Class 10 Mathematics Practice Assignments: CBSE Class 10 Mathematics Statistics Assignment Set 14
Review targeted academic assignments with the CBSE Class 10 Mathematics Statistics Assignment Set 14. Built according to official CBSE standards for the 2026-27 term, these downloadable Class 10 Mathematics worksheets support effective daily practice for Chapter 13 Statistics.
Download Chapter 13 Statistics Assignment PDF with Solutions
Access the complete assignment PDF for Class 10 Mathematics below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school evaluations.
Mean of Grouped Data
| Number of plants | 0 - 2 | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 | 10 - 12 | 12 - 14 |
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Answer:
Since the values of \( f_i \) and \( x_i \) are small, we use the direct method.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 0 - 2 | 1 | 1 | 1 |
| 2 - 4 | 2 | 3 | 6 |
| 4 - 6 | 1 | 5 | 5 |
| 6 - 8 | 5 | 7 | 35 |
| 8 - 10 | 6 | 9 | 54 |
| 10 - 12 | 2 | 11 | 22 |
| 12 - 14 | 3 | 13 | 39 |
| Total | \( \sum f_i = 20 \) | \( \sum f_ix_i = 162 \) |
Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1 \).
Therefore the mean number of plants per house is 8.1.
In simple words: The numbers are small, so the direct method is the quickest.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Daily wages (in Rs.) | 500 - 520 | 520 - 540 | 540 - 560 | 560 - 580 | 580 - 600 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Answer:
The values of \( x_i \) and \( f_i \) are large, so we use the step-deviation method.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 550 \), class size \( h = 20 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 500 - 520 | 12 | 510 | -2 | -24 |
| 520 - 540 | 14 | 530 | -1 | -14 |
| 540 - 560 | 8 | 550 | 0 | 0 |
| 560 - 580 | 6 | 570 | 1 | 6 |
| 580 - 600 | 10 | 590 | 2 | 20 |
| Total | \( \sum f_i = 50 \) | \( \sum f_iu_i = -12 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 550 + 20 \times \frac{-12}{50} = 545.2 \).
Thus the mean daily wage of the workers is Rs. 545.20.
In simple words: Large class marks make the step-deviation method easiest.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Daily pocket allowance (in Rs.) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Answer:
| Class interval | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 11 - 13 | 7 | 12 | 84 |
| 13 - 15 | 6 | 14 | 84 |
| 15 - 17 | 9 | 16 | 144 |
| 17 - 19 | 13 | 18 | 234 |
| 19 - 21 | f | 20 | \( 20f \) |
| 21 - 23 | 5 | 22 | 110 |
| 23 - 25 | 4 | 24 | 96 |
| Total | \( \sum f_i = 44 + f \) | \( \sum f_ix_i = 752 + 20f \) |
Mean \( = \frac{752 + 20f}{44 + f} = 18 \Rightarrow 752 + 20f = 792 + 18f \Rightarrow 2f = 40 \Rightarrow f = 20 \).
In simple words: Write the mean formula with f unknown and solve the equation.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Number of heart beats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Answer:
Use the assumed mean method.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take the assumed mean \( a = 75.5 \) and \( d_i = x_i - a \).
| Class | \( f_i \) | \( x_i \) | \( d_i = x_i - a \) | \( f_id_i \) |
|---|---|---|---|---|
| 65 - 68 | 2 | 66.5 | -9 | -18 |
| 68 - 71 | 4 | 69.5 | -6 | -24 |
| 71 - 74 | 3 | 72.5 | -3 | -9 |
| 74 - 77 | 8 | 75.5 | 0 | 0 |
| 77 - 80 | 7 | 78.5 | 3 | 21 |
| 80 - 83 | 4 | 81.5 | 6 | 24 |
| 83 - 86 | 2 | 84.5 | 9 | 18 |
| Total | \( \sum f_i = 30 \) | \( \sum f_id_i = 12 \) |
Mean \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 75.5 + \frac{12}{30} = 75.9 \).
The mean number of heart beats per minute is 75.9.
In simple words: The class marks are moderate, so the assumed mean method keeps the numbers small.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Answer:
The class intervals are not continuous: there is a gap of 1 between two classes. So subtract 0.5 from each lower limit and add 0.5 to each upper limit, giving the classes 49.5 - 52.5, 52.5 - 55.5, and so on.
The values of \( x_i \) and \( f_i \) are large, so we use the step-deviation method.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 57 \), class size \( h = 3 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 49.5 - 52.5 | 15 | 51 | -2 | -30 |
| 52.5 - 55.5 | 110 | 54 | -1 | -110 |
| 55.5 - 58.5 | 135 | 57 | 0 | 0 |
| 58.5 - 61.5 | 115 | 60 | 1 | 115 |
| 61.5 - 64.5 | 25 | 63 | 2 | 50 |
| Total | \( \sum f_i = 400 \) | \( \sum f_iu_i = 25 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 57 + 3 \times \frac{25}{400} = 57.19 \).
Therefore the mean number of mangoes in a packing box is 57.19.
In simple words: Make the classes continuous first, then use the step-deviation method.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Daily expenditure (in Rs.) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
| Number of households | 4 | 5 | 12 | 2 | 2 |
Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 225 \), class size \( h = 50 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 100 - 150 | 4 | 125 | -2 | -8 |
| 150 - 200 | 5 | 175 | -1 | -5 |
| 200 - 250 | 12 | 225 | 0 | 0 |
| 250 - 300 | 2 | 275 | 1 | 2 |
| 300 - 350 | 2 | 325 | 2 | 4 |
| Total | \( \sum f_i = 25 \) | \( \sum f_iu_i = -7 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 225 + 50 \times \frac{-7}{25} = 211 \).
The mean daily expenditure on food of a household is Rs. 211.
In simple words: The class size 50 is common, so the step-deviation method is convenient.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Concentration of SO\( _2 \) (in ppm) | 0.00-0.04 | 0.04-0.08 | 0.08-0.12 | 0.12-0.16 | 0.16-0.20 | 0.20-0.24 |
| Frequency | 4 | 9 | 9 | 2 | 4 | 2 |
Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Concentration (in ppm) | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 0.00 - 0.04 | 4 | 1/50 | 2/25 |
| 0.04 - 0.08 | 9 | 3/50 | 27/50 |
| 0.08 - 0.12 | 9 | 1/10 | 9/10 |
| 0.12 - 0.16 | 2 | 7/50 | 7/25 |
| 0.16 - 0.20 | 4 | 9/50 | 18/25 |
| 0.20 - 0.24 | 2 | 11/50 | 11/25 |
| Total | \( \sum f_i = 30 \) | \( \sum f_ix_i = 2.96 \) |
Mean \( = \frac{\sum f_ix_i}{\sum f_i} = \frac{2.96}{30} = 0.099 \) ppm (to three decimal places).
Therefore the mean concentration of SO\( _2 \) in the air is 0.099 ppm.
In simple words: The numbers are small decimals, so the direct method is best.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Number of days | 0 - 6 | 6 - 10 | 10 - 14 | 14 - 20 | 20 - 28 | 28 - 38 | 38 - 40 |
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Answer:
The class sizes are unequal, so we use the direct method.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 0 - 6 | 11 | 3 | 33 |
| 6 - 10 | 10 | 8 | 80 |
| 10 - 14 | 7 | 12 | 84 |
| 14 - 20 | 4 | 17 | 68 |
| 20 - 28 | 4 | 24 | 96 |
| 28 - 38 | 3 | 33 | 99 |
| 38 - 40 | 1 | 39 | 39 |
| Total | \( \sum f_i = 40 \) | \( \sum f_ix_i = 499 \) |
Mean \( = \frac{499}{40} = 12.48 \). Therefore the mean number of days a student was absent is 12.48.
In simple words: Unequal class sizes rule out the step-deviation method, so use the direct method.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
| Literacy rate (in %) | 45 - 55 | 55 - 65 | 65 - 75 | 75 - 85 | 85 - 95 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Answer:
The class marks are large, so take \( a = 70 \) and \( h = 10 \).
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 70 \), class size \( h = 10 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 45 - 55 | 3 | 50 | -2 | -6 |
| 55 - 65 | 10 | 60 | -1 | -10 |
| 65 - 75 | 11 | 70 | 0 | 0 |
| 75 - 85 | 8 | 80 | 1 | 8 |
| 85 - 95 | 3 | 90 | 2 | 6 |
| Total | \( \sum f_i = 35 \) | \( \sum f_iu_i = -2 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 70 + 10 \times \frac{-2}{35} = 69.43 \).
Therefore the mean literacy rate is 69.43%.
In simple words: Use the step-deviation method with a = 70 and h = 10.
Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.
Free study material for Mathematics
Chapter Assignment & Practice Material for Class 10 Mathematics Chapter 13 Statistics
Class 10 Mathematics Chapter 13 Statistics Printable Assignments
Explore reliable practice questions for Chapter 13 Statistics tailored for Class 10 learners. Use these structured worksheets to evaluate preparedness and strengthen core problem-solving skills.
Why Practice Class 10 Mathematics Assignments?
- Exam Alignment: Questions strictly follow contemporary CBSE sample papers and grading schemes.
- Comprehensive Coverage: Features objective drills, case studies, and structured descriptive problems for Chapter 13 Statistics.
- Pacing & Precision: Regular problem-solving builds critical calculation speed and test-taking accuracy.
Steps to Complete Chapter 13 Statistics Assignments Successfully
- Textbook Review: Always study the core NCERT book for Class 10 Mathematics prior to beginning the assignment.
- Independent Attempt: Solve Chapter 13 Statistics questions on your own initially before cross-checking with expert solutions.
- Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.
FAQs
You can download free PDF assignments for Class 10 Mathematics Chapter 13 Statistics from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.
Yes, our teachers have given solutions for all questions in the Class 10 Mathematics Chapter 13 Statistics assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.
Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 13 Statistics.
Practicing topicw wise assignments will help Class 10 students understand every sub-topic of Chapter 13 Statistics. Daily practice will improve speed, accuracy and answering competency-based questions.
Yes, all printable assignments for Class 10 Mathematics Chapter 13 Statistics are available for free download in mobile-friendly PDF format.