Assertion (A): For the distribution 0-10:5, 10-20:8, 20-30:15, 30-40:12, 40-50:10, the median is 25. Reason (R): The median class is 20-30 because the cumulative frequency just greater than N/2 (which is 25) is 28.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

The mean of the following frequency distribution is 57.6 and the sum of all frequencies is 50. Find the missing frequencies f1 and f2. Classes: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequency: 7, f1, 12, f2, 8, 5. Also, identify the median class of the completed distribution.

f1 = 8, f2 = 10; median class is 40-60.

The following table gives the life-time (in hours) of 400 neon lamps: Life-time (in hours): more than 500, more than 600, more than 700, more than 800, more than 900, more than 1000 Number of lamps: 400, 340, 240, 120, 40, 0 Find the median life-time. Also, find the mode.

Median life-time = 733.33 hours, Mode life-time = 733.33 hours

The mean of the following frequency distribution is 25. The total frequency is 80. Find the missing frequencies f1 and f2. Hence, find the median of the distribution. Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50 Frequencies: 12, 18, f1, f2, 15

f1 = 23, f2 = 12, Median = 24.35 (approx.)

Class 10 Maths Chapter 13: Statistics — Important Questions & Sample Paper

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Statistics in CBSE Class 10 extends your understanding of data analysis beyond simple averages. You learn to compute measures of central tendency—mean, median, mode—for grouped data using various methods. For the mean, three methods are covered: direct, assumed mean, and step deviation, each suited to different data complexities. Mode and median are calculated using specific formulas that involve identifying the modal or median class and applying cumulative frequencies. The chapter also introduces graphical representation through cumulative frequency curves (ogives) and teaches you to estimate the median by intersecting 'less than' and 'more than' ogives. Exam questions typically ask you to compute the mean using the assumed mean or step deviation method, find the mode or median from given frequency distributions, or interpret ogives to determine the median or other percentiles. You may also encounter problems requiring comparison of two datasets or solving missing frequency problems. The mode formula, for instance, uses the frequencies of the modal class and its neighbors, while the median formula relies on cumulative frequency just above half the total observations. The assumed mean method simplifies calculations by reducing the size of numbers, and the step deviation method further streamlines the process when class sizes are equal. Graphical questions often ask you to draw an ogive and then find the median or other quartiles from it. Mastering these concepts requires careful handling of class intervals and frequency tables, but the systematic approaches make them manageable. Regular practice with varied problems—like finding the mode of leaf lengths, median of lamp lifetimes, or mean marks—builds confidence and accuracy, essential for scoring well in the CBSE board exam.

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Mathematics — Statistics

Class 10Time: 3 hrsMax Marks: 80

SECTION A

1.
The mean of the following distribution using assumed mean method (assumed mean = 25) is: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | | Frequency | 10 | 20 | 40 | 20 | 10 |
(a) 25(b) 24(c) 26(d) 30
1
2.
For a given data, mean = 53 and median = 55. The approximate mode is:
(a) 51(b) 59(c) 57(d) 53
1
3.
The mode of the data 2, 3, 3, 4, 5, 3, 6 is:
(a) 3(b) 2(c) 4(d) 6
1

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Marks distribution & blueprint

In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

Real, exam-style questions to practise and revise — each with its answer. Generate a full paper for unlimited more.

Q1. The mean of the following distribution using assumed mean method (assumed mean = 25) is: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | | Frequency | 10 | 20 | 40 | 20 | 10 |
1 mark
Multiple Choice (MCQ)
(A) 25(B) 24(C) 26(D) 30
▸ Answer▾ Answer
25
Q2. For a given data, mean = 53 and median = 55. The approximate mode is:
1 mark
Multiple Choice (MCQ)
(A) 51(B) 59(C) 57(D) 53
▸ Answer▾ Answer
59
Q3. The mode of the data 2, 3, 3, 4, 5, 3, 6 is:
1 mark
Multiple Choice (MCQ)
(A) 3(B) 2(C) 4(D) 6
▸ Answer▾ Answer
3
Q4. The median of the numbers 3, 7, 1, 9, 5 is:
1 mark
Multiple Choice (MCQ)
(A) 3(B) 5(C) 7(D) 9
▸ Answer▾ Answer
5
Q5. Assertion (A): For the distribution 0-10:5, 10-20:8, 20-30:15, 30-40:12, 40-50:10, the median is 25. Reason (R): The median class is 20-30 because the cumulative frequency just greater than N/2 (which is 25) is 28.
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
A is false but R is true.
Q6. The mean of 6 observations is 15. If five of the observations are 12, 13, 17, 14, and 15, find the sixth observation.
2 marks
Short Answer
▸ Answer▾ Answer
19
Q7. Find the mean of the following data using the assumed mean method. Use A = 50. Classes: 0-20, 20-40, 40-60, 60-80, 80-100; Frequencies: 12, 18, 15, 10, 5.
2 marks
Short Answer
▸ Answer▾ Answer
42.67
Q8. The mean of the following frequency distribution is 57.6 and the sum of all frequencies is 50. Find the missing frequencies f1 and f2. Classes: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequency: 7, f1, 12, f2, 8, 5. Also, identify the median class of the completed distribution.
3 marks
Short Answer
▸ Answer▾ Answer
f1 = 8, f2 = 10; median class is 40-60.
Q9. The following distribution shows the distance travelled by 80 scooters in a day (in km): Distance (km) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 Number of scooters | 7 | 12 | 24 | 28 | 9 Find the median distance.
3 marks
Short Answer
▸ Answer▾ Answer
37.5 km
Q10. The following table gives the life-time (in hours) of 400 neon lamps: Life-time (in hours): more than 500, more than 600, more than 700, more than 800, more than 900, more than 1000 Number of lamps: 400, 340, 240, 120, 40, 0 Find the median life-time. Also, find the mode.
5 marks
Long Answer
▸ Answer▾ Answer
Median life-time = 733.33 hours, Mode life-time = 733.33 hours
Q11. The mean of the following frequency distribution is 25. The total frequency is 80. Find the missing frequencies f1 and f2. Hence, find the median of the distribution. Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50 Frequencies: 12, 18, f1, f2, 15
5 marks
Long Answer
▸ Answer▾ Answer
f1 = 23, f2 = 12, Median = 24.35 (approx.)
Q12. The following table shows the daily wages of workers in a small factory. The mean wage is known to be ₹50. Wages (₹): 0-20, 20-40, 40-60, 60-80, 80-100 Number of workers: 5, f, 12, 8, 5
4 marks
Case Study
1. (i) Find the missing frequency f.2 marks
2. (ii) Calculate the median wage.2 marks
▸ Answer▾ Answer
f = 8, Median = ₹50

Frequently asked questions

How do I find the mode of grouped data in CBSE Class 10 Statistics?

Use the formula Mode = l + (f1 - f0) / (2f1 - f0 - f2) * h, where l is the lower limit of the modal class, f1 is the frequency of the modal class, f0 is the frequency of the class preceding the modal class, f2 is the frequency of the class succeeding the modal class, and h is the class width. Identify the modal class as the one with the highest frequency.

What is the difference between the direct method and the assumed mean method for calculating the mean?

The direct method uses the formula Mean = Σ(fi * xi) / Σfi, where xi is the class mark. The assumed mean method simplifies calculations by choosing an assumed mean 'a', then computing di = xi - a, and Mean = a + Σ(fi * di) / Σfi. The assumed mean method reduces arithmetic errors when xi values are large.

How can I find the median from a cumulative frequency curve (ogive)?

Draw both 'less than' and 'more than' ogives on the same graph. The x-coordinate of their intersection point gives the median. Alternatively, for a single 'less than' ogive, locate Σfi/2 on the cumulative frequency axis, draw a horizontal line to the curve, and drop a perpendicular to the x-axis to read the median.

When should I use the step deviation method in CBSE Class 10 Statistics?

The step deviation method is an extension of the assumed mean method and is most convenient when the class sizes are equal and large. You calculate ui = (xi - a)/h, then Mean = a + h * (Σfi * ui) / Σfi. It further simplifies calculations, especially in exams when time is limited.

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