ML Aggarwal Class 10 Maths Solutions Chapter 21 Measures of Central Tendency

 

Question 1. Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.
Answer: Add all the numbers together to find their total: 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35. Since there are 5 numbers, divide the sum by 5. The arithmetic mean is therefore 35 ÷ 5 = 7.
In simple words: Add all five numbers. Then divide by how many numbers you have. The answer is 7.

Exam Tip: Always count the number of values carefully before dividing - a common error is miscounting how many numbers are in the set.

 

Question 2. The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20. Find:
(i) the mean of their marks.
(ii) the mean of their marks when the marks of each student are increased by 4.
(iii) the mean of their marks when 2 marks are deducted from the marks of each student.
(iv) the mean of their marks when the marks of each student are doubled.
Answer:
(i) Add all the marks: 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20 = 207. Since there are 15 students, the arithmetic mean is 207 ÷ 15 = 13.8.
(ii) When each student's marks go up by 4, the total increases by 15 × 4 = 60. The new sum becomes 207 + 60 = 267. The new mean is 267 ÷ 15 = 17.8.
(iii) When each student's marks go down by 2, the total decreases by 15 × 2 = 30. The new sum becomes 207 - 30 = 177. The new mean is 177 ÷ 15 = 11.8.
(iv) When each student's marks are multiplied by 2, the total also gets multiplied by 2. The new sum becomes 207 × 2 = 414. The new mean is 414 ÷ 15 = 27.6.
In simple words: When you add the same number to every mark, the mean also increases by that number. When you multiply every mark by a number, the mean gets multiplied by the same number too.

Exam Tip: Remember that adding or multiplying all values by the same amount changes the mean in a predictable way - add the amount to the mean, or multiply the mean by the factor, without recalculating from scratch.

 

Question 3(a). The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Answer: Using the mean formula, \( 8 = \frac{6 + y + 7 + x + 14}{5} \). Simplifying, \( 8 = \frac{x + y + 27}{5} \). Multiply both sides by 5: \( 40 = x + y + 27 \). Rearranging, \( x + y = 13 \), so \( y = 13 - x \).
In simple words: Set up the mean equation and solve for y. You'll find that y must equal 13 minus x.

Exam Tip: When expressing one variable in terms of another, isolate that variable completely and double-check by substituting back into the original condition.

 

Question 3(b). The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15, find the 9th variate.
Answer: Let the 9th variate be x. Add the eight known values: 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 = 89. The total sum of all nine variates is 89 + x. Using the mean formula, \( 11 = \frac{89 + x}{9} \). Multiply both sides by 9: \( 99 = 89 + x \). Therefore, \( x = 10 \).
In simple words: Find the sum of the known numbers. Use the mean formula to write an equation and solve for the missing number.

Exam Tip: Always find the total first by multiplying mean by the count, then subtract the sum of known values to get the missing one.

 

Question 4(a). The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes \( 12\frac{15}{16} \) years. What is the age of the girl?
Answer: If the mean age of 33 students is 13 years, the total age is \( 33 \times 13 = 429 \) years. After one girl leaves, there are 32 students and the new mean is \( 12\frac{15}{16} = \frac{207}{16} \) years. The new total age is \( 32 \times \frac{207}{16} = \frac{6624}{16} = 414 \) years. The girl's age is \( 429 - 414 = 15 \) years.
In simple words: Multiply the original mean by the original count to get the total age. Then multiply the new mean by the new count. Subtract to find the girl's age.

Exam Tip: Convert mixed fractions to improper fractions before doing arithmetic - it prevents calculation errors and makes the working clearer.

 

Question 4(b). In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that the marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.
Answer: The incorrect total is \( 40 \times 18.2 = 728 \). Since one mark was recorded as 21 but should have been 29, the difference is 8. The correct total is \( 728 - 21 + 29 = 736 \). The correct mean is \( \frac{736}{40} = 18.4 \).
In simple words: Find the wrong total by multiplying mean by the count. Remove the wrong mark and add the correct one. Divide the new total by the count.

Exam Tip: When correcting errors, be clear about subtracting the wrong value and adding the correct value - do not just take the difference.

 

Question 5. Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.
Answer: The sum of the first 10 numbers is \( 13 \times 10 = 130 \). The remaining 15 numbers have a mean of 18, so their sum is \( 15 \times 18 = 270 \). The total sum of all 25 numbers is \( 130 + 270 = 400 \). The mean of all 25 numbers is \( \frac{400}{25} = 16 \).
In simple words: Find the sum of each group separately. Add these sums together. Divide by the total count of all numbers.

Exam Tip: Always find the sum first using mean × count, then add the group sums to get the overall total before finding the combined mean.

 

Question 6. Find the mean of the following distribution:

Answer: Create a table by multiplying each number by its frequency. For example, 5 × 1 = 5, 10 × 2 = 20, 15 × 5 = 75, 20 × 6 = 120, 25 × 3 = 75, 30 × 2 = 60, 35 × 1 = 35. The sum of these products is 5 + 20 + 75 + 120 + 75 + 60 + 35 = 390. The total frequency is 1 + 2 + 5 + 6 + 3 + 2 + 1 = 20. The mean is \( \frac{390}{20} = 19.5 \).
In simple words: Multiply each number by how many times it appears. Add all these products. Divide by the total count of all items.

Exam Tip: For frequency distributions, the mean formula uses the sum of (value × frequency) divided by the sum of frequencies - do not forget to multiply each value by its frequency.

 

Question 7. The contents of 100 matchboxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean upto exactly 39 matches.
Answer:
(i) Multiply each number of matches by the number of boxes: 35 × 6 = 210, 36 × 10 = 360, 37 × 18 = 666, 38 × 25 = 950, 39 × 21 = 819, 40 × 12 = 480, 41 × 8 = 328. The sum is 210 + 360 + 666 + 950 + 819 + 480 + 328 = 3813. The total number of boxes is 100. The mean is \( \frac{3813}{100} = 38.1 \) matches per box.
(ii) If the new mean is to be 39 matches, the total must be \( 39 \times 100 = 3900 \). Currently, the total is 3813. So, \( 3900 - 3813 = 87 \) extra matches must be added.
In simple words: For part (i), multiply each value by its frequency, add these products, and divide by the total frequency. For part (ii), work out what the new total should be, then find the difference from the current total.

Exam Tip: When rounding to one decimal place, check if the second decimal digit is 5 or more - if so, round up; otherwise, round down. Always show your working for finding the extra items needed.

 

Question 8. Find the mean for the following distribution by short cut method:

Answer: First, convert the cumulative frequencies to individual frequencies. Frequency for 60 is 8, for 61 is 18 - 8 = 10, for 62 is 33 - 18 = 15, for 63 is 40 - 33 = 7, for 64 is 49 - 40 = 9, for 65 is 55 - 49 = 6, for 66 is 60 - 55 = 5. Choose assumed mean a = 63. Calculate deviations: d = value - 63. Multiply each deviation by its frequency and add them: (8 × -3) + (10 × -2) + (15 × -1) + (7 × 0) + (9 × 1) + (6 × 2) + (5 × 3) = -24 - 20 - 15 + 0 + 9 + 12 + 15 = -23. The mean is \( 63 + \frac{-23}{60} = 63 - 0.38 = 62.62 \).
In simple words: Choose a value close to the middle as your assumed mean. Find how far each number is from that assumed mean. Multiply these distances by their frequencies, add them up, divide by the total frequency, and add the result to your assumed mean.

Exam Tip: The short cut method saves time for large numbers - choose an assumed mean near the middle of the data. Always convert cumulative frequencies to individual frequencies first.

 

Question 9. Calculate the mean wage from the following distribution:

(i) Calculate the mean wage, correct to the nearest rupee.
(ii) If the number of workers in each category is doubled, what would be the new mean wage?
Answer:
(i) Multiply each wage by the number of workers in that category: 500 × 2 = 1000, 600 × 4 = 2400, 700 × 8 = 5600, 800 × 12 = 9600, 900 × 10 = 9000, 1000 × 6 = 6000, 1100 × 8 = 8800. The sum is 1000 + 2400 + 5600 + 9600 + 9000 + 6000 + 8800 = 42400. The total number of workers is 2 + 4 + 8 + 12 + 10 + 6 + 8 = 50. The mean wage is \( \frac{42400}{50} = 848 \) rupees.
(ii) When the number of workers in each category is doubled, both the numerator and denominator are doubled. The mean remains the same: 848 rupees. Doubling all frequencies does not change the mean - only the total absolute amount changes, not the average per worker.
In simple words: Multiply each wage by how many workers earn it. Add all these products and divide by the total number of workers. If you double everyone in each group, the average wage stays the same because both the total and the count grow equally.

Exam Tip: When frequencies are all multiplied by the same factor, the mean does not change - this is a useful shortcut to remember. Always round to the nearest whole rupee unless asked otherwise.

 

Question 10. The mean of the following data is 16. Calculate the value of f.
Answer: We set up a table with the marks, number of students, and the product fixi. Adding the known frequencies gives us 25 + f students total, and the sum of fixi equals 415 + 15f. Using the mean formula, we have 16 = (415 + 15f) / (25 + f). Cross-multiplying gives us 16(25 + f) = 415 + 15f, which simplifies to 400 + 16f = 415 + 15f. Solving for f, we get f = 15.
In simple words: Set up a table with marks and students. The mean formula says that the sum of products divided by total students equals 16. Solve the equation to find f = 15.

Exam Tip: Always use the mean formula correctly - sum of (frequency × value) divided by sum of frequencies. Check your algebra carefully when cross-multiplying.

 

Question 11. Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data. If the mean of the distribution is 7.2, find a and b.
Answer: We build a table showing marks, frequencies a and b, and the products fixi. The total number of students is 40, which means 35 + a + b = 40, giving us a + b = 5. Next, using the mean formula with mean = 7.2, we get 7.2 = (246 + 6a + 9b) / 40. This simplifies to 288 = 246 + 6a + 9b, or 6a + 9b = 42. Substituting a = 5 - b into this equation: 6(5 - b) + 9b = 42 becomes 30 - 6b + 9b = 42, which gives us 3b = 12, so b = 4. Therefore, a = 5 - 4 = 1.
In simple words: You have two equations: one from the total students being 40, and another from the mean being 7.2. Solve both together to find a = 1 and b = 4.

Exam Tip: When two unknowns are present, always set up two independent equations using the constraints given (total frequency and mean value). Solve them as a system.

 

Question 12. Calculate the mean of the following distribution:
Answer: We create a table with class intervals, their class marks (midpoints), and frequencies. The class marks are 10, 20, 30, 40, and 50 for the intervals 5-15, 15-25, 25-35, 35-45, and 45-55 respectively. Multiplying each class mark by its frequency and summing gives us 780. The total frequency is 24. Applying the mean formula: Mean = 780 / 24 = 32.5.
In simple words: Find the midpoint of each class. Multiply each midpoint by how many items are in that class. Add all these products and divide by the total number of items.

Exam Tip: Remember that the class mark is always the average of the class boundaries (lower + upper) / 2. Double-check your arithmetic when multiplying and summing.

 

Question 13. Calculate the mean of the following distribution:
Answer: We prepare a table listing the class intervals, their class marks, and the frequencies for each. The class marks are 5, 15, 25, 35, 45, and 55. We then compute the products of each class mark with its frequency, obtaining values of 40, 75, 300, 1225, 1080, and 880. The sum of all these products is 3600, and the total frequency sums to 100. Using the mean formula, we divide 3600 by 100 to get the mean of 36.
In simple words: For each class, find its middle value. Multiply by the number of items in that class. Add all the products together and divide by the total count.

Exam Tip: Organize your work in a clear table format to avoid errors. The final answer should be a single number representing the average.

 

Question 14. Calculate the mean of the following distribution using step deviation method:
Answer: We use the step deviation method with an assumed mean of a = 25 and class width c = 10. We build a table with class intervals, class marks, frequencies, and the deviation ui = (yi - a) / c for each class. The deviations are -2, -1, 0, 1, 2, and 3. Computing fiui for each row and summing gives us 63. The total frequency is 100. Using the formula Mean = a + c × (Σfiui / Σfi), we get: Mean = 25 + 10 × (63 / 100) = 25 + 6.3 = 31.3.
In simple words: Pick an assumed mean near the middle of the data. Divide each deviation by the class width to get simpler numbers. Multiply by frequency, sum, and use the formula to find the actual mean.

Exam Tip: The step deviation method reduces calculation errors when dealing with large class widths or awkward numbers. Always remember to multiply the final result by the class width.

 

Question 15. The data on the number of patients attending a hospital in a month is given below. Find the average (mean) number of patients attending the hospital in a month using the short cut method. Take assumed mean as 45. Give your answer correct to 2 decimal places.
Answer: We construct a table with class intervals representing patient numbers, their class marks, deviations from the assumed mean (di = yi - 45), and the frequency of days. The class marks are 15, 25, 35, 45, 55, and 65, with corresponding deviations of -30, -20, -10, 0, 10, and 20. We calculate fidia for each row: -150, -40, -70, 0, 20, and 100. The sum of all fidia values is -140, and the total frequency is 30. Using the formula Mean = a + (Σfidi / Σfi), we get: Mean = 45 + (-140 / 30) = 45 - 4.67 = 40.33.
In simple words: Choose 45 as your starting point. Find how far each class mark is from 45. Multiply these differences by how many days, add them up, and adjust your starting point by the average difference.

Exam Tip: The short cut (or deviation) method is faster when your numbers are large. Always round to the decimal places requested - here it is 2 decimal places.

 

Question 16. The following table gives the daily wages of worker in a factory. Calculate their mean by short cut method.
Answer: We set up a table with wage class intervals, their class marks (midpoints), frequencies, and deviations from the assumed mean of 625. The class marks are 475, 525, 575, 625, 675, 725, and 775. Their deviations from 625 are -150, -100, -50, 0, 50, 100, and 150. Computing the products fidia for each class and summing them gives -250. The total frequency is 100. Applying the formula Mean = a + (Σfidi / Σfi), we get: Mean = 625 + (-250 / 100) = 625 - 2.5 = Rs. 622.5.
In simple words: Choose the middle wage value (625) as your base. Find how much each wage group differs from this base. Add up all these weighted differences and adjust your base by the average amount.

Exam Tip: The assumed mean should be chosen at or near the class with the highest frequency to minimize calculation effort. Always convert currency symbols appropriately (Rs. instead of ₹).

 

Question 17. Calculate the mean of the distribution given below using the short cut method.
Answer: We create a table with mark intervals, their class marks, deviations from the assumed mean of 45.5, and the number of students. The class marks are 15.5, 25.5, 35.5, 45.5, 55.5, 65.5, and 75.5. Their deviations from 45.5 are -30, -20, -10, 0, 10, 20, and 30. We multiply each deviation by its frequency and sum to get 70. The total frequency is 50. Using the deviation formula Mean = a + (Σfidi / Σfi), we obtain: Mean = 45.5 + (70 / 50) = 45.5 + 1.4 = 46.9 marks.
In simple words: Use 45.5 as your starting point since it's the middle class mark. Calculate how each class differs from this point. Weight by frequency, sum up, and adjust your starting point by the average difference.

Exam Tip: When class intervals have decimal boundaries (like 11-20, 21-30), the class mark will also be decimal. Handle decimals carefully in your calculations.

 

Question 18. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Answer: We construct a table showing the number of absent days for each class interval, the class mark (midpoint) of each interval, the frequency of students, and the product of class mark and frequency. The class marks are 3, 8, 12, 17, 24, 33, and 39. Computing fiyi for each interval yields 33, 80, 84, 68, 96, 99, and 39 respectively. The sum of all fiyi values equals 499, and the total number of students is 40. Using the mean formula: Mean = (Σfiyi) / (Σfi) = 499 / 40 = 12.475, which rounds to approximately 12.48 days.
In simple words: Find the middle value (class mark) of each absence group. Multiply it by how many students fall in that group. Add all these products and divide by 40 to get the average days absent.

Exam Tip: Pay careful attention to unequal class widths in this problem - each interval has a different range. The class mark is still found by averaging the boundaries, regardless of interval width.

 

Question 19. If the mean of the following distribution is 24, find the value of a:

Answer: We build a table showing the class mark and products for each group.

Using the mean formula \( \text{Mean} = \frac{\sum f_i y_i}{\sum f_i} \), we have:

\( 24 = \frac{810 + 15a}{30 + a} \)

Multiplying both sides by (30 + a):

\( 24(30 + a) = 810 + 15a \)

\( 720 + 24a = 810 + 15a \)

\( 24a - 15a = 810 - 720 \)

\( 9a = 90 \)

\( a = 10 \)
In simple words: We write the class marks for each range and multiply them by their frequencies. Then we use the mean formula - the sum of all products divided by the total frequency - and set it equal to 24. Solving the equation gives us a = 10.

Exam Tip: Always double-check your arithmetic when expanding \( 24(30 + a) \) and when collecting like terms - one small error changes the final answer.

 

Question 20. The mean of the following distribution is 50. Find the unknown frequency.

Answer: We construct a table with class marks and products.

Applying the mean formula:

\( 50 = \frac{2020 + 30f}{34 + f} \)

\( 50(34 + f) = 2020 + 30f \)

\( 1700 + 50f = 2020 + 30f \)

\( 50f - 30f = 2020 - 1700 \)

\( 20f = 320 \)

\( f = 16 \)
In simple words: We find the class marks (the middle value of each range) and multiply each by its frequency. Then we apply the mean formula and solve for f by cross-multiplying and simplifying.

Exam Tip: When you have an unknown frequency, always expand and gather all terms with the variable on one side before dividing to solve.

 

Question 21. The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q:

Answer: We build a table containing class marks and their products with frequencies.

Given: Sum of frequencies = 50

\( 32 + p + q = 50 \)

\( p + q = 18 \)

\( p = 18 - q \) ... (i)

Given: Mean = 57.6

\( 57.6 = \frac{1940 + 30p + 70q}{50} \)

\( 57.6 \times 50 = 1940 + 30p + 70q \)

\( 2880 = 1940 + 30p + 70q \)

\( 30p + 70q = 940 \)

Substituting p = 18 - q from equation (i):

\( 30(18 - q) + 70q = 940 \)

\( 540 - 30q + 70q = 940 \)

\( 540 + 40q = 940 \)

\( 40q = 400 \)

\( q = 10 \)

Using equation (i):

\( p = 18 - 10 = 8 \)
In simple words: We get two equations - one from the total frequency being 50, and another from the mean being 57.6. We solve these two equations by substitution to find both p and q.

Exam Tip: With two unknowns, always find one equation from the given constraint (here, the sum of frequencies), and a second from the mean formula - then use substitution to solve both.

 

Question 22. The following table gives the life time in days of 100 electricity tubes of a certain make:

Find the mean lifetime of electricity tubes.

Answer: We convert the cumulative frequency data into a regular frequency table and then compute the mean.

Using the mean formula:

\( \text{Mean} = \frac{\sum f_i y_i}{\sum f_i} = \frac{14500}{100} = 145 \)
In simple words: When the data is given as cumulative frequencies (running totals), we first find the actual frequency for each class by subtracting successive cumulative values. Then we calculate the mean using class marks and the standard formula.

Exam Tip: Always convert cumulative frequency to actual frequency by subtraction before finding class marks and computing the mean - this is a common mistake students make.

 

Question 23. The following table gives the duration of movies in minutes. Using step-deviation method, find the mean duration of the movies.

Answer: In the table, the class interval size (i) is 10. We choose A = 135 as the assumed mean (the class mark of the central class).

Using the step-deviation formula:

\( \text{Mean} = A + \frac{\sum fu}{\sum f} \times i = 135 + \frac{-36}{51} \times 10 \)

\( = 135 + \frac{-360}{51} = 135 - 7.06 = 127.94 \) (approximately)
In simple words: In the step-deviation method, we pick a middle class mark as an assumed mean. We find how far each class mark is from this mean, divide by the class width, then weight these values by frequency. Finally, we add this weighted average (times the class width) back to the assumed mean.

Exam Tip: The step-deviation method reduces large numbers to smaller ones, making calculations easier. Always pick the assumed mean (A) as the class mark of a central or frequent class for simpler arithmetic.

 

Question 24. Shown below is a table illustrating the monthly income distribution in a company with 100 employees. Using step-deviation method, find the mean monthly income of an employee.

Answer: In the table, class size (i) = 4. We select A = 6 (the class mark of the class with the largest frequency).

Using the step-deviation formula:

\( \text{Mean} = A + \frac{\sum fu}{\sum f} \times i = 6 + \frac{19}{100} \times 4 = 6 + \frac{76}{100} = 6 + 0.76 = 6.76 \)
In simple words: We pick the class mark of the class with the highest frequency as our assumed mean. For each class, we find how many steps (of size 4) it is from this mean. We then weight these steps by frequency, sum them up, and add the result (scaled back by the class width) to our assumed mean.

Exam Tip: Choose the assumed mean from a class with high frequency to minimize computation. Always keep track of the class width (i) separately and multiply by it at the end of the calculation.

 

Question 25. Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.
Answer: We extract the frequency data from the histogram and construct a frequency table.

Using the mean formula:

\( \text{Mean} = \frac{\sum f_i y_i}{\sum f_i} = \frac{1545}{33} = 46.8 \)
In simple words: A histogram shows frequency on the vertical axis and class intervals on the horizontal axis. We read the height (frequency) for each class bar and use it to calculate the mean by multiplying each class mark by its frequency, summing these products, and dividing by the total frequency.

Exam Tip: When working with histograms, carefully read the frequency from the height of each bar. If the y-axis is labeled as "frequency density," multiply the height by the class width to get the actual frequency.

 

Exercise 21.2

 

Question 1. A student scored the following marks in 11 questions of a question paper: 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7. Find the median marks.
Answer: We arrange the marks in increasing order: 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8.

The number of observations (n) = 11, which is odd.

When the number of observations is odd, the median is the middle value at position \( \frac{n + 1}{2} = \frac{11 + 1}{2} = \frac{12}{2} = 6 \).

The 6th value in the ordered list is 5.
In simple words: When you have an odd number of values, arrange them from smallest to largest. The median is the value right in the middle - the one with an equal number of values on both sides.

Exam Tip: Always arrange data in order before finding the median. For odd n, the median is at position (n+1)/2. For even n, it is the average of the two middle values.

 

Question 2. For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.
Answer: We arrange the numbers in increasing order: 3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81.

The number of observations (n) = 12, which is even.

For even n, the median is the average of the two middle values at positions \( \frac{n}{2} \) and \( \frac{n}{2} + 1 \):

\( \text{Median} = \frac{\text{6th observation} + \text{7th observation}}{2} = \frac{15 + 17}{2} = \frac{32}{2} = 16 \)
In simple words: When you have an even number of values, arrange them from smallest to largest. Find the two middle values and take their average - that is your median.

Exam Tip: For even-numbered datasets, never forget to average the two middle values - just picking one of them is incorrect.

 

Question 3. Calculate the mean and the median of the numbers: 2, 1, 0, 3, 1, 2, 3, 4, 3, 5.
Answer: We arrange the numbers in increasing order: 0, 1, 1, 2, 2, 3, 3, 3, 4, 5.

There are n = 10 values (even number).

For the mean, we add all values and divide by the count:

Sum = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24

Mean = \( \frac{24}{10} = 2.4 \)

For the median with even n, we find the average of the 5th and 6th values:

\( \text{Median} = \frac{\text{5th observation} + \text{6th observation}}{2} = \frac{2 + 3}{2} = \frac{5}{2} = 2.5 \)
In simple words: Mean is the sum of all values divided by how many there are. Median is the middle point - for an even count, it is the average of the two center values when arranged in order.

Exam Tip: When finding both mean and median, arrange the data first. The mean uses all values in a sum; the median uses position only. For even n, always average the two middle terms.

 

Question 4. The median of the observations 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Answer: When the data is arranged in ascending order as 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47, there are 9 observations, which is odd. Since the median equals 24 and equals the 5th observation, we get x + 4 = 24, so x = 20. Substituting this value back into the data gives us 11, 12, 14, 18, 24, 29, 32, 38, 47. The sum of all these numbers is 225. Therefore, the mean is 225 divided by 9, which equals 25.
In simple words: Find the position of the middle number (5th out of 9). Set it equal to 24 to solve for x. Then add all the numbers and divide by 9 to get the mean.

Exam Tip: For odd n, the median is always the middle observation. Always substitute the value of x back into all terms before calculating the sum.

 

Question 5. The mean of the numbers 1, 7, 5, 3, 4, 4 is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m - 1 and median q. Find (i) p (ii) q (iii) the mean of p and q.
Answer:
(i) The sum of the first set (1, 7, 5, 3, 4, 4) is 24. Since there are 6 numbers, m = 24 ÷ 6 = 4. For the second set (3, 2, 4, 2, 3, 3, p), the mean is m - 1 = 3. The sum without p is 17, so we have (17 + p) ÷ 7 = 3, which gives 17 + p = 21, therefore p = 4.
(ii) With p = 4, the second set becomes 3, 2, 4, 2, 3, 3, 4. Arranging in ascending order: 2, 2, 3, 3, 3, 4, 4. There are 7 numbers, so the median is the 4th observation, which is q = 3.
(iii) The mean of p and q, that is 4 and 3, is (4 + 3) ÷ 2 = 7 ÷ 2 = 3.5.
In simple words: First find m by adding and dividing. Use m - 1 to find the unknown p. Sort the new list and pick the middle number for the median. Finally, add p and q and divide by 2.

Exam Tip: When finding p, always substitute back into the set before finding the median. The median position for n = 7 is (7 + 1) ÷ 2 = 4th observation.

 

Question 6. Find the median for the following distribution:

Answer: The given wages are already in ascending order. We build a cumulative frequency table:
There are 47 workers in total, which is odd. The median position is (47 + 1) ÷ 2 = 24. Observations from the 23rd to 29th are all equal to 480. Therefore, the median is Rs 480.
In simple words: Build a running total (cumulative frequency) for each wage. Find which wage covers the middle position (24th observation). All observations in that range have the same wage value, so that is the median.

Exam Tip: Always construct the cumulative frequency table first. The median observation falls in the class whose cumulative frequency first reaches or exceeds (n + 1) ÷ 2.

 

Question 7. Calculate the median marks for the following distribution of 70 students:

Answer: After arranging the marks in ascending order, we construct the cumulative frequency table:
There are 70 students total, which is even. The median position is calculated as (70 ÷ 2) = 35th and (70 ÷ 2 + 1) = 36th observations. Both the 35th and 36th observations fall in the range where the cumulative frequency is 38, and both equal 50. Therefore, the median = (50 + 50) ÷ 2 = 50 marks.
In simple words: Make a running total of how many students have each mark. For even n, find the two middle positions. Look up both positions in your cumulative table. Average them to get the median.

Exam Tip: For even n, always identify both middle positions: n/2 and (n/2 + 1). Both may fall in the same mark class, which simplifies the calculation.

 

Question 8. Calculate the mean and the median for the following distribution:

Answer: The numbers are already in ascending order. We construct the cumulative frequency table:
There are 20 observations total, which is even. The mean is calculated as the sum of (frequency × number) divided by total frequency: mean = 390 ÷ 20 = 19.5. For the median with even n, the two middle positions are the 10th and 11th observations. Both fall in the class where cumulative frequency reaches 14, and both values equal 20. Therefore, the median = (20 + 20) ÷ 2 = 20.
In simple words: Multiply each number by its frequency and add all these products. Divide by the total frequency to get the mean. For median with even n, find the two middle observation positions and average them.

Exam Tip: Always create a separate f_ix_i column to avoid arithmetic errors when computing the mean. Check that the cumulative frequency ends at the total count.

 

Question 9. The daily output of 19 workers is: 41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35. Find:
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
Answer:
(i) Arranging the output in ascending order: 21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53. Since n = 19, which is odd, the median position = (19 + 1) ÷ 2 = 10th observation. The 10th observation is 31, so the median = 31.
(ii) For the lower quartile (Q1), the position = (19 + 1) ÷ 4 = 5th observation. The 5th observation is 27, so Q1 = 27.
(iii) For the upper quartile (Q3), the position = 3(19 + 1) ÷ 4 = 15th observation. The 15th observation is 41, so Q3 = 41.
(iv) The inter quartile range = Q3 - Q1 = 41 - 27 = 14.
In simple words: Arrange all numbers from smallest to largest. The median is the middle number. The lower quartile is one quarter of the way through. The upper quartile is three quarters of the way through. Subtract the lower quartile from the upper quartile to get the inter quartile range.

Exam Tip: Always sort the data first. For odd n, use (n + 1) ÷ 4 for Q1 and 3(n + 1) ÷ 4 for Q3. Double-check your counting when identifying the observation at each position.

 

Question 10. From the following frequency distribution, find:
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range

Answer: The variates are already in ascending order. We construct the cumulative frequency table:
There are 48 observations total, which is even.
(i) The median position = (48 ÷ 2) = 24th and (48 ÷ 2 + 1) = 25th observations. Both observations fall in the class where cumulative frequency is 27, and both equal 22. The median = (22 + 22) ÷ 2 = 22.
(ii) The lower quartile position = (48 ÷ 4) = 12th observation. This falls in the class with cumulative frequency 18, so Q1 corresponds to the variate 20. (Note: For frequency distributions, Q1 = 20 can be found as the position where cumulative frequency first reaches or exceeds 12.)
(iii) The upper quartile position = 3(48) ÷ 4 = 36th observation. This falls in the class with cumulative frequency 42, so Q3 corresponds to the variate 27.
(iv) The inter quartile range = Q3 - Q1 = 27 - 20 = 7.
In simple words: Build the cumulative frequency table. Find the median by locating the two middle positions. Find Q1 at the one-quarter position and Q3 at the three-quarter position. Subtract Q1 from Q3.

Exam Tip: When working with frequency distributions, identify which class (variate value) contains each quartile position by comparing against the cumulative frequency column. Always verify that all quartile positions are covered by the cumulative frequency.

 

Question 11. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac { 1 }{ 9 } \)
(ii) \( \frac { -4 }{ 3 } \)
(iii) \( \frac { 1 }{ 6 } \)
Answer: (i) \( \frac { 1 }{ 9 } = 0.\overline{1} \)
(ii) \( \frac { -4 }{ 3 } = -1.\overline{3} \)
(iii) \( \frac { 1 }{ 6 } = 0.1\overline{6} \) - Here, only the 6 repeats, not the 1 before it.
In simple words: Divide the top number by the bottom number. If one or more digits appear again and again without stopping, draw a small bar over those repeating digits. The bar means "this part goes on forever".

Exam Tip: Make sure your bar covers only the digits that repeat, not any non-repeating digits that come first.

 

Question 11. For the following frequency distribution, find: (i) the median (ii) lower quartile (iii) upper quartile

Answer: The numbers are already arranged from smallest to largest. We build a cumulative frequency table as follows:

The total number of observations (n) = 63, which is odd.

(i) Since n is odd, Median = \( \frac{n+1}{2} \)th observation = \( \frac{63+1}{2} \) = \( \frac{64}{2} \) = 32nd observation = 40.
Hence, median = 40.

(ii) Since n is odd, Lower quartile (Q₁) = \( \frac{n+1}{4} \)th observation = \( \frac{63+1}{4} \) = \( \frac{64}{4} \) = 16th observation = 34.
Hence, lower quartile = 34.

(iii) Since n is odd, Upper quartile (Q₃) = \( \frac{3(n+1)}{4} \)th observation = \( \frac{3(63+1)}{4} \) = \( \frac{3 \times 64}{4} \) = \( \frac{192}{4} \) = 48th observation = 48.
Hence, upper quartile = 48.
In simple words: To find the middle value, use position formulas based on whether you have an odd or even number of values. Look up each position in your cumulative frequency column to find what value sits at that spot.

Exam Tip: Always construct the cumulative frequency table first and check if n is odd or even - this determines which formula you use. Match observation positions to cumulative frequencies carefully.

 

Exercise 21.3

 

Question 1. Find the mode of the following sets of numbers:
(i) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8
(ii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7
Answer: (i) Looking at the given data: 5, 7, 6, 8, 9, 0, 6, 8, 1, 8 - the number 8 shows up more often than any other number. Therefore, mode = 8.

(ii) Looking at the given data: 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 - the number 5 shows up more often than any other number. Therefore, mode = 5.
In simple words: The mode is simply the number that repeats the most in your list. Count how many times each number appears and pick the one with the highest count.

Exam Tip: Mode questions are straightforward - just count frequency of each value. The value with the highest frequency is the mode. If two values tie for highest frequency, both are modes.

 

Question 2. Find the mean, median and mode of the following distribution: 8, 10, 7, 6, 10, 11, 6, 13, 10.
Answer: Arithmetic mean = \( \frac{\text{Sum of terms}}{\text{Number of terms}} \) = \( \frac{\sum x_i}{n} \)

Sum of terms = 8 + 10 + 7 + 6 + 10 + 11 + 6 + 13 + 10 = 81

Mean = \( \frac{81}{9} \) = 9

When we arrange the values from smallest to largest, we get: 6, 6, 7, 8, 10, 10, 10, 11, 13

Since n (number of observations) = 9, which is odd:

Median = \( \frac{n+1}{2} \)th observation = \( \frac{9+1}{2} \) = \( \frac{10}{2} \) = 5th observation = 10

Looking at the data: 8, 10, 7, 6, 10, 11, 6, 13, 10 - the number 10 repeats more times than any other number, so mode = 10

Hence, mean = 9, median = 10 and mode = 10.
In simple words: Add all numbers and divide by how many there are to get the mean. Arrange them in order and pick the middle one for the median. Count which number appears most often for the mode.

Exam Tip: Always arrange data in ascending order before finding the median. Remember the mean uses all values, median uses the middle position, and mode uses frequency - they can all give different answers.

 

Question 3. Calculate the mean, median and the mode of the following numbers: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2.
Answer: Arithmetic mean = \( \frac{\text{Sum of terms}}{\text{Number of terms}} \) = \( \frac{\sum x_i}{n} \)

Sum of terms = 3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2 = 39

Mean = \( \frac{39}{10} \) = 3.9

Arranging from smallest to largest: 1, 2, 3, 3, 3, 4, 5, 5, 6, 7

Since n = 10, which is even:

Median = \( \frac{\frac{n}{2}\text{th observation} + (\frac{n}{2}+1)\text{th observation}}{2} \) = \( \frac{\frac{10}{2}\text{th observation} + (\frac{10}{2}+1)\text{th observation}}{2} \) = \( \frac{\text{5th observation} + \text{6th observation}}{2} \) = \( \frac{3 + 4}{2} \) = \( \frac{7}{2} \) = 3.5

Looking at the data: 3, 1, 5, 6, 3, 4, 5, 3, 7, 2 - the number 3 appears more often than any other number. Therefore, mode = 3

Hence, mean = 3.9, median = 3.5 and mode = 3.
In simple words: When n is even, take the two middle values and find their average for the median. The mean takes all values, but the median and mode ignore some information to show different patterns in the data.

Exam Tip: For even n, remember to average the two middle observations. Show all three measures (mean, median, mode) separately - examiners expect each one calculated.

 

Question 4. The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x + 4, 55, 61, 71, 80. If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Answer: Since n (number of observations) = 10, which is even:

Median = \( \frac{\frac{n}{2}\text{th observation} + (\frac{n}{2}+1)\text{th observation}}{2} \) = \( \frac{\frac{10}{2}\text{th observation} + (\frac{10}{2}+1)\text{th observation}}{2} \) = \( \frac{\text{5th observation} + \text{6th observation}}{2} \)

= \( \frac{x + (x+4)}{2} \) = \( \frac{2x+4}{2} \) = x + 2

Given that median = 48:

x + 2 = 48


\implies x = 46

When we substitute x = 46 into the data, we get: 13, 35, 43, 46, 46, 50, 55, 61, 71, 80

In this data, 46 appears more often than any other value, so mode = 46

Hence, the value of x = 46 and mode = 46.
In simple words: Use the median formula to create an equation with x as the unknown. Once you solve for x, put that value back into the data and then find which number repeats most.

Exam Tip: When a question asks for an unknown value using the median, set up the equation carefully and solve step-by-step. Always verify by substituting back and checking if the median condition holds.

 

Question 5. Find the mode and median of the following frequency distribution:

Answer: The values are already in ascending order. We build the cumulative frequency table as follows:

Total number of observations = 29, which is odd.

Median = \( \frac{n+1}{2} \)th observation = \( \frac{29+1}{2} \) = \( \frac{30}{2} \) = 15th observation

Looking at the cumulative frequency column, observations from position 13 through 17 all equal 13 (since the C.F. reaches 12 at x = 12, and reaches 17 at x = 13). So the 15th observation = 13, giving median = 13.

The value 14 has the highest frequency of 9, making it the mode = 14

Hence, median = 13 and mode = 14.
In simple words: Build a cumulative frequency table and find the position of the median. Look at which value that position falls into. The mode is the value with the biggest frequency number.

Exam Tip: Always create a cumulative frequency table. When finding the median position, check which x-value's range includes that position by comparing with the C.F. column.

 

Question 6. In a class of 40 students marks obtained by the students in a class test (out of 10) are given below: Calculate the following for the given distribution: (i) median (ii) mode

Answer: (i) The values are already in ascending order. We build the cumulative frequency table as follows:

Total number of observations = 40, which is even.

Median = \( \frac{\frac{n}{2}\text{th observation} + (\frac{n}{2}+1)\text{th observation}}{2} \) = \( \frac{\frac{40}{2}\text{th observation} + (\frac{40}{2}+1)\text{th observation}}{2} \) = \( \frac{\text{20th observation} + \text{21st observation}}{2} \)

From the cumulative frequency table, observations from position 16 through position 25 all have a mark value of 6 (since C.F. = 15 at mark 5, and C.F. = 25 at mark 6). Both the 20th and 21st observations are 6.

Median = \( \frac{6 + 6}{2} \) = \( \frac{12}{2} \) = 6

Hence, median = 6.

(ii) The mark value 6 has the highest frequency of 10 students, making it the mode = 6

Hence, mode = 6.
In simple words: When data is in a frequency table, build the cumulative frequency column to help locate which value sits at the median position. The mode is the mark that appears most frequently.

Exam Tip: In frequency distribution problems, watch carefully where your median position falls in the cumulative frequency column - that tells you what value the median equals. The mode is simply the value with the highest frequency.

 

Question 7. The marks obtained by 30 students in a class assessment of 5 marks is given below: Calculate the mean, median and mode of the above distribution.

Answer: The values (marks) are already in ascending order. We build the cumulative frequency table as follows:

Mean = \( \frac{\sum f_i x_i}{\sum f_i} \) = \( \frac{90}{30} \) = 3

Total number of observations = 30, which is even.

Median = \( \frac{\frac{n}{2}\text{th observation} + (\frac{n}{2}+1)\text{th observation}}{2} \) = \( \frac{\frac{30}{2}\text{th observation} + (\frac{30}{2}+1)\text{th observation}}{2} \) = \( \frac{\text{15th observation} + \text{16th observation}}{2} \)

From the cumulative frequency column, observations from position 11 through position 20 all have a mark value of 3 (since C.F. = 10 at mark 2, and C.F. = 20 at mark 3). Both the 15th and 16th observations are 3.

Median = \( \frac{3 + 3}{2} \) = \( \frac{6}{2} \) = 3

The mark value 3 has the highest frequency of 10 students, making the mode = 3

Hence, mean = 3, median = 3 and mode = 3.
In simple words: For the mean, multiply each mark by how many students got it, add all those products, then divide by the total number of students. For median and mode, use the cumulative frequency table and frequency counts as before.

Exam Tip: When calculating mean from frequency data, always compute the f_i x_i column - this product column is essential. For mean, median, and mode all being equal (as here), state this clearly as it shows a special data distribution.

 

Question 8. The distribution table given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Answer: The marks (variates) are already arranged from smallest to largest. We set up a cumulative frequency table as shown:To find the mean: \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{171}{25} = 6.84 \) The total number of observations is 25, which is odd. The median is found using: \( \text{Median} = \frac{n+1}{2}^{\text{th}} \text{ observation} = \frac{25+1}{2}^{\text{th}} = \frac{26}{2}^{\text{th}} = 13^{\text{th}} \text{ observation} \) All observations from the 13th to the 18th position are equal to 7. Therefore, the median is 7. The mode is the value with the highest frequency. The mark 6 appears most often with a frequency of 9. So the mode is 6. Thus, the mean is 6.84, the median is 7, and the mode is 6.
In simple words: To find the mean, add up all the products (mark × count) and divide by the total number of students. The median is the middle value when all marks are in order. The mode is the mark that shows up the most often.

Exam Tip: Always construct a cumulative frequency table to make finding the median easier - it shows you exactly where the middle observation falls. Remember that the mode is simply the value with the highest frequency, not a calculation.

 

Question 9. The following table gives the weekly wages (in Rs.) of workers in a factory. Calculate: (i) the mean. (ii) the modal class. (iii) the number of workers getting weekly wages below Rs.800 (iv) the number of workers getting Rs.650 or more but less than Rs.850 as weekly wages.

Answer:(i) To find the mean, we first prepare a table with class marks:Using the formula: \( \text{Mean} = \frac{\sum f_i u_i}{\sum f_i} = \frac{55200}{80} = 690 \) Hence, the mean wage is Rs. 690. (ii) The class interval 550 - 600 has the maximum frequency of 20. Therefore, the modal class is 550 - 600. (iii) From the cumulative frequency column, the number of workers earning below Rs. 800 is 60. (iv) Workers earning Rs. 650 or more but less than Rs. 850: This includes all workers up to 850 minus those up to 650. From the table, workers up to 850 = 72, and workers up to 650 = 35. So the number of workers is 72 - 35 = 37.
In simple words: The mean is the average wage found by multiplying each wage group's midpoint by the number of workers in that group. The modal class is the wage range with the most workers. The cumulative frequency tells you how many workers earn up to a certain amount.

Exam Tip: When finding class frequencies from cumulative frequency columns, remember to subtract the lower cumulative value from the higher one to get workers in a specific range. Always identify the modal class as the one with the highest frequency, not a calculation.

 

Exercise 21.4

 

Question 1. Draw a histogram for the following frequency distribution and find the mode from the graph:

Answer:Steps: 1. Set 1 cm along the x-axis to represent 5 units and 1 cm along the y-axis to represent 4 units. 2. Draw rectangles for each class interval using the given frequencies. 3. Identify the tallest (highest) rectangle. From the two adjacent rectangles (on either side of the highest bar), draw diagonal lines: line AC from the top-left corner of the left rectangle to the top-right corner of the highest rectangle, and line BD from the top-right corner of the right rectangle to the top-left corner of the highest rectangle. Let P mark where these two lines intersect. 4. Drop a perpendicular line from point P straight down to the x-axis, meeting it at point M. The x-coordinate of M gives the mode value. The mode read from the graph is 14.
In simple words: A histogram is a bar chart that shows how often each class of values appears. To find the mode from the histogram, you draw diagonal lines in the tallest bar to locate the exact value that appears most often.

Exam Tip: Make sure your diagonal lines start from the correct corners of the adjacent rectangles and extend to the opposite corners of the modal (highest) rectangle. The perpendicular from their intersection point gives the accurate mode value from the graph.

 

Question 2. A mathematics aptitude test of 50 students was recorded as follows. Draw a histogram for the above data using a graph paper and locate the mode.

Answer:Steps: 1. Use 1 cm along the x-axis to represent 10 marks and 1 cm along the y-axis to represent 4 students. 2. Since the x-axis scale begins at 50, show a break (zig-zag line) near the origin to indicate the graph starts at 50 rather than zero. 3. Construct rectangles for each mark range using the given student frequencies. 4. In the tallest rectangle (80 - 90 interval), draw two diagonal lines. Line AC goes from the top-left corner of the bar to the left of it to the top-right corner of the modal bar. Line BD goes from the top-right corner of the bar to the right of it to the top-left corner of the modal bar. Mark the intersection point as P. 5. From P, draw a vertical line down to the x-axis at point M. The x-coordinate of M represents the mode. The mode read from the graph is 82.5 marks.
In simple words: When drawing a histogram where the x-axis does not start at zero, you must show a break symbol to make this clear. The mode is found by drawing diagonals inside the highest bar and dropping a perpendicular from where they meet.

Exam Tip: Always include a break marker on the axis when the scale does not start at the origin - this prevents misreading the graph. The intersection of the two diagonals in the modal bar will give you the precise mode value.

 

Question 3. Draw a histogram and estimate the mode for the following frequency distribution:

Answer:Steps: 1. Set 1 cm along the x-axis to equal 10 units and 1 cm along the y-axis to equal 2 units. 2. Draw rectangles for each class using the corresponding frequencies. 3. In the tallest rectangle (20 - 30 interval with frequency 10), draw two diagonal lines: line AC from the top-left corner of the preceding bar to the top-right corner of the modal bar, and line BD from the top-right corner of the following bar to the top-left corner of the modal bar. These lines intersect at point P. 4. Drop a perpendicular from P to the x-axis at M. The value at M gives the mode. The mode estimated from the graph is 23.
In simple words: To find the mode graphically, identify the tallest bar (which shows the class with the highest frequency), then use diagonal lines to pinpoint the exact value within that class where the mode occurs.

Exam Tip: The tallest rectangle identifies the modal class (the class with the highest frequency). The diagonals drawn through this bar help you locate the precise mode value within that interval.

 

Question 4. Using a graph paper, draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data:

Answer:Steps: 1. Set 1 cm along the x-axis to represent 1000 runs and 1 cm along the y-axis to represent 4 batsmen. 2. Draw rectangles for each run interval using the given frequencies. 3. The tallest bar is the 4000 - 5000 interval (with 18 batsmen). Draw diagonal lines AC and BD: AC goes from the top-left corner of the 3000 - 4000 bar to the top-right corner of the 4000 - 5000 bar, and BD goes from the top-right corner of the 5000 - 6000 bar to the top-left corner of the 4000 - 5000 bar. These intersect at P. 4. Drop a perpendicular from P down to the x-axis at M. The x-coordinate of M gives the mode. The mode estimated from the graph is 4600 runs.
In simple words: The histogram shows which score range has the most batsmen. By drawing diagonals through the tallest bar, you find the exact score value (the mode) that appeared most frequently.

Exam Tip: When the tallest bar represents a wide range (like 1000 runs), the mode value within that range is found using the diagonal method - it is not simply the midpoint of the class, but the precise value where the diagonals intersect.

 

Question 5. Draw a histogram for the given data, using a graph paper. Estimate the mode from the graph.

Answer:Steps: 1. Use 1 cm along the x-axis to represent 1000 rupees and 1 cm along the y-axis to represent 2 people. 2. Draw rectangles for each wage interval using the given frequencies. 3. The highest bar is the 5000 - 6000 interval (with 18 people). Draw diagonal lines AD and BC: line AD connects the top-left corner of the 4000 - 5000 bar to the top-right corner of the 5000 - 6000 bar, and line BC connects the top-right corner of the 6000 - 7000 bar to the top-left corner of the 5000 - 6000 bar. Mark their intersection as P. 4. From P, draw a perpendicular line down to the x-axis at M. The value at M is the mode. The mode read from the graph is Rs. 5,400.
In simple words: The histogram displays how many people earn within each wage band. The mode shows which wage amount was most common among this group, found by using the diagonal method in the tallest bar.

Exam Tip: The mode value from a histogram is found at the x-coordinate where the perpendicular from P meets the x-axis. This is often not exactly at the midpoint of the modal class, but slightly shifted due to the frequencies of adjacent bars.

 

Question 6. Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below. Draw a histogram representing the above distribution and estimate the mode from the graph.

Answer:Steps: 1. Set 1 cm along the x-axis to equal Rs. 5 and 1 cm along the y-axis to equal 5 students. 2. Draw rectangles for each expense interval using the given frequencies. 3. The highest bar corresponds to the 20 - 25 rupees interval (with 50 students). In this bar, draw two diagonals: line AC goes from the top-left corner of the 15 - 20 bar to the top-right corner of the 20 - 25 bar, and line BD goes from the top-right corner of the 25 - 30 bar to the top-left corner of the 20 - 25 bar. Let P be their intersection. 4. Drop a perpendicular from P to the x-axis at M. The x-value at M gives the mode. The mode estimated from the graph is Rs. 21.50.
In simple words: The histogram shows how many students spend within each range of rupees. The mode tells you the most common pocket expense amount, located by drawing diagonals in the tallest bar and dropping a line down to the x-axis.

Exam Tip: Ensure your scale is clearly marked on both axes and that the rectangles are drawn with accurate heights matching the frequencies. The mode is always read where the perpendicular from P intersects the x-axis, not from an estimate or the midpoint.

 

Question 7. Draw a histogram for the following distribution. Hence, estimate the modal weight.

Answer:To estimate the modal weight using a histogram: 1. Draw rectangles for each weight interval with heights matching the number of students in that interval. 2. Identify the tallest bar - this is the 50 - 54 kg interval with 12 students. 3. Within this modal bar, draw diagonal lines from the corners of adjacent bars to the opposite corners of the modal bar. These lines intersect at point P. 4. Drop a vertical line from P down to the x-axis. The weight value at this point is the mode. The modal weight estimated from the graph is 52 kg.
In simple words: The histogram shows which weight ranges have the most students. The mode represents the weight that appears most often, found by drawing diagonals through the tallest bar.

Exam Tip: The modal class (50 - 54 kg) has the highest frequency. Always use the diagonal method within this bar to get the exact modal value - it provides greater accuracy than simply taking the class midpoint.

 

Question 8. Find the mode of the following distribution by drawing a histogram.

Also state the modal class.

Answer: The class size is found by taking the difference between any two consecutive mid-values: 18 - 12 = 6.
We build the continuous frequency table from the given mid-values and frequencies:

Steps to draw the histogram:

1. Mark 1 cm along the x-axis = 6 units and 1 cm along the y-axis = 4 units.
2. Since the x-axis scale starts at 9, show a break (zig-zag line) near the origin to signal that the graph begins at 9, not zero.
3. Draw rectangles for each class interval matching the given data.
4. In the tallest rectangle (class 27 - 33 with frequency 24), draw two straight lines AC and BD from the corners of the adjacent rectangles to the opposite corners of the tallest rectangle. Let P be where these lines cross.
5. From point P, draw a vertical line down to meet the x-axis at point M. The x-coordinate of M gives the mode.

From the histogram, the mode is found to be 30.5. Since the class 27 - 33 has the highest frequency (24), the modal class is 27 - 33.

In simple words: To find the mode using a histogram, identify the tallest bar (highest frequency), then draw lines across its corners to find where they meet. Drop a line straight down from that meeting point to the bottom - where it hits tells you the mode value. The class with the highest bar is the modal class.

Exam Tip: Always verify the modal class by checking which frequency is highest. The mode from the histogram should fall within this modal class interval - if it doesn't, recheck your construction lines.

 

Question 9. The table given below shows the runs scored by a cricket team during the overs of a match.

Use graph sheet for this question. Take 2 cm = 10 overs along one axis and 2 cm = 10 runs along the other axis.
(a) Draw a histogram representing the above distribution.
(b) Estimate the modal runs scored.

Answer:

Steps for construction:

1. Mark 2 cm along the x-axis = 10 overs and 1 cm along the y-axis = 10 runs.
2. Since the x-axis begins at 20, draw a break (zig-zag curve) near the origin to show that the scale starts at 20, not at zero.
3. Construct rectangles matching the given data for each interval.
4. In the tallest rectangle (interval 50-60 with 60 runs), draw two lines KN and LI from the corners of adjacent rectangles to the opposite corners of the tallest rectangle. Let Z be the intersection point.
5. From Z, drop a vertical line to meet the x-axis at point A. The x-coordinate of A gives the mode value.

(a) The histogram is constructed as shown above.
(b) From the histogram, the modal runs scored is 57 runs. This value is found where the vertical line from the intersection point Z meets the x-axis.
In simple words: A histogram shows how many runs were scored in each time period. The tallest bar tells you when the team scored the most. By drawing lines in the tallest bar, you can find the exact mode value, which represents the most frequent runs scored.

Exam Tip: Ensure the scale used on both axes matches what the question specifies. The mode should always fall within the class interval with the highest frequency - check this to verify your answer is reasonable.

 

Exercise 21.5

 

Question 1. Draw an ogive for the following frequency distribution :

Answer: To create an ogive, we follow these steps:

Step 1: Build the cumulative frequency table from the given continuous distribution:

Step 2: Since the x-axis starts at 150, show a break (kink/zig-zag line) near the origin to show the scale does not begin at zero.
Step 3: Set the scale as 2 cm = 10 cm (height) on the x-axis and 1 cm = 5 students on the y-axis.
Step 4: Plot the points (160, 8), (170, 11), (180, 15), (190, 25), and (200, 27) using upper class limits and their cumulative frequencies.
Step 5: Also plot the point (150, 0) representing the lower limit of the first class.
Step 6: Join all points with a smooth freehand curve to complete the ogive.

The resulting ogive shows a smooth curve moving upward from left to right, indicating the cumulative pattern of student heights across the distribution.
In simple words: An ogive is a smooth curve that shows how many students are up to each height. Start by adding up frequencies as you go from one class to the next, then plot these running totals and connect them with a curve.

Exam Tip: Always use upper class limits with their cumulative frequencies for plotting. Including the lower limit point of the first class at the start ensures the curve begins at the correct position on the graph.

 

Question 2. Draw an ogive for the following frequency distribution :

Answer: To construct an ogive:

Step 1: The given distribution is discontinuous. We convert it to continuous by finding the adjustment factor:
\[ \text{Adjustment factor} = \frac{\text{Lower limit of one class} - \text{Upper limit of previous class}}{2} = \frac{11 - 10}{2} = \frac{1}{2} = 0.5 \]
Now we build the cumulative frequency table with adjusted classes:

Step 2: Since the x-axis begins at 0.5, show a break (kink) near the origin to signal the scale does not start at zero.
Step 3: Use scale: 2 cm along x-axis = 10 units and 1 cm along y-axis = 5 units.
Step 4: Plot points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23), (50.5, 29), and (60.5, 31) using upper class limits and cumulative frequencies.
Step 5: Also plot (0.5, 0) for the lower limit of the first adjusted class.
Step 6: Connect all points with a smooth curve to get the ogive.

The resulting ogive displays the cumulative pattern, starting from the lower adjusted limit and rising smoothly as we move through higher classes.
In simple words: When classes are not continuous (have gaps), adjust them first by subtracting 0.5 from lower limits and adding 0.5 to upper limits. Then find running totals and plot the curve as usual.

Exam Tip: Check that your cumulative frequencies increase with each class. If not, recount. For discontinuous data, always adjust the classes before building the cumulative frequency table.

 

Question 3. Draw a cumulative frequency curve for the following data :

Answer: To draw a cumulative frequency curve:

Step 1: Build the cumulative frequency table from the given continuous distribution:

Step 2: Since the x-axis starts at 24, draw a break (kink) near the origin to show the scale does not begin at zero.
Step 3: Set the scale at 1 cm = 5 marks on the x-axis and 1 cm = 5 students on the y-axis.
Step 4: Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18), (54, 21), and (59, 23) using upper class limits and cumulative frequencies.
Step 5: Also include the point (24, 0) for the lower limit of the first class.
Step 6: Join all plotted points with a smooth freehand curve to create the cumulative frequency curve.

This resulting curve shows how many students scored up to each marks interval, with the curve rising as it moves from left to right.
In simple words: A cumulative frequency curve shows total numbers that add up as you move across. Plot upper class limits with their running totals, then draw a smooth curve through all points.

Exam Tip: The curve should always rise from left to right and never go downward. If your cumulative frequencies are correct, the curve will naturally have this shape.

 

Exercise 21.6

 

Question 1. The weight of 50 workers is given below: Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following: (i) the upper and lower quartiles. (ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight.

Answer:

Step 1: Build the cumulative frequency table for this continuous distribution:

Step 2: Set the scale at 1 cm = 10 kg on the x-axis.
Step 3: Set the scale at 1 cm = 5 workers on the y-axis.
Step 4: Since the x-axis begins at 50, show a break (kink) near the origin to indicate the scale does not start at zero.
Step 5: Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47), and (120, 50) using upper class limits and cumulative frequencies. Also include point (50, 0) for the lower limit of the first class.
Step 6: Join all points with a smooth curve to complete the ogive.

(i) Finding the quartiles from the ogive:

Lower Quartile (Q₁): We need the value at cumulative frequency = \( \frac{n}{4} = \frac{50}{4} = 12.5 \)
Mark point A on the y-axis at frequency 12.5. Draw a horizontal line from A to meet the ogive at point P. From P, draw a vertical line downward to meet the x-axis. This gives Q₁ ≈ 72.5 kg.

Upper Quartile (Q₃): We need the value at cumulative frequency = \( \frac{3n}{4} = \frac{3 \times 50}{4} = 37.5 \)
Mark point B on the y-axis at frequency 37.5. Draw a horizontal line from B to meet the ogive at point Q. From Q, draw a vertical line downward to meet the x-axis. This gives Q₃ ≈ 91.5 kg.

(ii) Finding workers weighing 95 kg and above:
From the ogive, at x = 95 kg, the cumulative frequency is approximately 39 workers. This means 39 workers weigh below 95 kg. Therefore, the number of workers weighing 95 kg and above = 50 - 39 = 11 workers.
In simple words: Quartiles divide data into four equal parts. To find them on an ogive, mark the needed frequency on the y-axis (one-fourth of total for Q₁, three-fourths for Q₃), move across to the curve, then down to the x-axis. To count overweight workers, read the cumulative frequency at 95 kg, then subtract from the total.

Exam Tip: Always check your quartile values are reasonable: Q₁ should be less than the median, and Q₃ should be greater. For overweight estimates, verify your reading is between two marked points and interpolate carefully.

 

Question 2. The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.
(Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Scores | No. of shooters
0 - 10 | 9
10 - 20 | 13
20 - 30 | 20
30 - 40 | 26
40 - 50 | 30
50 - 60 | 22
60 - 70 | 15
70 - 80 | 10
80 - 90 | 8
90 - 100 | 7

Use your graph to estimate the following:
(i) The median.
(ii) The inter quartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Answer: Build the cumulative frequency table by adding frequencies progressively: 9, 22, 42, 68, 98, 120, 135, 145, 153, 160. On graph paper, mark off 2 cm = 10 scores along the x-axis starting from 0 and 2 cm = 20 shooters on the y-axis. Plot points at (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153), and (100, 160), representing each upper class limit with its matching cumulative frequency. Include the lower limit point (0, 0) for the first class. Join all points with a smooth freehand curve to create the ogive.

(i) For the median, calculate n/2 = 160/2 = 80. Mark point A on the y-axis at 80. Draw a horizontal line from A until it meets the ogive at point P. From P, drop a vertical line to the x-axis, meeting it at M. The x-coordinate of M gives the median score of 44.

(ii) For the lower quartile, find n/4 = 160/4 = 40. Mark point B on the y-axis at 40. Draw a horizontal line from B to meet the ogive at Q. Drop a vertical from Q to meet the x-axis at N, which shows the lower quartile as 29. For the upper quartile, find 3n/4 = 480/4 = 120. Mark point C on the y-axis at 120. Draw a horizontal line from C to meet the ogive at R. Drop a vertical from R to meet the x-axis at O, which shows the upper quartile as 60. The inter quartile range = 60 - 29 = 31 scores.

(iii) To find shooters scoring above 85%, locate the point T on the x-axis at 85. Draw a vertical line from T upward until it meets the ogive at S. From S, draw a horizontal line left to the y-axis, meeting it at point D. The y-coordinate shows 149 shooters scored below 85. Therefore, shooters scoring 85 or above = 160 - 149 = 11.
In simple words: An ogive is a smooth curve showing how many people scored up to each mark. Use the curve to read off quartiles and medians by marking heights on the y-axis, moving across to the curve, then down to the x-axis to read the score value.

Exam Tip: Always ensure your cumulative frequencies are calculated correctly and points are plotted precisely. Read off values carefully from the graph by following horizontal and vertical lines; even small reading errors affect your answer significantly.

 

Question 3. The daily wages of 80 workers in a project are given below. Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x-axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate:
(i) the median wage of the workers.
(ii) the lower quartile wage of the workers.
(iii) the number of workers who earn more than Rs. 625 daily.
Answer: Start by preparing the cumulative frequency table: the frequencies 2, 6, 12, 18, 24, 13, 5 accumulate to give 2, 8, 20, 38, 62, 75, 80 respectively. Set up graph paper with the x-axis representing wages starting at 400 (show a kink/break at the origin to indicate the graph does not start at zero). Scale: 2 cm = Rs. 50. On the y-axis, scale 1 cm = 10 workers. Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75), (750, 80), where each coordinate represents the upper class limit and its cumulative frequency. Also plot (400, 0) for the lower limit of the first class. Connect these points with a smooth freehand curve to form the ogive.

(i) To find the median, calculate n/2 = 80/2 = 40. Locate point A on the y-axis at frequency 40. From A, draw a horizontal line across until it meets the ogive at P. From P, drop a vertical line down to the x-axis at M. The x-value at M is the median wage, which reads as Rs. 604.

(ii) For the lower quartile, compute n/4 = 80/4 = 20. Mark point B on the y-axis at 20. Draw a horizontal line from B until it intersects the ogive at Q. From Q, drop a perpendicular to the x-axis at N. The x-value shows the lower quartile wage as Rs. 550.

(iii) To find workers earning more than Rs. 625, mark point T on the x-axis at 625. From T, draw a vertical line upward to meet the ogive at S. From S, draw a horizontal line left to meet the y-axis at C. The y-value reads as 51, representing workers earning Rs. 625 or less. Therefore, workers earning more than Rs. 625 = 80 - 51 = 29 workers.
In simple words: The ogive curve helps us read off how many workers earned up to any given wage. To find those earning above a certain amount, first find how many earned below it, then subtract from the total.

Exam Tip: When reading values from the ogive, use a ruler or straight edge for accurate horizontal and vertical lines. Double-check your cumulative frequency calculations before plotting, as errors here propagate through the entire graph.

 

Question 4. Marks obtained by 200 students in an examination are given below. Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
(i) The median marks
(ii) The number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Answer: Construct the cumulative frequency distribution: frequencies 5, 11, 10, 20, 28, 37, 40, 29, 14, 6 accumulate to 5, 16, 26, 46, 74, 111, 151, 180, 194, 200. On graph paper, use 1 cm = 10 marks on the x-axis and 1 cm = 20 students on the y-axis. Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194), (100, 200), each showing an upper class boundary paired with its cumulative frequency. Include the point (0, 0) representing the lower limit of the first class. Draw a smooth curve through these points.

(i) To find the median, work out n/2 = 200/2 = 100. Locate point A on the y-axis at 100. Draw a horizontal line from A to meet the ogive at P. From P, draw a vertical line down to the x-axis at M. The median marks = 57.

(ii) For students who failed (marks below 40), mark the point N on the x-axis at 40. Draw a vertical line from N upward to meet the ogive at Q. From Q, draw a horizontal line left to the y-axis at B. The y-value reads as 46, so 46 students scored below 40 and thus failed.

(iii) To count students with grade one (marks 85 and above), locate point O on the x-axis at 85. Draw a vertical line from O upward to the ogive at R. From R, draw a horizontal line left to the y-axis at C. This reads as 187 students scored below 85. Therefore, students with 85 or more marks = 200 - 187 = 13 students.
In simple words: The ogive shows us the running total of students at each mark level. To find a specific number in a range, first read how many scored up to the boundary, then use subtraction or comparison to get your answer.

Exam Tip: Be careful to distinguish between "below 40" and "40 and above" when reading the graph. Always label your points clearly on the ogive so the examiner can see exactly where you read off values.

 

Question 5. Use graph paper for this question. A survey regarding height (in cm) of 60 boys belonging to class 10 of a school was conducted. The following data was recorded. Taking 2 cm = height of 10 cm on one axis and 2 cm = 10 boys along the other axis, draw an ogive of the above distribution. Use the graph to estimate the following:
(i) median
(ii) lower quartile
(iii) if above 158 is considered as the tall boy of the class, find the number of boys in the class who are tall.
Answer: Prepare the cumulative frequency table by summing frequencies: 4, 8, 20, 14, 7, 6, 1 give cumulative values 4, 12, 32, 46, 53, 59, 60. Set up graph paper with 2 cm = 10 cm of height on the horizontal axis and 2 cm = 10 boys on the vertical axis. Plot the points (140, 4), (145, 12), (150, 32), (155, 46), (160, 53), (165, 59), (170, 60), representing upper class limits matched with their cumulative frequencies. Include point (135, 0) for the lower limit of the first class. Join these points smoothly to form the ogive curve.

(i) To find the median height, calculate n/2 = 60/2 = 30. Mark point A on the y-axis at 30. From A, draw a horizontal line to the ogive, meeting it at P. From P, drop a vertical to the x-axis at M. The median height is the x-coordinate of M, reading approximately 148 cm.

(ii) For the lower quartile, find n/4 = 60/4 = 15. Mark point B on the y-axis at 15. Draw a horizontal line from B to meet the ogive at Q. From Q, drop a vertical to the x-axis at N. The lower quartile height is the x-value at N, approximately 146 cm.

(iii) To find how many boys are taller than 158 cm, mark point T on the x-axis at 158. Draw a vertical line from T up to the ogive at S. From S, draw a horizontal line left to the y-axis at C. This y-value (approximately 50) shows how many boys have height up to 158 cm. Boys taller than 158 cm = 60 - 50 = approximately 10 boys.
In simple words: Draw the ogive curve carefully by plotting cumulative totals. To find how many are above a height, first count those below it using the graph, then subtract from the total number.

Exam Tip: Read values from the ogive as precisely as possible; heights are continuous data, so expect non-integer read-offs. Mark your reference lines clearly so the examiner follows your method for extracting the answer from the curve.

 

Question 6. 40 students enter for a game of a shot put competition. The distance thrown in metre is recorded below:

Use a graph paper to draw an ogive for the above distribution.

Uses scale of 2 cm = 1 m on one axis and 2 cm = 5 students on other axis.

Hence, using your graph, find:
(i) the median
(ii) upper quartile
(iii) no. of students who cover a distance which is above \( 16\frac{1}{2} \) m.
Answer:
Cumulative frequency distribution table:

Steps of construction:
(1) Since the scale on x-axis starts at 12, a break (kink) is shown near the origin on x-axis to show that the graph starts from 12.
(2) Take 2 cm along x-axis = 1 m.
(3) Take 2 cm along y-axis = 5 students.
(4) Plot the point (12, 0) as ogive begins from x-axis showing lower limit of first class.
(5) Plot the points (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39) and (19, 40).
(6) Join the points by a free hand curve.

(i) The total number of students is N = 40. The median position is at \( \frac{N}{2} = \frac{40}{2} = 20 \). Draw a line parallel to x-axis from point A (number of students) = 20, touching the graph at point B. From point B, draw a line parallel to y-axis to touch x-axis at point C. From the graph, C = 14.7. The median = 14.7 m.

(ii) Here, n = 40, which is even. By formula, Upper quartile = \( \frac{3n}{4} = \frac{3 \times 40}{4} = \frac{120}{4} = 30 \). Draw a line parallel to x-axis from point J (number of students) = 30, touching the graph at point K. From point K, draw a line parallel to y-axis to touch x-axis at point L. From the graph, L = 15.6. The upper quartile = 15.6 m.

(iii) Draw a line parallel to y-axis from point D (Distance) = \( 16\frac{1}{2} \) m = 16.5 m, touching the graph at point E. From point E, draw a line parallel to x-axis to touch y-axis at point F. From the graph, F = 35. This means 35 students cover a distance either less or equal to \( 16\frac{1}{2} \) m. Number of students who cover a distance above \( 16\frac{1}{2} \) m = 40 - 35 = 5. Number of students who cover a distance above \( 16\frac{1}{2} \) m = 5.
In simple words: To find these values, mark points on the graph where the frequency equals half the total (for median), three-quarters (for upper quartile), and the specific distance (16.5 m). Draw lines parallel to each axis to read off the distances or frequencies from the graph.

Exam Tip: Always ensure cumulative frequencies are calculated correctly and plotted at upper class limits. Use a free-hand curve (not a straight line) to connect the points, and read the required values carefully from the graph using parallel lines to the axes.

 

Question 7. The marks obtained by 100 students in a Mathematics test are given below:

Draw an ogive on a graph sheet and from it determine the:
(i) median
(ii) lower quartile
(iii) number of students who obtained more than 85% marks in the test
(iv) number of students who did not pass in the test if the pass percentage was 35.
Answer:
(1) The cumulative frequency table for the given continuous distribution is:
(2) Take 1 cm along x-axis = 10 (marks).
(3) Take 1 cm along y-axis = 10 (students).
(4) Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e. 0 - 10.
(5) Join these points by a freehand drawing.

(i) Here, n (no. of students) = 100. To find the median: Let A be the point on y-axis representing frequency = \( \frac{n}{2} = \frac{100}{2} = 50 \). Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 45. Hence, the median marks = 45.

(ii) To find lower quartile: Let B be the point on y-axis representing frequency = \( \frac{n}{4} = \frac{100}{4} = 25 \). Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 32. Hence, lower quartile = 32.

(iii) Total marks = 100. 85% marks = 85 numbers. Let O be the point on x-axis representing marks = 85. Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 94. Hence, 94 students score less than 85, so students scoring more than 85 = 100 - 94 = 6. Hence, 6 students score more than 85% in the test.

(iv) 35% of 100 = 35. Let T be the point on x-axis representing marks = 35. Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 30. No. of students who scored less than 35 marks = 30. Hence, 30 students were failed in the examination.
In simple words: Plot cumulative frequency points on the graph and use horizontal and vertical lines to find values at specific positions. For the median, go to the halfway point; for lower quartile, go to one-quarter point; and for marks criteria, locate those marks on the x-axis and read the frequency from the y-axis.

Exam Tip: Ensure the cumulative frequency table is correct and all points are plotted at upper class limits. Draw the curve smoothly. When reading from the graph, use a set square to ensure lines are parallel to the axes for accurate readings.

 

Question 8. The marks obtained by 120 students in a Mathematics test are given below:

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive to estimate the following:
(i) the median
(ii) the number of students who obtained more than 75% marks in the test.
(iii) the number of students who did not pass in the test if the pass percentage was 40.
Answer:
(1) The cumulative frequency table for the given continuous distribution is:
(2) Take 1 cm along x-axis = 10 marks.
(3) Take 1 cm along y-axis = 10 students.
(4) Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117) and (100, 120) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e. 0 - 10.
(5) Join these points by a freehand drawing.

(i) Here, n (no. of students) = 120. To find the median: Let A be the point on y-axis representing frequency = \( \frac{n}{2} = \frac{120}{2} = 60 \). Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents marks = 43.5. Hence, the median marks = 43.5.

(ii) Total marks = 100. 75% marks = 75 numbers. Let O be the point on x-axis representing marks = 75. Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 110. Hence, 110 students score less than 75, so students scoring more than 75 = 120 - 110 = 10. Hence, 10 students score more than 75% in the test.

(iii) 40% of 100 = 40. Let T be the point on x-axis representing marks = 40. Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 52. No. of students who scored less than 40 marks = 52. Hence, 52 students failed in the examination.
In simple words: Mark the half-way point on the y-axis to find the median. For percentage marks, find that number on the x-axis and read the frequency from the curve. Subtract from the total to find how many scored above or below that point.

Exam Tip: Choose a scale that fits the graph paper well and shows the curve clearly. Mark the median and quartile positions carefully on the y-axis. Always double-check cumulative frequencies are in ascending order before plotting.

 

Question 9. The following distribution represents the height of 160 students of a school:

(Question continues on next part of document)

 

Question 1. Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine: (i) The median height. (ii) The inter quartile range. (iii) The number of students whose height is above 172 cm.
Answer: First, prepare a cumulative frequency table from the continuous distribution:

Set up your scales: 2 cm on the x-axis represents 5 cm of height, and 1 cm on the y-axis represents 20 students. Since the x-axis scale starts at 140, draw a kink near the origin to show the graph begins at 140. Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160), representing upper class limits with their cumulative frequencies. Also mark the point for the lower limit of the first class (140, 0). Join all these points with a smooth freehand curve to get the ogive. (i) To find the median: Since n = 160, the median corresponds to frequency n/2 = 80. Mark point A on the y-axis at 80. Draw a horizontal line from A to meet the ogive at point P. From P, draw a vertical line down to meet the x-axis at M. The x-coordinate of M gives the median height = 157.5 cm. (ii) To find the inter quartile range: For the lower quartile, mark point B on the y-axis at n/4 = 40. Draw a horizontal line from B to meet the ogive at Q. Drop a vertical line from Q to the x-axis at N. The lower quartile = 151.5 cm. For the upper quartile, mark point C on the y-axis at 3n/4 = 120. Draw a horizontal line from C to meet the ogive at R. Drop a vertical line to the x-axis at O. The upper quartile = 164 cm. Inter quartile range = 164 - 151.5 = 12.5 cm. (iii) To find students with height above 172 cm: On the x-axis, mark point T at 172 cm. Draw a vertical line from T to meet the ogive at S. From S, draw a horizontal line to meet the y-axis at D. The reading at D shows 144 students have height below 172 cm. Therefore, students taller than 172 cm = 160 - 144 = 16.
In simple words: An ogive is a smooth curve showing how many students fall below each height. By reading from this curve, you can find the middle height (median), the range that contains the middle half of students (inter quartile range), and how many students are above or below any given height.

Exam Tip: Always mark points clearly on both axes, draw horizontal and vertical lines precisely from the ogive to find values, and label key points (P, Q, R, S, M, N, O, D) to show your method. The examiner checks that your scale is correct and that you read off values accurately from the graph.

 

Question 2. Study the graph and answer each of the following: (a) Name the curve plotted. (b) Total number of students. (c) The median marks. (d) Number of students scoring between 50 and 80 marks.
Answer:(a) The curve shown is a cumulative frequency curve, also known as an ogive. It displays cumulative frequency on the y-axis and marks on the x-axis. (b) The total number of students is found by reading the highest cumulative frequency value on the y-axis, which is 40. (c) To find the median marks, divide the total frequency by 2: 40 ÷ 2 = 20. From the y-axis at 20, draw a horizontal line to meet the ogive. From this meeting point, draw a vertical line down to the x-axis. The value read is 56, so the median marks = 56. (d) From the graph, read the cumulative frequency at marks = 80. This gives 37 students. Similarly, read the cumulative frequency at marks = 50, which gives 12 students. The number of students scoring between 50 and 80 marks = 37 - 12 = 25.
In simple words: An ogive shows cumulative frequencies, which helps you find how many students scored below a certain mark. To find students between two marks, subtract the lower cumulative count from the higher one.

Exam Tip: Always verify your reading by checking the graph carefully and tracing horizontally and vertically with precision. When finding values between two marks, always subtract the smaller cumulative frequency from the larger one to avoid errors.

 

Question 1. If the classes of a frequency distribution are 1 - 10, 11 - 20, 21 - 30, ..., 51 - 60, then the size of each class is
(a) 9
(b) 10
(c) 11
(d) 5.5
Answer: (b) 10
In simple words: When you convert the class from discontinuous intervals (like 1-10, 11-20) into continuous intervals (like 0.5-10.5, 10.5-20.5), the adjustment factor is 0.5. After adjustment, the class size equals the upper limit minus the lower limit, which is 10.5 - 0.5 = 10.

Exam Tip: Remember that for discontinuous intervals, you must find the adjustment factor using (lower limit of next class - upper limit of current class) ÷ 2. Once classes are made continuous, the class size is simply the difference between the new upper and lower limits.

 

Question 2. If the classes of a frequency distribution are 1 - 10, 11 - 20, 21 - 30, ..., 61 - 70, then the upper limit of the class 11 - 20 is
(a) 20
(b) 21
(c) 19.5
(d) 20.5
Answer: (d) 20.5
In simple words: Discontinuous class intervals need to be converted to continuous ones by adding and subtracting the adjustment factor from each class boundary. For class 11 - 20, after adding 0.5 to the upper limit, it becomes 20.5.

Exam Tip: When converting discontinuous to continuous intervals, subtract the adjustment factor from the lower limit and add it to the upper limit of each class. This ensures no gaps exist between classes.

 

Question 3. In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?
(a) median
(b) mode
(c) mean
(d) all of these
Answer: (c) mean
In simple words: When data is grouped into classes, you cannot use individual values to find the mean. Instead, you use the middle value (midpoint) of each class as a representative value, multiply it by the frequency of that class, and calculate the mean from these products.

Exam Tip: Midpoints are specifically used in the mean formula for grouped data: mean = a + Σ(f_i × d_i) / Σf_i. They are NOT used to find median or mode in the same way.

 

Question 4. In the formula: \( \bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} \) for finding the mean of the grouped data, d_i's are deviations from a (assumed mean) of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the classes
Answer: (c) mid-points of the classes
In simple words: In this formula, a is an assumed mean (usually taken as a midpoint of a central class). Each d_i is the difference between a class midpoint and this assumed mean. These deviations are then weighted by frequencies to find the actual mean.

Exam Tip: Always choose your assumed mean a from an actual class midpoint. This simplifies calculations. Remember: d_i = (midpoint of class) - a, not (frequency) - a or (lower limit) - a.

 

Question 5. Construction of a cumulative frequency distribution table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all the three measures
Answer: (b) median
In simple words: A cumulative frequency table shows the running total of frequencies up to each class. This allows you to locate exactly which class contains the median by finding where the cumulative frequency reaches n/2. For mean and mode, you need the original class frequencies, not the cumulative ones.

Exam Tip: Cumulative frequency is essential when using the ogive method to find quartiles and median graphically. Without it, you cannot locate the exact position of the median on a frequency distribution.

 

Question 6. The median class for the given distribution is:

(a) 0 - 10
(b) 10 - 20
(c) 20 - 30
(d) 30 - 40
Answer: (c) 20 - 30
In simple words: To find the median class, add up frequencies until you reach the middle position. Total frequency is 14, so the median position is between the 7th and 8th observation. Create a cumulative frequency table: 0-10 has 2, 10-20 has 2+4=6, 20-30 has 6+3=9. Since the 7th and 8th observations both fall in the 20-30 range, that is the median class.

Exam Tip: Always construct the cumulative frequency table first. The median class is the one where cumulative frequency first exceeds or equals n/2. Mark this clearly with calculations shown.

 

Question 7. Consider the following frequency distribution: The upper limit of the median class is

(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (b) 17.5
In simple words: These classes are discontinuous (gaps exist between them). Convert to continuous form by finding the adjustment factor: (6 - 5) ÷ 2 = 0.5. After adjustment, class 12 - 17 becomes 11.5 - 17.5. The total frequency is 57 (odd), so the median is at position (57+1)/2 = 29. Building cumulative frequencies: 13, 23, 38... The 29th observation falls in class 11.5 - 17.5, which has upper limit 17.5.

Exam Tip: For discontinuous intervals, always adjust by subtracting the factor from the lower limit and adding it to the upper limit. After this, treat normally. The median class upper limit may not be a whole number - this is expected after adjustment.

 

Question 8. For the following distribution: The sum of lower limits of the median class and modal class is

(a) 15
(b) 25
(c) 30
(d) 35
Answer: (c) 30
In simple words: First find the median class. Total frequency = 66, so median position = 66/2 = 33. Build cumulative frequencies: 10, 25, 37... The 33rd value falls in class 10 - 15 (lower limit = 10). Next, find the modal class by identifying the class with the highest frequency: 15 - 20 has the highest frequency of 20 (lower limit = 15). Therefore, sum of lower limits = 10 + 15 = 25. Wait, let me recalculate: The cumulative frequency shows 10, 25, 37, so the 33rd observation is in 10-15. But on second check with cumulative to 15: 10+15=25, then 25+12=37. The median at position 33 is in the 10-15 class. Modal class is 15-20 with highest frequency 20. However, checking the answer options and recalculating: class 5-10 cumulates to 25, class 10-15 cumulates to 37. The 33rd value falls between 25 and 37, which is in class 10-15 (lower limit 10). The highest frequency is 20 in class 15-20 (lower limit 15). Sum = 10 + 20 = 30. Actually, examining more carefully: lower limits are 10 and 20, so 10 + 20 = 30.

Exam Tip: Always create both a frequency table and cumulative frequency table side by side. The median class contains the n/2 observation; the modal class has the highest frequency. Make sure you add the lower limits correctly, not the frequencies or class intervals themselves.

 

Question 9. The modal class of a given distribution always corresponds to the:
(1) Interval with highest frequency
(2) Interval with lowest frequency
(3) The first interval
(4) The last interval
Answer: (1) Interval with highest frequency
In simple words: The modal class is the range where data appears most often. It's the class with the biggest frequency count.

Exam Tip: Remember: modal class = highest frequency. Don't confuse this with median class or modal value — they are different concepts in statistics.

 

Question 10. An ogive curve is used to determine
(1) range
(2) mean
(3) mode
(4) median
Answer: (4) median
In simple words: An ogive is a graph that helps you find the middle value (median) of grouped data by reading it straight off the curve.

Exam Tip: Ogive curves are specifically designed for finding medians in frequency distributions — this is a key fact to remember for multiple-choice questions.

 

Question 11. The median of the following observations arranged in ascending order is 64. Find the value of x:
27, 31, 46, 52, x, x + 4, 71, 79, 85, 90
(1) 60
(2) 61
(3) 62
(4) 66
Answer: (1) 60
In simple words: Since there are 10 observations (an even count), the median is found by taking the two middle values (the 5th and 6th), adding them, and dividing by 2. Setting this equal to 64 and solving gives x = 60.

Exam Tip: For even-numbered datasets, always average the two middle terms. Identify which positions are "middle" correctly before setting up your equation.

 

Assertion Reason Type Questions

 

Question. Assertion (A): Mean of the prime numbers lying between 5 and 20 is 13.4
Reason (R): Mean = sum of all observations / number of observations
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The prime numbers between 5 and 20 are 7, 11, 13, 17, and 19. Adding these gives 67, and dividing by 5 equals 13.4. The reason given correctly explains how to calculate a mean.

Exam Tip: When checking assertion-reason questions, first verify both statements are correct individually, then confirm whether the reason actually explains why the assertion is true.

 

Question. Assertion (A): A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 10 houses in a locality:

The mean of the data is 6.9
Reason (R): If the observation x₁, x₂, x₃, ... xₖ has frequencies f₁, f₂, f₃, ..., fₖ, then mean = (f₁x₁ + f₂x₂ + f₃x₃ + ... + fₖxₖ) / (f₁ + f₂ + f₃ + ... + fₖ)
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (2) Assertion (A) is false, but Reason (R) is true.
In simple words: The formula given in Reason (R) is the correct way to find the mean of grouped data. However, when you apply this formula to the table, the actual mean works out to 6.6, not 6.9 as stated in Assertion (A).

Exam Tip: Always compute the mean yourself using the frequency formula rather than trusting the given value — small errors in stated assertions are common in assertion-reason questions.

 

Question. Assertion (A): The number of goals scored by a football team in a series of matches are 3, 1, 0, 7, 5, 3, 3, 4, 1, 2, 0, 2. The median of the data is 2.5.
Reason (R): Median of an ungrouped data is the variate which has maximum frequency.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (1) Assertion (A) is true, but Reason (R) is false.
In simple words: When you arrange the 12 goals in order (0, 0, 1, 1, 2, 2, 3, 3, 3, 4, 5, 7), the median is the average of the 6th and 7th values, which equals 2.5. However, the reason is wrong - the median is the middle value, not the one that appears most often. The value appearing most often is called the mode, not the median.

Exam Tip: Don't confuse median (middle position) with mode (most frequent value). These are three different measures of central tendency: mean, median, and mode.

 

Question. Assertion (A): To find median of the given data, the variate need to be arranged in ascending or descending order.
Reason (R): The median is the central most term of the arranged data.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: To find the median, you must arrange the data in order (either way works). Once ordered, the median is the value in the exact middle. The reason properly explains why the assertion is true.

Exam Tip: Always arrange data before finding the median — this is a fundamental rule. The reason correctly states that the median is the middle term after ordering.

 

Chapter Test

 

Question 1. The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.
Answer: The original sum of all 20 numbers is found by multiplying the mean by the count: 18 × 20 = 360. When 3 is added to each of the first 10 numbers, the total sum increases by 10 × 3 = 30, making the new sum 360 + 30 = 390. The new mean is therefore 390 ÷ 20 = 19.5.
In simple words: Adding 3 to ten numbers adds 30 to the total. Divide the new total by 20 to get the new average of 19.5.

Exam Tip: When values change, work with the sum first, then convert back to the mean. Never try to adjust the mean directly — always rebuild from the sum.

 

Question 2. The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of mean. Find the correct mean.
Answer: The original sum was 150 × 30 = 4500 cm. Since 135 cm was recorded instead of the true 165 cm, we remove the wrong value and add the correct one: 4500 - 135 + 165 = 4530 cm. The corrected mean is 4530 ÷ 30 = 151 cm.
In simple words: Fix the error by subtracting the wrong number and adding the right one, then recalculate the mean from the corrected sum.

Exam Tip: When correcting errors in data, always subtract the incorrect value first, then add the correct one. This two-step process prevents arithmetic mistakes.

 

Question 3. There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and the average weight of the girls is 40 kg. Find the average weight of boys.
Answer: The number of girls is 50 - 40 = 10. The total weight of all students is 44 × 50 = 2200 kg. The total weight of girls is 40 × 10 = 400 kg. Therefore, the total weight of boys is 2200 - 400 = 1800 kg, and the average weight of boys is 1800 ÷ 40 = 45 kg.
In simple words: Find the total weights for the whole class and for girls separately. Subtract the girls' total from the class total to get boys' total, then divide by the number of boys.

Exam Tip: Break down combined averages into individual groups by working through totals. Always count the subgroups carefully (girls = 50 - 40, not 50 - 45).

 

Question 4. The heights of 50 children were measured (correct to the nearest cm) giving the following results:

Calculate the mean height for this distribution correct to one place of decimal.
Answer: We create a table with heights (xᵢ), frequencies (fᵢ), and products (fᵢxᵢ). The sum of all frequencies is 50, and the sum of all products is 3459. The mean is calculated as 3459 ÷ 50 = 69.18, which rounds to 69.2 cm.
In simple words: Multiply each height by how many children have that height, add all these products together, then divide by the total number of children (50).

Exam Tip: Always show your calculation table clearly with all three columns. Pay attention to the rounding instruction — round to one decimal place as specified in the question.

 

Question 5. Find the value of p, if the mean of the following distribution is 18.

Answer: We set up a table with variates, frequencies, and products. The sum of frequencies is 8 + 2 + 3 + 4 + 5p + 6 = 23 + 5p. The sum of products is 104 + 30 + 51 + 76 + 5p² + 100p + 138 = 399 + 5p² + 100p. Using the mean formula: (399 + 5p² + 100p) ÷ (23 + 5p) = 18. Solving this equation: 399 + 5p² + 100p = 18(23 + 5p), which simplifies to 399 + 5p² + 100p = 414 + 90p. Rearranging: 5p² + 10p - 15 = 0, or p² + 2p - 3 = 0. Factoring: (p + 3)(p - 1) = 0. Since frequency must be positive, p = 1.
In simple words: Set up the mean equation using the unknown p, then solve the resulting quadratic equation. Choose the solution that makes sense (positive frequency).

Exam Tip: When the mean formula involves an unknown variable, cross-multiply to get an equation, then solve carefully. Always check that your answer gives positive frequencies — reject any solution that doesn't.

 

Question 6. Find the mean age in years from the frequency distribution given below:

Answer: The given data is discontinuous. To convert it into continuous data, we apply an adjustment factor of 0.5 (calculated as (30 - 29) ÷ 2). After adjusting the class boundaries, we construct a table with class marks and calculate the sum of frequencies times class marks. Using the mean formula \( \text{Mean} = \frac{\sum f_i u_i}{\sum f_i} = \frac{2755}{70} \), we get the mean age as 39.36 years.
In simple words: When the class intervals are not continuous, we add 0.5 to the upper limit and subtract 0.5 from the lower limit to make them continuous. Then we multiply each class mark by its frequency, add all these products, and divide by the total frequency to find the mean.

Exam Tip: Always check if class intervals are discontinuous and apply the adjustment factor before calculating the mean. Verify your final answer by ensuring the mean falls within the range of the data.

 

Question 7. The mean of the following frequency distribution is 62.8. Find the value of p:

Answer: We set up a table with class marks and calculate \( \sum f_i u_i \). Since the mean is 62.8, we apply the formula \( 62.8 = \frac{2640 + 50p}{40 + p} \). Cross-multiplying gives \( 62.8(40 + p) = 2640 + 50p \), which simplifies to \( 2512 + 62.8p = 2640 + 50p \). Solving for p: \( 12.8p = 128 \), so \( p = 10 \).
In simple words: Multiply the mean by the total frequency, then use this to create an equation. Solve the equation to find the missing frequency.

Exam Tip: Always set up the mean equation carefully and cross-multiply to avoid algebraic errors. Double-check by substituting p back into the original formula to verify the mean equals 62.8.

 

Question 8. The daily expenditure on milk and vegetables of 100 families are given below. Calculate f₁ and f₂, if the mean daily expenditure is Rs. 188.

Answer: Since there are 100 families total, we have \( 35 + f_1 + f_2 = 100 \), which gives \( f_1 + f_2 = 65 \). From this, \( f_1 = 65 - f_2 \). Applying the mean formula: \( 188 = \frac{6150 + 190f_1 + 210f_2}{100} \). Substituting \( f_1 = 65 - f_2 \) and simplifying: \( 18800 = 18500 + 20f_2 \), so \( f_2 = 15 \). Therefore, \( f_1 = 65 - 15 = 50 \).
In simple words: Use the fact that all frequencies add to 100 to write one unknown in terms of the other. Then use the mean equation to solve for the second unknown.

Exam Tip: When two frequencies are unknown, always use both the total frequency and the mean formula together. Set up your equations carefully before substituting to avoid calculation mistakes.

 

Question 9. The median of the following numbers, arranged in ascending order, is 25. Find x:
11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46.
Answer: There are 10 observations (an even number). The median is the average of the 5th and 6th observations: \( \text{Median} = \frac{(x + 2) + (x + 4)}{2} = \frac{2x + 6}{2} = x + 3 \). Since the median equals 25, we have \( x + 3 = 25 \), giving \( x = 22 \).
In simple words: When there is an even count of numbers, the median is found by taking the average of the two middle numbers. Set this average equal to the given median and solve.

Exam Tip: Always count the observations first to determine if n is odd or even. For even n, remember to average the middle two values; for odd n, take the middle value directly.

 

Question 10. If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.
Answer: First, arrange the numbers in ascending order: 3, 4, 5, x, 8, 9, 11. With 7 observations (odd number), the median is the 4th observation. Since \( \text{Median} = \frac{n + 1}{2} = \frac{7 + 1}{2} = 4 \), the median is the 4th value. Given that the median equals 6, the 4th observation must be x, so \( x = 6 \).
In simple words: Arrange all numbers in order from smallest to largest. Find the middle position using \( \frac{n + 1}{2} \). The value at that position is the median.

Exam Tip: Always sort the data first before finding the median. For odd-numbered datasets, the median is always one of the actual data values, not an average.

 

Question 11. The marks scored by 16 students in a class test are: 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12. Find:
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range.
Answer:
(i) Arranging in ascending order: 1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23. With 16 observations (even), the median is \( \frac{12 + 13}{2} = 12.5 \).
(ii) The lower quartile is the \( \frac{n}{4} \) th observation. For n = 16, this is the 4th observation, which is 6.
(iii) The upper quartile is the \( \frac{3n}{4} \) th observation. For n = 16, this is the 12th observation, which is 18.
(iv) The inter quartile range is \( 18 - 6 = 12 \).
In simple words: The median splits data in half. The lower quartile splits the lower half in half. The upper quartile splits the upper half in half. The inter quartile range measures the spread of the middle 50% of data.

Exam Tip: For quartiles, use \( \frac{n}{4} \) for Q1 and \( \frac{3n}{4} \) for Q3. When these positions fall between two values, interpolate; when they land exactly on a value, use that value directly.

 

Question 12. Calculate the mean, the median and the mode of the following distribution:

Answer: We compute \( \sum f_i x_i = 374 \) and \( \sum f_i = 25 \). The mean is \( \frac{374}{25} = 14.96 \). For the median, with 25 observations (odd), the position is \( \frac{25 + 1}{2} = 13 \). The 13th observation falls in the cumulative frequency range of 11 to 16, which corresponds to age 15, so the median is 15. The mode is the value with the highest frequency. Since 6 students are 15 years old (the highest count), the mode is 15.
In simple words: The mean is the average of all ages weighted by how many students are that age. The median is the middle value when arranged in order. The mode is the age that appears most often.

Exam Tip: In a frequency distribution, always compute cumulative frequencies to find the median easily. The mode is simply the value (or variate) with the maximum frequency - no calculation needed.

 

Question 13. The daily canteen bill of 30 employees in an establishment is distributed as follows:

Estimate the modal daily canteen bill for this distribution by a graphical method.
Answer: Draw a histogram using the scale: 2 cm along the x-axis represents Rs. 10 and 1 cm along the y-axis represents 1 employee. Build rectangles for each class. In the tallest rectangle (corresponding to class 20 - 30 with frequency 10), draw two diagonal lines AC and BD connecting the upper corners of the adjacent rectangles to the opposite corners of the tallest rectangle. These lines intersect at point P. From P, drop a vertical line perpendicular to the x-axis, meeting it at M. The x-coordinate of M gives the mode. From the graph, the mode is found to be Rs. 23.
In simple words: Draw a histogram from the frequency table. Find the tallest bar. Draw diagonal lines from its adjacent corners to opposite corners as shown. Where these lines cross, drop a perpendicular line down to the x-axis. That point shows the mode.

Exam Tip: This graphical method for finding the mode is useful for grouped data. The mode is always located within the modal class (the class with the highest frequency). Ensure your scale is clear and diagonals are drawn accurately for precise results.

 

Question 14. Draw a cumulative frequency curve for the following data:

Hence, determine:
(i) the median
(ii) the pass marks if 85% of the students pass.
Answer: Construct a cumulative frequency table with upper-class boundaries on the x-axis and cumulative frequencies on the y-axis. The cumulative frequencies are: 8, 18, 40, 80, 100. Plot these points and draw a smooth curve (ogive). For the median, locate \( \frac{N}{2} = \frac{100}{2} = 50 \) on the y-axis. Trace horizontally to the curve, then drop a perpendicular to the x-axis. The median is approximately 31 marks. For the pass mark at 85% passing, find 85% of 100 = 85 on the y-axis. Trace to the curve and drop a perpendicular to the x-axis. The pass mark is approximately 40 marks.
In simple words: Cumulative frequency means adding up frequencies as you go. When you plot this on a graph, the curve shows how many students scored up to any given mark. The median is where exactly half the students lie below that mark.

Exam Tip: Always use upper-class boundaries for the x-axis when drawing an ogive. The y-axis should show cumulative frequency. To read values, always trace horizontally from the y-axis to the curve, then vertically down to the x-axis for accuracy.

 

Question. (iii) the marks which 45% of the students exceed.
Answer: To find the marks that 45% of students exceed, first calculate the remaining students: 100 - 45 = 55 students. Mark point C on the y-axis at frequency 55. Draw a horizontal line from C to meet the ogive at point R. From R, draw a vertical line down to the x-axis at point O. The x-coordinate of point O is 34 marks. Therefore, 45% of students exceed 34 marks.
In simple words: When 45% of students do better than a certain mark, that means 55% scored at or below that mark. Find 55 on the y-axis, follow across to the curve, then down to find the mark value - which is 34.

Exam Tip: Always subtract the given percentage from 100 to find the corresponding frequency on the ogive - this helps you read the correct value from the graph.

 

Question 15. The given graph with a histogram represents the number of plants of different heights grown in a school campus. Study the graph carefully and answer the following questions:
(a) Make a frequency table with respect to the class boundaries and their corresponding frequencies.
(b) State the modal class.
(c) Identify and note down the mode of the distribution.
(d) Find the number of plants whose height range is between 80 cm to 90 cm.
Answer:
(a) Frequency table:

(b) By examining the histogram, the modal class is the class with the highest frequency. The class 60-70 has 12 plants, which is the highest frequency. Therefore, the modal class is 60-70.
(c) The mode is the midpoint of the modal class. For the class 60-70, the mode = (60 + 70) / 2 = 65. However, from a more precise reading of the graph, the mode is 64.
(d) From the frequency table, the number of plants with height between 80 cm and 90 cm is 3.
In simple words: The modal class is where most plants are found - that is 60-70 cm. The mode is around 64 cm, which is near the middle of that class. Only 3 plants fall in the 80-90 cm range.

Exam Tip: The modal class has the tallest bar in the histogram. The mode itself is usually close to the midpoint of the modal class, though you may need to read it more carefully from the graph for a precise answer.