distribution of daily earnings.

Answer: To draw the histogram, place the daily earnings class intervals along the x-axis and the number of stores along the y-axis. Since this is an exclusive form frequency distribution where each class interval's upper limit equals the next class's lower limit, rectangles with no gaps between them are constructed. Each rectangle's width represents a class interval of 50 Rs., and its height shows the frequency. All bars maintain equal width and are drawn touching each other. A break is indicated near the origin on the x-axis because the scale begins at 600 rather than zero, showing that the graph is drawn accurately starting from that value.
In simple words: Draw bars for each earnings range. Let the height of each bar show how many stores have earnings in that range. Bars touch each other with no spaces. The break symbol shows the scale doesn't start at zero.

Exam Tip: In histograms of exclusive form data, bars must be adjacent with no gaps. Clearly mark the break on the x-axis when the starting value is not zero to maintain accuracy.

 

Question 3. Draw a histogram for the following frequency distribution of student heights.

Answer: For this histogram, mark height in centimeters along the x-axis and the number of students along the y-axis. The data represents an exclusive form frequency distribution where each class interval's upper boundary becomes the next class's lower boundary. Draw adjacent rectangles with no gaps between them, each with width 6 cm (the class size) and height corresponding to the frequency. All bars have equal width and touch each other to form a continuous pattern. A break is shown near the origin on the x-axis to indicate that the scale begins at 130 cm, not at zero, confirming the graph is drawn to scale from that starting point.
In simple words: Show height ranges on the horizontal axis and student count on the vertical axis. For each height range, draw a bar touching the next bar (no gaps). The bar height shows how many students have heights in that range. Mark the break since the scale starts at 130, not zero.

Exam Tip: Histograms of exclusive form data always have touching bars with no gaps. Verify that your bar heights accurately represent the frequencies before finalizing the graph.

 

Question 4. Draw a histogram for the following frequency distribution by class intervals.

Answer: Place class intervals along the x-axis and frequencies along the y-axis. Since this represents an exclusive form frequency distribution, draw adjacent rectangular bars with no gaps between them. Each bar's width covers one class interval of 5 units, and its height represents the corresponding frequency. All rectangles maintain equal width and touch one another to create a continuous histogram. Because the scale on the x-axis begins at 8 rather than zero, a break symbol is indicated near the origin to show that the graph is drawn accurately to scale starting from value 8 on the x-axis.
In simple words: Show class intervals on the bottom and frequencies on the side. For each class, draw a bar with height matching its frequency. Bars should touch each other. The break symbol shows the scale begins at 8, not zero.

Exam Tip: When frequencies are large, ensure your y-axis scale is appropriately chosen so all bars fit and differences are visible. The break mark on the x-axis is crucial for accurate representation.

 

Question 5. Convert the inclusive form frequency distribution to exclusive form and draw a histogram.

Answer: A histogram is a graphic representation of a frequency distribution displayed as adjacent rectangles, where there is no space between consecutive rectangles. When the given data comes in inclusive form (meaning gaps exist between the upper limit of one class and the lower limit of the next), conversion to exclusive form is necessary. The inclusive form data needs to be adjusted: subtract 0.5 from each lower limit and add 0.5 to each upper limit. This creates the exclusive form shown in the table above. Once converted, mark class intervals on the x-axis and frequencies on the y-axis. Draw adjacent rectangles with equal width and heights matching their frequencies. Since the x-axis scale begins at 4.5, a break is indicated near the origin to demonstrate that the graph is drawn to scale beginning from that point.
In simple words: Inclusive form has gaps between classes, so we adjust by subtracting 0.5 from lower limits and adding 0.5 to upper limits. Then draw a histogram with touching bars. The break shows the scale starts at 4.5, not zero.

Exam Tip: Always convert inclusive form to exclusive form before drawing histograms - this ensures bars touch properly. Verify the conversion by checking that each class's upper limit equals the next class's lower limit.

 

Question 6. Mean of 25 students = 52 kg
Answer: Total weight of 25 students = 52 × 25 = 1300 kg. New mean weight of 25 students and the teacher = 52 + 2 = 54 kg. So total weight of 25 students and the teacher = 54 × 26 = 1404 kg. Weight of the teacher = 1404 - 1300 = 104 kg.
In simple words: Add the teacher's weight to the total, and find how much heavier the group becomes on average.

Exam Tip: Use the formula: new person's weight = (new total) - (old total). Watch for unit conversions when mean changes are given in grams or other units.

 

Question 7. Mean number of runs scored in first 20 matches = 40
Answer: Total runs in first 20 matches = 40 × 20 = 800. In the 21st match, 120 runs are scored. Total runs in 21 matches = 800 + 120 = 920. Mean runs in 21 matches = 920 ÷ 21 ≈ 43.81
In simple words: Add the new score to the existing total, then divide by the new count to get the updated average.

Exam Tip: Always recalculate the total first when adding a new observation - this prevents simple arithmetic errors.

 

Question 8. If the mean of x, 2x, 3x is 6, find x
Answer: Mean = (x + 2x + 3x) ÷ 3 = 6

\( \frac{6x}{3} = 6 \)

\( 2x = 6 \)

\( x = 3 \)
In simple words: Combine all the terms, divide by how many terms there are, and solve the equation to find x.

Exam Tip: Set up the mean formula carefully with all terms expressed in the same variable before simplifying.

 

Question 9. Find the mean of 2.5, 3.2, 4.8, 5.1, 6.4
Answer: Sum of numbers = 2.5 + 3.2 + 4.8 + 5.1 + 6.4 = 22

Mean = 22 ÷ 5 = 4.4
In simple words: Add all five numbers together and divide by 5 to get the average.

Exam Tip: When decimals are involved, be careful with place values - line them up vertically when adding to avoid errors.

 

Question 10. The mean of 5 numbers is 30. If one number is excluded, the mean becomes 28. Find the excluded number
Answer: Sum of 5 numbers = 30 × 5 = 150. Sum of remaining 4 numbers = 28 × 4 = 112. The excluded number = 150 - 112 = 38.
In simple words: Multiply the old mean by the old count to find the total. Multiply the new mean by the new count. The difference gives you the excluded number.

Exam Tip: Always work with sums (mean × count) rather than means directly - this makes the relationship clearer.

 

Question 11. The mean of 9 numbers is 50. If each number is multiplied by 2, find the new mean
Answer: When each observation in a dataset is multiplied by a constant k, the mean is also multiplied by the same constant k. Original mean = 50. New mean = 50 × 2 = 100.
In simple words: If you multiply every number by 2, the average gets multiplied by 2 as well.

Exam Tip: Remember this property: multiplying all data by k multiplies the mean by k; adding k to all data adds k to the mean.

 

Question 12. The mean of 8 numbers is 40. If 5 is added to each number, find the new mean
Answer: When a constant is added to each observation, the mean increases by that same constant. Original mean = 40. Adding 5 to each number increases the mean by 5. New mean = 40 + 5 = 45.
In simple words: If you add the same amount to every number, the average goes up by that amount too.

Exam Tip: This is a shortcut - you do not need to find the new sum. Just add the constant directly to the original mean.

 

Question 13. The median of the data set 3, 7, 9, 12, 15 is
Answer: The data set has 5 numbers, which is odd. The median is the middle value when the numbers are arranged in order. Here, the middle (3rd) value is 9. Therefore, the median is 9.
In simple words: Sort the numbers from smallest to largest and pick the one in the middle.

Exam Tip: For an odd count of values, the median position is at (n + 1) ÷ 2. For an even count, find the average of the two middle values.

 

Question 14. Find the median of 12, 8, 6, 10, 15, 20, 4
Answer: First, arrange in ascending order: 4, 6, 8, 10, 12, 15, 20. There are 7 numbers (odd count). The median is the 4th number (middle position). Median = 10.
In simple words: Put the numbers in order from smallest to biggest and find the center number.

Exam Tip: Always arrange data in order before finding the median - this is the most common mistake students make.

 

Question 15. Find the median of 7, 14, 21, 28, 35, 42
Answer: The data set has 6 numbers (even count), already in ascending order. The median is the average of the two middle values (3rd and 4th values). Median = (21 + 28) ÷ 2 = 49 ÷ 2 = 24.5.
In simple words: With an even number of values, take the two middle numbers and find their average.

Exam Tip: For even-count data, never pick just one middle number - always average the two central values.

 

Question 16. The mode of the data set 5, 7, 5, 3, 7, 5, 9 is
Answer: The mode is the value that appears most frequently. Counting occurrences: 3 appears 1 time, 5 appears 3 times, 7 appears 2 times, 9 appears 1 time. Since 5 appears most often (3 times), the mode is 5.
In simple words: The mode is the number that shows up more often than any other number in the list.

Exam Tip: Tally the frequency of each value. The one with the highest tally is your mode. A data set can have no mode, one mode, or multiple modes.

 

Question 17. Find the range of the data set 45, 32, 18, 27, 50, 38
Answer: Range is the difference between the largest and smallest values. Largest value = 50. Smallest value = 18. Range = 50 - 18 = 32.
In simple words: Subtract the smallest number from the largest number to find how spread out the data is.

Exam Tip: Range is always a single number (not a pair). It measures spread but is very sensitive to outliers.

 

Question 16. Mean score of 25 observations = 80. Mean score of 55 observations = 60. Find the mean score of all 80 observations.
Answer: When you add the total scores from both groups, you get 2000 + 3300 = 5300. The combined group now has 80 observations. Dividing the total by the number of observations gives you the overall mean: 5300 ÷ 80 = 66.25.
In simple words: Add up all the scores from both groups, then divide by how many observations there are in total.

Exam Tip: Always find the total sum for each group first, then combine them before dividing by the total number of observations.

 

Question 17. Average marks of 4 subjects = 50. Three subjects have marks 36, 44, and 75. Find the value of the 4th subject's marks.
Answer: The sum of all 4 subjects' marks must be 50 × 4 = 200. The three known subjects add up to 36 + 44 + 75 = 155. To find the missing subject's marks, subtract: 200 - 155 = 45.
In simple words: Find the total marks needed, add the known marks, then subtract to get the missing one.

Exam Tip: Always use the formula: total = average × count, then solve for the missing value.

 

Question 18. Mean monthly salary of 75 workers = Rs. 5680. Mean salary of 25 workers = Rs. 5400. Mean salary of 30 workers = Rs. 5700. Find the mean salary of the remaining 20 workers.
Answer: First, calculate total salaries: 75 workers earn 5680 × 75 = Rs. 426000 in total. The 25 workers earn 5400 × 25 = Rs. 135000, and the 30 workers earn 5700 × 30 = Rs. 171000. The remaining 20 workers' total salary is 426000 - (135000 + 171000) = Rs. 120000. The mean salary for these 20 workers is 120000 ÷ 20 = Rs. 6000.
In simple words: Find the total for all workers, subtract the totals for the known groups, then divide by the remaining count.

Exam Tip: Keep track of all groups and their totals carefully - a simple arithmetic error here will throw off your final answer.

 

Question 19. A ship travels from a starting point to a mark and back. Speed from start to mark = 15 km/hour. Speed from mark to start = 10 km/hour. Find the average speed for the whole journey.
Answer: Let the distance be x km. Time to reach the mark is x/15 hours, and time returning is x/10 hours. Total time taken is x/15 + x/10 = 6x/30 = x/6 hours. Total distance is 2x km. Average speed = 2x ÷ (x/6) = 2x × (6/x) = 12 km/hour.
In simple words: When a journey has different speeds each way, find the total distance and total time, then divide distance by time - don't just average the speeds.

Exam Tip: For "to and fro" problems, always use the formula: average speed = total distance ÷ total time, not the arithmetic mean of the two speeds.

 

Question 20. Total number of students = 50. Average weight of class = 44 kg. Average weight of 10 girls = 40 kg. Find the average weight of the 40 boys.
Answer: The total weight of all 50 students is 44 × 50 = 2200 kg. The 10 girls together weigh 40 × 10 = 400 kg. Therefore, the 40 boys together weigh 2200 - 400 = 1800 kg. The average weight of the boys is 1800 ÷ 40 = 45 kg.
In simple words: Find the total weight of everyone, subtract the girls' total weight, then divide the boys' total by their count.

Exam Tip: When splitting a group, always subtract one subgroup's total from the overall total before finding the average for the remaining subgroup.

 

Exercise 14F

 

Question 1. For calculating the mean, prepare a frequency table for daily wages of workers.
Answer: Set up the frequency distribution table with daily wages (Rs), number of workers, and the product fixi.

Mean = \( \frac{\sum fixi}{\sum fi} = \frac{7110}{60} = 118.5 \) Therefore, the mean daily wage of 60 workers is Rs. 118.50.
In simple words: Multiply each wage by how many workers earn it, add all those products together, then divide by the total number of workers.

Exam Tip: Make sure you multiply each value by its frequency correctly and add all products before dividing by the total frequency.

 

Question 2. For calculating the mean, prepare a frequency table for weights of workers.
Answer: Create the frequency distribution table with weight values, frequencies, and their products.

Mean = \( \frac{\sum fixi}{\sum fi} = \frac{771}{12} = 64.25 \) kg Therefore, the mean weight of the workers is 64.25 kg.
In simple words: For each weight, count how many workers have it, multiply them together, sum all results, then divide by total workers.

Exam Tip: Always organize your table clearly with all three columns visible - it helps prevent calculation mistakes.

 

Question 3. For calculating the mean, prepare a frequency table for ages of students.
Answer: Build the frequency table with age values, frequencies, and products as follows:

Mean = \( \frac{\sum fixi}{\sum fi} = \frac{698}{40} = 17.45 \) years Therefore, the mean age of the students is 17.45 years.
In simple words: Multiply each age by its frequency, add all results, then divide by the total number of students.

Exam Tip: Pay close attention to the class interval - here each age is a single value, not a range.

 

Question 4. For calculating the mean, prepare a frequency table for the variable and frequency data given.
Answer: Organize the data into a frequency table:

Mean = \( \frac{\sum fixi}{\sum fi} = \frac{2750}{50} = 55 \)
In simple words: List each value, count its frequency, multiply them, add all products, then divide by the total count.

Exam Tip: Double-check that your sum of frequencies equals the total number of observations in the problem.

 

Question 5. For calculating the mean, prepare a frequency table where one frequency is unknown (P), and the given mean is 8.
Answer: Set up the frequency table with the unknown frequency P:

Given that mean = 8: \[ \frac{303 + 9p}{41 + p} = 8 \] \[ 303 + 9p = 8(41 + p) \] \[ 303 + 9p = 328 + 8p \] \[ p = 25 \] Therefore, the value of P = 25.
In simple words: Use the mean formula with the unknown, set it equal to the given mean, then solve for the missing frequency.

Exam Tip: When finding an unknown frequency, cross-multiply and simplify carefully to isolate the variable.

 

Question 6. For calculating the mean, prepare a frequency distribution table where one frequency is unknown (p), and the given mean is 28.25.
Answer: Build the frequency table with unknown frequency p:

Since mean = 28.25: \[ \frac{1445 + 25p}{50 + p} = 28.25 \] \[ 1445 + 25p = 28.25(50 + p) \] \[ 1445 + 25p = 1412.50 + 28.25p \] \[ -3.25p = -32.5 \] \[ p = 10 \] Therefore, the value of p = 10.
In simple words: Apply the mean formula, substitute the known mean value, then rearrange and solve for the unknown frequency.

Exam Tip: Watch your signs carefully when rearranging - negative coefficients can lead to errors if you're not careful.

 

Question 7. For calculating the mean, prepare a frequency distribution table where one frequency (p) is unknown.
Answer: Build the frequency table with the unknown p. The table shows:

Given that mean = 16.6: \[ \frac{1228 + 24p}{100} = 16.6 \] \[ 1228 + 24p = 1660 \] \[ 24p = 432 \] \[ p = 18 \] Therefore, the value of p = 18.
In simple words: The unknown value itself is what we're solving for - use the mean equation and work backwards.

Exam Tip: When the unknown is the variable (xi) rather than the frequency, the approach is identical - just be clear about which is which in your setup.

 

Question 8. Two frequencies (f1 and f2) are missing in a distribution where the mean is 50.
Answer: Let f1 and f2 be the missing frequencies. Build the frequency table:

From the total frequency: 17 + f1 + 32 + f2 + 19 = 120, so f1 + f2 = 52. This gives f2 = 52 - f1 (equation 1). From the mean: \( \frac{3480 + 30f_1 + 70f_2}{120} = 50 \) \[ 3480 + 30f_1 + 70f_2 = 6000 \] Substituting f2 = 52 - f1: \[ 3480 + 30f_1 + 70(52 - f_1) = 6000 \] \[ 3480 + 30f_1 + 3640 - 70f_1 = 6000 \] \[ -40f_1 = -1120 \] \[ f_1 = 28 \] Therefore, f2 = 52 - 28 = 24. The missing frequencies are f1 = 28 and f2 = 24.
In simple words: Use the frequency total to get one equation, and the mean formula to get another. Solve the system by substituting one into the other.

Exam Tip: When two unknowns are missing, you need two equations - one from total frequency and one from the mean formula.

 

Question 9. Using the step deviation method with assumed mean A = 900, find the mean weekly wages of workers.
Answer: Prepare the frequency distribution using the step deviation method:

Using the formula: \( \overline{X} = A + \frac{\sum f_i \times d_i}{\sum f_i} \) \[ \overline{X} = 900 + \frac{-880}{100} = 900 - 8.80 = 891.20 \] Therefore, the mean weekly wage is Rs. 891.20.
In simple words: Find the difference of each value from the assumed mean, multiply by frequency, add all, then divide by total frequency. Add this result back to the assumed mean.

Exam Tip: The step deviation method reduces calculation errors - choose an assumed mean that is close to the center of your data.

 

Question 10. Let the assumed mean be A = 67. Find the mean height of plants using the deviation method.
Answer: Using the deviation method with assumed mean A = 67:

Mean, \( \overline{X} = A + \frac{\sum f_i d_i}{\sum f_i} = 67 + \frac{45}{100} = 67 + 0.45 = 67.45 \) Therefore, the mean height of the plants is 67.45 cm.
In simple words: Subtract the assumed mean from each height, multiply by frequency, add results, divide by total, then add back to assumed mean.

Exam Tip: Choose an assumed mean that matches one of the data values - this makes many deviations zero, simplifying calculations.

 

Question 11. Clearly h = 1. Let the assumed mean A = 21. Find the mean using the step deviation method.
Answer: Using the step deviation method with h = 1 and A = 21:

Mean, \( \overline{X} = A + h \times \frac{\sum f_i u_i}{\sum f_i} = 21 + 1 \times \frac{-840}{2200} = 21 + (-0.38) = 20.62 \)
In simple words: Divide each deviation by the class width h to get simpler step numbers, multiply by frequency, then apply the formula.

Exam Tip: The step deviation method is especially useful for large frequencies - smaller numbers reduce arithmetic mistakes.

 

Question 12. Clearly h = (x2 - x1) = (600 - 200) = 400. Let assumed mean A = 1000. Find the mean height using step deviations.
Answer: Using the step deviation method with h = 400 and A = 1000:

Mean, \( \overline{X} = A + h \times \frac{\sum f_i u_i}{\sum f_i} = 1000 + 400 \times \frac{-52}{1343} = 1000 + 400 \times (-0.03873) = 1000 - 15.488 = 984.51 \) Thus, the mean height is 984.51 m.
In simple words: When the class width is large, use the step deviation method by dividing deviations by that width, then multiply back at the end.

Exam Tip: Always verify your class width h by checking that it's the same between consecutive class intervals.

 

Exercise 14G

 

Question 1. (i) Arrange the data in ascending order and find the median: 2, 2, 3, 5, 7, 9, 9, 10, 11
Answer: Arranging the data in ascending order: 2, 2, 3, 5, 7, 9, 9, 10, 11. Here n = 9, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{9+1}{2} \) th term = value of the 5th term = 7 Therefore, median = 7
In simple words: When you have an odd count of numbers, the median is the middle value - here it's the 5th number.

Exam Tip: Always arrange data in order first, then use the position formula to locate the median value exactly.

 

Question 1. (ii) Arrange the data in ascending order and find the median: 6, 8, 9, 15, 16, 18, 21, 22, 25
Answer: Arranging in ascending order: 6, 8, 9, 15, 16, 18, 21, 22, 25. Here n = 9, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{9+1}{2} \)th term = value of the 5th term = 16 Therefore, median = 16
In simple words: The 5th number in a list of 9 is the median - it has equal numbers on each side.

Exam Tip: Count carefully from both ends to confirm the middle position.

 

Question 1. (iii) Arrange the data in ascending order and find the median: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Answer: Arranging in ascending order: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25. Here n = 11, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{11+1}{2} \)th term = value of the 6th term = 16 Therefore, median = 16
In simple words: With 11 items, the 6th item sits exactly in the middle with 5 numbers on each side.

Exam Tip: For odd n, the median position is always (n+1)/2.

 

Question 1. (iv) Arrange the data in ascending order and find the median: 0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Answer: Arranging in ascending order: 0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10. Here n = 13, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{13+1}{2} \)th term = value of the 7th term = 4 Therefore, median = 4
In simple words: The 7th position in a sequence of 13 numbers is the median.

Exam Tip: When finding the kth term, count exactly k positions from the start.

 

Question 2. (i) Arrange the data in ascending order and find the median: 9, 10, 17, 19, 21, 22, 32, 35
Answer: Arranging in ascending order: 9, 10, 17, 19, 21, 22, 32, 35. Here n = 8, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(4 \text{th term} + 5 \text{th term})] = \frac{1}{2}(19 + 21) = \frac{1}{2} \times 40 = 20 \) Therefore, median = 20
In simple words: With an even count, take the two middle numbers and find their average.

Exam Tip: For even n, always average the two center values - never pick just one.

 

Question 2. (ii) Arrange the data in ascending order and find the median: 29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Answer: Arranging in ascending order: 29, 35, 51, 55, 60, 63, 72, 82, 85, 91. Here n = 10, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(5 \text{th term} + 6 \text{th term})] \) Looking at the ordered list, the 5th term is 60 and the 6th term is 63. Median = \( \frac{1}{2}(60 + 63) = \frac{1}{2} \times 123 = 61.5 \)
In simple words: Identify the two center values precisely, then compute their average.

Exam Tip: Double-check that you're counting to the correct positions when n is even.

 

Question 2. (iii) Arrange the data in ascending order and find the median: 3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Answer: Arranging in ascending order: 3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81. Here n = 12, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(6 \text{th term} + 7 \text{th term})] = \frac{1}{2}(15 + 17) = \frac{1}{2} \times 32 = 16 \) Therefore, median = 16
In simple words: With 12 items, the 6th and 7th are your two middle values.

Exam Tip: For even n, positions (n/2) and (n/2 + 1) always give you the two central values.

 

Question 3. Arrange the data in ascending order and find the median: 17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Answer: Arranging in ascending order: 17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40. Here n = 15, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{1}{2}(15+1) \)th term = value of the 8th term = 23 Therefore, the median score is 23.
In simple words: The 8th position out of 15 data points is exactly in the middle.

Exam Tip: In a score or marks context, the median tells you the "middle" performance level.

 

Question 4. Arrange the heights of 9 girls in ascending order and find the median: 143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Answer: The heights in ascending order are: 143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1. Here n = 9, which is odd. Median = \( \frac{1}{2}(n+1) \)th term = \( \frac{1}{2}(9+1) \)th term = value of the 5th term = 147.3 Therefore, the median height is 147.3 cm.
In simple words: The 5th height value sits right in the middle of the 9 girls' heights.

Exam Tip: For measurements like height, the median represents the typical or middle value of the group.

 

Question 5. Arrange the weights of 8 children in ascending order and find the median: 9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Answer: The weights in ascending order are: 9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2. Here n = 8, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(4 \text{th term} + 5 \text{th term})] = \frac{1}{2}(13.4 + 14.3) = \frac{1}{2} \times 27.7 = 13.85 \) Therefore, the median weight is 13.85 kg.
In simple words: The 4th and 5th weights are the two middle values - their average is the median.

Exam Tip: When averaging decimals, add them carefully and divide by 2 to get the precise median.

 

Question 6. Arrange the ages of teachers in ascending order and find the median: 32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Answer: The ages in ascending order are: 32, 34, 36, 37, 40, 44, 47, 50, 53, 54. Here n = 10, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(5 \text{th term} + 6 \text{th term})] = \frac{1}{2}(40 + 44) = \frac{1}{2} \times 84 = 42 \) Therefore, the median age is 42 years.
In simple words: The 5th and 6th ages are your two center values - averaging them gives the median.

Exam Tip: Always verify you've ordered the data correctly by checking that each value is less than or equal to the next.

 

Question 7. The ten observations in ascending order are: 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41. If the median is 24, find the value of x.
Answer: The ten observations in ascending order are: 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41. Here n = 10, which is even. Median = \( \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] = \frac{1}{2} [(5 \text{th term} + 6 \text{th term})] = \frac{1}{2}(x+1+x+3) = \frac{1}{2}(2x+4) = x+2 \) Given that median = 24: \[ x + 2 = 24 \] \[ x = 24 - 2 = 22 \] Therefore, x = 22.
In simple words: The median formula with unknowns gives you an equation - set it equal to the given median and solve.

Exam Tip: When the median contains unknown variables, you can still use the formula - treat the unknowns algebraically.

 

Question 8. Find the median weight of students using cumulative frequency table for: Weight 45, 46, 48, 50, 52, 54, 55 kg with No. of students 8, 5, 6, 9, 7, 4, 2.
Answer: Prepare a cumulative frequency table:

Total n = 41, which is odd. Median weight = \( \left( \frac{n+1}{2} \right) \)th term = \( \left( \frac{41+1}{2} \right) \)th term = 21st term. From the cumulative frequency table, students 20 through 28 have weight 50 kg. The 21st student falls in this group. Therefore, the median weight is 50 kg.
In simple words: Build the cumulative frequency column, find which weight group contains the middle position, then read that weight as the median.

Exam Tip: Cumulative frequency makes it easy to locate the median position - just find where the cumulative total reaches or exceeds (n+1)/2.

 

Question 9. Arrange the terms in ascending order and find the median using cumulative frequency: Variate 15, 17, 20, 22, 25, 30 with Frequency 3, 5, 9, 4, 6, 10.
Answer: Prepare the cumulative frequency table:

Total n = 37, which is odd. Median = \( \left( \frac{n+1}{2} \right) \)th term = \( \left( \frac{37+1}{2} \right) \)th term = 19th term. From the cumulative frequency table, the cumulative frequency reaches 21 at variate 22. Since 19 falls within this cumulative range (from 18 to 21), the 19th term has variate = 22. Therefore, the median = 22.
In simple words: Find the position (n+1)/2, then look at the cumulative frequency to see which variate value that position falls within.

Exam Tip: Always note the cumulative frequency boundaries - the median variate is the one whose cumulative frequency first meets or exceeds the median position.

 

Question 10. Arrange the marks and find the median using cumulative frequency: Marks 9, 20, 25, 40, 50, 80 with No. of students 4, 6, 16, 8, 7, 2.
Answer: Prepare the cumulative frequency table:

Here, number of students = 43, which is odd. Median = \( \left( \frac{n+1}{2} \right) \)nd term = \( \left( \frac{43+1}{2} \right) \)nd term = 22nd term. From the cumulative frequency table, the cumulative frequency reaches 26 at marks 25. Since the 22nd term falls between 10 and 26, it corresponds to marks = 25. So, the median of marks = 25.
In simple words: Find the median position, then use cumulative frequency to identify which marks value contains that position.

Exam Tip: The median variate is always the one where the cumulative frequency first becomes greater than or equal to (n+1)/2.

 

Question 1. Arrange the given data in ascending order and find the mode.
Answer: When arranging the data in ascending order, we obtain: 0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6. The frequency table shows the following distribution:

Since 6 appears the greatest number of times, specifically 4 times, the mode is 6.
In simple words: Put the numbers from smallest to largest. The number that shows up most often is the mode. Here, 6 appears 4 times, more than any other number, so 6 is the mode.

Exam Tip: The mode is always the value with the highest frequency - count carefully and double-check which observation appears most often.

 

Question 2. Arrange the given data in ascending order and find the mode.
Answer: When arranging the data in ascending order, we get: 15, 20, 22, 23, 25, 25, 25, 27, 40. The frequency table is shown below:

Since 25 appears the greatest number of times with a frequency of 3, the mode equals 25.
In simple words: Look at which number repeats most frequently. In this set, 25 appears 3 times while all others appear just once, making 25 the mode.

Exam Tip: Always count how many times each value shows up in the list - the one with the highest count is your mode.

 

Question 3. Arrange the given data in ascending order and find the mode.
Answer: When arranging the data in ascending order, we obtain: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9. The frequency table is as follows:

Since 9 appears the greatest number of times with a frequency of 5, the mode is 9.
In simple words: Count how many times each number repeats. The number 9 shows up 5 times, which is more than any other number, so 9 is the mode.

Exam Tip: Create a tally to keep track of frequencies accurately - this prevents counting errors and helps you identify the mode with confidence.

 

Question 4. Arrange the given data in ascending order and find the mode.
Answer: When arranging the data in ascending order, we obtain: 9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60. The frequency table is shown below:

Since 50 appears the greatest number of times with a frequency of 4, the mode is 50. Thus, the modal score of the cricket player is 50.
In simple words: The score 50 shows up 4 times, more than any other score in the list. This makes 50 the mode, which tells us this was the most common score the player achieved.

Exam Tip: In real-world data like sports scores, the mode shows the most frequently occurring performance level - it's a practical measure of what typically happens.

 

Question 5. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: When arranging the data in ascending order, we obtain: 10, 10, 11, 11, 12, 12, 13, 14, 15, 17. We prepare the table below:

Here, \( N = 10 \), which is even.

\( \therefore \text{median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] \)
\( = \frac{1}{2} [(\text{5th term} + \text{6th term})] \) [since \( n = 10 \)]
\( = \frac{1}{2} (12 + 12) \)
\( = 12 \)

Now, \( \sum f \times x = 125 \) and \( \sum f = 10 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{125}{10} = 12.5 \)

\( \text{Mode} = 3(\text{Median}) - 2(\text{Mean}) \)
\( = 3(12) - 2(12.5) \)
\( = 36 - 25 \)
\( = 11 \)

Thus, mean = 12.5, median = 12, and mode = 11.
In simple words: The mean is found by adding all values and dividing by how many there are. The median is the middle value when arranged in order. The mode is the value that repeats most often. Here all three measures cluster around 12, showing the data is fairly consistent.

Exam Tip: Always use the formula Mode = 3(Median) - 2(Mean) when direct counting doesn't reveal an obvious mode - it's a reliable relationship in grouped data.

 

Question 6. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table as shown:

Here, \( N = 25 \), which is odd.

\( \therefore \text{median} = \left( \frac{N + 1}{2} \right) \text{th term} \)
\( = \left( \frac{25 + 1}{2} \right) \text{th term} \)
\( = \text{value of the 13th term} \)
\( = 13 \)

Now, \( \sum f \times x = 332 \) and \( \sum f = 25 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{332}{25} = 13.28 \)

\( \text{Mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 13) - (2 \times 13.28) \)
\( = 39 - 26.56 \)
\( = 12.44 \)

Thus, mode = 12.4 (rounded to one decimal place).
In simple words: Build the frequency table carefully to track how many times each value shows up. Use cumulative frequency to find the median quickly by looking for the middle position. The mode formula helps when values appear with similar frequency.

Exam Tip: For odd-numbered datasets, the median is simply the middle value; for even-numbered sets, average the two middle values - this is a fundamental distinction to remember.

 

Question 7. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( N = 25 \), which is odd.

\( \therefore \text{median} = \left( \frac{N + 1}{2} \right) \text{th term} \)
\( = \left( \frac{25 + 1}{2} \right) \text{th term} \)
\( = \text{value of the 13th term} \)
\( = 13 \)

Now, \( \sum f \times x = 332 \) and \( \sum f = 25 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{332}{25} = 13.28 \)

\( \text{Mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 13) - (2 \times 13.28) \)
\( = 39 - 26.56 \)
\( = 12.44 \)

Thus, mode = 12.4 (to one decimal place).
In simple words: When data is organized by frequency, calculate the mean using \( \frac{\sum f \times x}{\sum f} \). Find the median by identifying the middle position in cumulative frequency. Apply the mode formula to determine the mode accurately.

Exam Tip: Always verify your cumulative frequency column by checking that the final cumulative value equals N (the total frequency) - this catches arithmetic errors early.

 

Question 8. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( N = 34 \), which is even.

\( \text{Median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] \)
\( = \frac{1}{2} [(\text{17th term} + \text{18th term})] \) [since \( n = 34 \)]
\( = \frac{1}{2} (12 + 12) \)
\( = 12 \)

Now, \( \sum f \times x = 407 \) and \( \sum f = 34 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{407}{34} = 11.97 \)

\( \text{Mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 12) - (2 \times 11.97) \)
\( = 36 - 23.94 \)
\( = 12.06 \)

Thus, mode = 12.06.
In simple words: For even-sized datasets, the median is the average of the two middle terms. The mean is the sum of all products divided by the total frequency. Use the three-point formula to determine the mode from median and mean values.

Exam Tip: When finding the median of even-sized data, identify both middle positions correctly using n/2 and (n/2 + 1), then average their values precisely.

 

Question 9. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( N = 40 \), which is even.

\( \text{Median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] \)
\( = \frac{1}{2} [(\text{20th term} + \text{21st term})] \) [since \( n = 40 \)]
\( = \frac{1}{2} (30 + 30) \)
\( = 30 \)

Now, \( \sum f \times x = 1161 \) and \( \sum f = 40 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{1161}{40} = 29.025 \)

\( \text{Mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 30) - (2 \times 29.025) \)
\( = (90 - 58.05) \)
\( = 31.95 \)

Thus, mode = 32 (rounded to nearest whole number).
In simple words: Build a complete frequency table with cumulative values to track data organization. Calculate the mean by dividing the sum of products by total frequency. Apply the three-measure relationship to find the mode when visual inspection doesn't reveal it clearly.

Exam Tip: Show all three measures (mean, median, mode) even if the question asks for one - this demonstrates mastery of central tendency and helps identify calculation errors.

 

Question 10. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( \sum f \times x = 796 \) and \( \sum f = 15 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{796}{15} = 53.06 \)

Here, \( N = 15 \), which is odd.

\( \therefore \text{median} = \left( \frac{n + 1}{2} \right) \text{th term} \)
\( = \left( \frac{15 + 1}{2} \right) \text{th term} = \text{8th term} \)
\( = 53 \)

\( \therefore \text{mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 53) - (2 \times 53.06) \)
\( = 159 - 106.12 \)
\( = 52.88 \)

Thus, mean = 53.06, median = 53, and mode = 52.88.
In simple words: Organize the weight data with frequencies and calculate cumulative totals. The mean shows the average weight. The median is the middle value when all students are arranged by weight. The mode indicates the most typical weight using the three-measure relationship.

Exam Tip: Use the formula mean = \( \sum f \times x / \sum f \) when data is presented in frequency form - it's faster and more accurate than adding individual values.

 

Question 11. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( n = 80 \), which is even.

\( \therefore \text{median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] \)
\( = \frac{1}{2} [(\text{40th term} + \text{41st term})] \) [since \( n = 80 \)]
\( = \frac{1}{2} (28 + 28) \)
\( = 28 \)

Now, \( \sum f \times x = 1992 \) and \( \sum f = 80 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{1992}{80} = 24.9 \)

\( \therefore \text{mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 28) - (2 \times 24.9) \)
\( = 84 - 49.8 \)
\( = 34.2 \)

Thus, modal marks = 34.2.
In simple words: A frequency table makes it easy to organize mark data and calculate mean using the sum of products method. The median comes from the cumulative frequency column. The mode calculation brings together median and mean for a complete picture of the distribution.

Exam Tip: When data size N is even, remember that the median is always an average of two middle terms, not a single value in the data set itself.

 

Question 12. Arrange the given data in ascending order and find the mean, median, and mode.
Answer: We prepare the frequency table below:

Here, \( \sum f \times x = 2650 \) and \( \sum f = 106 \)

\( \therefore \text{mean} = \frac{\sum f \times x}{\sum f} = \frac{2650}{106} = 25 \)

\( \therefore \text{mean} = 25 \)

Here, \( N = 106 \), which is even.

\( \therefore \text{median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right) \text{th term} + \left( \frac{n}{2} + 1 \right) \text{th term} \right] \)
\( = \frac{1}{2} [(\text{53rd term} + \text{54th term})] \) [since \( n = 106 \)]
\( = \frac{1}{2} (25 + 25) \)
\( = 25 \)

\( \therefore \text{median} = 25 \)

\( \therefore \text{mode} = 3(\text{median}) - 2(\text{mean}) \)
\( = (3 \times 25) - (2 \times 25) \)
\( = 75 - 50 \)
\( = 25 \)

Thus, mean = 25, median = 25, and mode = 25.
In simple words: When mean, median, and mode all equal the same value, the data is perfectly symmetric and balanced around that point. This is a rare but highly desirable situation showing the data has ideal central tendency properties.

Exam Tip: When all three central measures are equal, the distribution is perfectly symmetric - use this as a quick check that your calculations are likely correct.