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Access comprehensive school assignments for Chapter 13 Statistics using the CBSE Class 10 Mathematics Statistics Assignment Set 19. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 10 Mathematics students reinforce core concepts and improve their problem-solving accuracy.
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Previous Year Problems
| Monthly consumption (in units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
Answer:
Add the frequencies from the top, using the upper class limits:
| Monthly consumption | Cumulative frequency |
|---|---|
| Less than 85 | 4 |
| Less than 105 | 9 |
| Less than 125 | 22 |
| Less than 145 | 42 |
| Less than 165 | 56 |
| Less than 185 | 64 |
| Less than 205 | 68 |
Teacher's Note:
1. The last cumulative frequency must equal the total, 68.
| Class | 0-20 | 20-40 | 40-60 | 60-80 |
| Frequency | 15 | 6 | 18 | 10 |
Answer:
The modal class is 40 - 60 (frequency 18): ℓ = 40, f₁ = 18, f₀ = 6, f₂ = 10, h = 20.
Mode = \( 40 + \dfrac{18 - 6}{36 - 6 - 10} \times 20 = 40 + 12 = \) 52.
Teacher's Note:
1. Mode formula with f₀ and f₂ the neighbouring frequencies.
| Weight | Less than 38 | Less than 40 | Less than 42 | Less than 44 | Less than 46 | Less than 48 | Less than 50 | Less than 52 |
| Number of students | 0 | 3 | 5 | 9 | 14 | 28 | 32 | 35 |
OR Find the missing frequencies in the following frequency distribution table, if N = 100 and the median is 32.
| Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
| No. of students | 10 | ? | 25 | 30 | ? | 10 | 100 |
Answer:
Frequencies: 3, 2, 4, 5, 14, 4, 3 for the classes 38-40, 40-42, …, 50-52 (N = 35). N/2 = 17.5, so the median class is 46 - 48 (c.f. before it = 14).
Median = \( 46 + \dfrac{17.5 - 14}{14} \times 2 = \) 46.5.
OR Let the missing frequencies be x (10 - 20) and y (40 - 50). 10 + x + 25 + 30 + y + 10 = 100, so x + y = 25.
The median 32 lies in 30 - 40: C = 10 + x + 25 = 35 + x, so \( 32 = 30 + \dfrac{50 - 35 - x}{30} \times 10 \Rightarrow 15 - x = 6 \Rightarrow x = 9 \), and y = 16. The missing frequencies are 9 and 16.
Teacher's Note:
1. A median in the interior of a class gives one equation for the unknown frequency.
| C.I. | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 8 | x | 10 | 11 | 9 |
Answer:
Class marks 5, 15, 25, 35, 45: Σfᵢ = 38 + x and Σfᵢxᵢ = 40 + 15x + 250 + 385 + 405 = 1080 + 15x.
\( 25.2 (38 + x) = 1080 + 15x \Rightarrow 957.6 + 25.2x = 1080 + 15x \Rightarrow 10.2x = 122.4 \Rightarrow \) x = 12.
Teacher's Note:
1. Form the equation Σfx = mean × Σf.
Question 5 [CBSE 2014, 1M]. Construction of a cumulative frequency table is useful in determining the
(A) Mean (B) Mode (C) Median (D) All the above three measures
Answer:
A cumulative frequency table gives the median class.
Answer: (C) Median.
Teacher's Note:
1. Median uses cumulative frequency.
| Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | Total |
| Number of students | 12 | 30 | 34 | 65 | 45 | 25 | 18 | 229 |
Answer:
Start from the total and subtract frequencies going down:
| Marks | Cumulative frequency |
|---|---|
| 10 and more | 229 |
| 20 and more | 217 |
| 30 and more | 187 |
| 40 and more | 153 |
| 50 and more | 88 |
| 60 and more | 43 |
| 70 and more | 18 |
Teacher's Note:
1. A more-than table starts with the total.
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
OR The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the value of p.
| 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 | p | 5 | 4 |
Answer:
Class marks: 51, 54, 57, 60, 63. Take a = 57 and h = 3, so uᵢ = −2, −1, 0, 1, 2.
Σfᵢ = 400 and Σfᵢuᵢ = −30 − 110 + 0 + 115 + 50 = 25.
Mean = \( 57 + \dfrac{25}{400} \times 3 = \) 57.1875 mangoes.
OR Class marks 12, 14, 16, 18, 20, 22, 24: Σfᵢ = 44 + p and Σfᵢxᵢ = 752 + 20p. \( 18(44 + p) = 752 + 20p \Rightarrow 792 + 18p = 752 + 20p \Rightarrow \) p = 20.
Teacher's Note:
1. Class marks are 51, 54, … for classes 50-52, 53-55, …
| Class interval | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
| Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Answer:
7 + 10 + x + 13 + y + 10 + 14 + 9 = 100, so x + y = 37 …(i)
The median 20.75 lies in 20 - 25: ℓ = 20, f = y, C = 7 + 10 + x + 13 = 30 + x, N/2 = 50.
\( 20.75 = 20 + \dfrac{50 - 30 - x}{y} \times 5 \Rightarrow 0.75y = 100 - 5x \). Using x = 37 − y: 0.75y = 100 − 185 + 5y, so 4.25y = 85.
y = 20 and x = 17.
Teacher's Note:
1. Eliminate x with x + y = 37.
| C.I. | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 |
| f | 4 | 7 | 1 | 1 | 6 | 5 |
Answer:
N = 24, so N/2 = 12. Cumulative frequencies: 4, 11, 12, 13, 19, 24. The class whose cumulative frequency is just greater than 12 is 150 - 155.
The highest frequency is 7, for the class 140 - 145.
Median class = 150 - 155 and modal class = 140 - 145.
Teacher's Note:
1. Take the first class whose cumulative frequency exceeds N/2.
| Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
| Frequency | 5 | 8 | p | 12 | 7 | 8 |
OR Find the mean of the following frequency distribution, using the step-deviation method.
| Classes | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 | 55-59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
Answer:
Class marks 10, 30, 50, 70, 90, 110: Σfᵢ = 40 + p and Σfᵢxᵢ = 50 + 240 + 50p + 840 + 630 + 880 = 2640 + 50p.
\( 62.8(40 + p) = 2640 + 50p \Rightarrow 12.8p = 128 \Rightarrow \) p = 10.
OR Class marks 27, 32, …, 57; take a = 42, h = 5, so uᵢ = −3, −2, −1, 0, 1, 2, 3. Σfᵢ = 70 and Σfᵢuᵢ = −42 − 44 − 16 + 0 + 5 + 6 + 12 = −79.
Mean = \( 42 + \dfrac{-79}{70} \times 5 = \) 36.357.
Teacher's Note:
1. Classes 25-29, 30-34, … have class marks 27, 32, …
| Class | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
| Frequency | 2 | 3 | 5 | 7 | 4 | 2 | 2 |
Answer:
The modal class is 35 - 45: ℓ = 35, f₁ = 7, f₀ = 5, f₂ = 4, h = 10.
Mode = \( 35 + \dfrac{7 - 5}{14 - 5 - 4} \times 10 = 35 + 4 = \) 39.
Teacher's Note:
1. Take the class before and after the modal class for f₀ and f₂.
| Classes | Frequency |
|---|---|
| 140-150 | 5 |
| 150-160 | x |
| 160-170 | 20 |
| 170-180 | y |
| 180-190 | 6 |
| 190-200 | 2 |
| Total | 52 |
Answer:
5 + x + 20 + y + 6 + 2 = 52, so x + y = 19 …(i)
Class marks 145, 155, 165, 175, 185, 195: Σfᵢxᵢ = 725 + 155x + 3300 + 175y + 1110 + 390 = 5525 + 155x + 175y. Mean × 52 = 8650.
155x + 175y = 3125 ⇒ 31x + 35y = 625. With x = 19 − y: 589 + 4y = 625, so y = 9 and x = 10. x = 10, y = 9.
Median: frequencies 5, 10, 20, 9, 6, 2, so cumulative frequencies are 5, 15, 35, … N/2 = 26 and the median class is 160 - 170: \( 160 + \dfrac{26 - 15}{20} \times 10 = \) 165.5.
Teacher's Note:
1. Solve for x and y first, then use them in the median.
| Length (in mm) | 118-126 | 126-134 | 134-142 | 142-150 | 150-158 | 158-166 |
| Number of leaves | 4 | 5 | 10 | 14 | 4 | 5 |
Answer:
The modal class is 142 - 150: ℓ = 142, f₁ = 14, f₀ = 10, f₂ = 4, h = 8.
Mode = \( 142 + \dfrac{14 - 10}{28 - 10 - 4} \times 8 = 142 + 2.2857 = \) 144.29.
Teacher's Note:
1. Apply the mode formula with h = 8.
| Percentage of female teachers | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 |
| No. of states/U.T. | 6 | 11 | 7 | 4 | 4 | 2 | 1 |
Answer:
Class marks 20, 30, …, 80. Take a = 50 and h = 10, so uᵢ = −3, −2, −1, 0, 1, 2, 3. Σfᵢ = 35.
Σfᵢuᵢ = −18 − 22 − 7 + 0 + 4 + 4 + 3 = −36.
Mean = \( 50 + \dfrac{-36}{35} \times 10 = \) 39.71.
Teacher's Note:
1. A negative Σfu means the mean is below the assumed mean.
| Runs scored | 3000-4000 | 4000-5000 | 5000-6000 | 6000-7000 | 7000-8000 | 8000-9000 | 9000-10000 | 10000-11000 |
| Number of batsmen | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 |
Answer:
The modal class is 4000 - 5000: ℓ = 4000, f₁ = 18, f₀ = 4, f₂ = 9, h = 1000.
Mode = \( 4000 + \dfrac{18 - 4}{36 - 4 - 9} \times 1000 = 4000 + 608.70 = \) 4608.70.
Teacher's Note:
1. Here f₀ = 4 comes from the class 3000 - 4000.
| Class-Interval | 0-60 | 60-120 | 120-180 | 180-240 | 240-300 |
| Frequency | 22 | 35 | 44 | 25 | 24 |
OR The mean of the following distribution is 52.5. Find the value of p.
| Classes | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 15 | 22 | 37 | p | 21 |
Answer:
Class marks 30, 90, 150, 210, 270. Take a = 150, h = 60, so uᵢ = −2, −1, 0, 1, 2. Σfᵢ = 150 and Σfᵢuᵢ = −44 − 35 + 0 + 25 + 48 = −6.
Mean = \( 150 + \dfrac{-6}{150} \times 60 = \) 147.6.
OR Class marks 10, 30, 50, 70, 90: Σfᵢ = 95 + p and Σfᵢxᵢ = 4550 + 70p. \( 52.5(95 + p) = 4550 + 70p \Rightarrow 17.5p = 437.5 \Rightarrow \) p = 25.
Teacher's Note:
1. A larger class width h reduces the size of the uᵢ values.
| Height (in cm) | less than 140 | less than 145 | less than 150 | less than 155 | less than 160 | less than 165 |
| Number of girls | 4 | 11 | 29 | 40 | 46 | 51 |
Answer:
Class frequencies: 4, 7, 18, 11, 6, 5 for the classes 135-140, …, 160-165.
N/2 = 25.5. Cumulative frequency 11 < 25.5 < 29, so the median class is 145 - 150: ℓ = 145, C = 11, f = 18, h = 5.
Median = \( 145 + \dfrac{25.5 - 11}{18} \times 5 = \) 149.03 cm.
Teacher's Note:
1. The median class is 145 - 150 because 29 is the first value above 25.5.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
| Frequency | 5 | x | 20 | 15 | y | 5 | 60 |
Answer:
5 + x + 20 + 15 + y + 5 = 60, so x + y = 15 …(i)
The median 28.5 lies in 20 - 30: ℓ = 20, f = 20, C = 5 + x, N/2 = 30.
\( 28.5 = 20 + \dfrac{30 - 5 - x}{20} \times 10 \Rightarrow 17 = 25 - x \Rightarrow x = 8 \). From (i), y = 7.
x = 8 and y = 7.
Teacher's Note:
1. Use the median formula for one equation and the total for the other.
Free study material for Mathematics
CBSE Class 10 Mathematics Assignments for Chapter 13 Statistics
Chapter Practice Questions for Class 10 Mathematics
Access structured practice assignments for Chapter 13 Statistics designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
Key Advantages of Solving Chapter 13 Statistics 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
- Initial Reading: Begin by reading the NCERT book for Class 10 Mathematics to build a baseline understanding.
- Independent Testing: Attempt assignment questions unassisted, then verify work using provided answer keys.
- Supplementary Aids: Leverage revision notes and worksheets whenever you encounter difficult topics.
FAQs
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