Looking for complete, step-by-step textbook solutions for SEBA Class 10 Mathematics Chapter 13 (Statistics – Exercise 13.2) for the 2026–27 academic session? Access verified, board-pattern solutions to all 6 textual questions on calculating the Mode and Mean of grouped frequency distributions. Master the standard mode formula (Mode = l + [(f1 – f0) / (2f1 – f0 – f2)] * h), identifying the modal class, finding the values of l, h, f1, f0, and f2, comparing central tendencies for hospital patients’ ages, electrical components lifetimes, monthly family expenditures, runs scored by top batsmen in ODI cricket, cars passing through a spot on a road, and student-teacher ratios across states, designed strictly according to the Assam Board (SEBA) HSLC curriculum.
1. The following table shows the ages of the patients admitted in a hospital during a year:
| Age (in years) | 5 – 15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Soln: (i) To find the Mode:
From the given data, the maximum class frequency is 23. Modal class = 35 – 45
Here, Lower limit of modal class (l) = 35 Class size (h) = 45 – 35 = 10 Frequency of modal class (f₁) = 23 Frequency of class preceding modal class (f₀) = 21 Frequency of class succeeding modal class (f₂) = 14
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 35 + [(23 – 21) / (2 × 23 – 21 – 14)] × 10
Mode = 35 + [2 / (46 – 35)] × 10
Mode = 35 + [2 / 11] × 10
Mode = 35 + 20 / 11
Mode = 35 + 1.82
Mode = 36.82 years (approx.)
(ii) To find the Mean:
Calculation Table:
| Class Interval | Frequency (fᵢ) | Class Mark (xᵢ) | dᵢ = xᵢ – a (a = 30) | fᵢdᵢ |
| 5 – 15 | 6 | 10 | -20 | -120 |
| 15 – 25 | 11 | 20 | -10 | -110 |
| 25 – 35 | 21 | 30 (a) | 0 | 0 |
| 35 – 45 | 23 | 40 | 10 | 230 |
| 45 – 55 | 14 | 50 | 20 | 280 |
| 55 – 65 | 5 | 60 | 30 | 150 |
| Total | Σfᵢ = 80 | Σfᵢdᵢ = 430 |
Assumed Mean (a) = 30
Σfᵢ = 80
Σfᵢdᵢ = (-120 – 110 + 0 + 230 + 280 + 150) = 430
Formula:
Mean (x̄) = a + (Σfᵢdᵢ / Σfᵢ)
Substitution:
Mean (x̄) = 30 + (430 / 80)
Mean (x̄) = 30 + 5.375
Mean (x̄) = 35.38 years (approx.)
(iii) Interpretation:
On an average, the age of a patient admitted in the hospital is 35.38 years (Mean).
Maximum number of patients admitted in the hospital are of the age 36.82 years (Mode).
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
| Lifetimes (in hours) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Soln: From the given data, the maximum class frequency is 61.
Modal class = 60 – 80
Here, Lower limit of modal class (l) = 60
Class size (h) = 80 – 60 = 20
Frequency of modal class (f₁) = 61
Frequency of class preceding modal class (f₀) = 52
Frequency of class succeeding modal class (f₂) = 38
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 60 + [(61 – 52) / (2 × 61 – 52 – 38)] × 20
Mode = 60 + [9 / (122 – 90)] × 20
Mode = 60 + [9 / 32] × 20
Mode = 60 + [9 × 5 / 8]
Mode = 60 + 45 / 8
Mode = 60 + 5.625
Mode = 65.625 hours
Hence, the modal lifetime of the components is 65.63 hours (or 65.625 hours).
SEBA Class 10 Maths Exercise 13.2 Solutions
3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
| Expenditure (in Rs) | Number of families |
| 1000 – 1500 | 24 |
| 1500 – 2000 | 40 |
| 2000 – 2500 | 33 |
| 2500 – 3000 | 28 |
| 3000 – 3500 | 30 |
| 3500 – 4000 | 22 |
| 4000 – 4500 | 16 |
| 4500 – 5000 | 7 |
Soln: (i) To find the Modal Monthly Expenditure:
From the given data, the maximum frequency is 40.
Modal class = 1500 – 2000
Here, Lower limit of modal class (l) = 1500
Class size (h) = 2000 – 1500 = 500
Frequency of modal class (f₁) = 40
Frequency of class preceding modal class (f₀) = 24
Frequency of class succeeding modal class (f₂) = 33
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 1500 + [(40 – 24) / (2 × 40 – 24 – 33)] × 500
Mode = 1500 + [16 / (80 – 57)] × 500
Mode = 1500 + [16 / 23] × 500
Mode = 1500 + 8000 / 23
Mode = 1500 + 347.83
Mode = Rs 1847.83 (approx.)
(ii) To find the Mean Monthly Expenditure:
Calculation Table:
| Expenditure (in Rs) | Number of families (fᵢ) | Class Mark (xᵢ) | uᵢ = (xᵢ – a)/h (a = 2750, h = 500) | fᵢuᵢ |
| 1000 – 1500 | 24 | 1250 | -3 | -72 |
| 1500 – 2000 | 40 | 1750 | -2 | -80 |
| 2000 – 2500 | 33 | 2250 | -1 | -33 |
| 2500 – 3000 | 28 | 2750 (a) | 0 | 0 |
| 3000 – 3500 | 30 | 3250 | 1 | 30 |
| 3500 – 4000 | 22 | 3750 | 2 | 44 |
| 4000 – 4500 | 16 | 4250 | 3 | 48 |
| 4500 – 5000 | 7 | 4750 | 4 | 28 |
| Total | Σfᵢ = 200 | Σfᵢuᵢ = -35 |
Assumed Mean (a) = 2750
Class size (h) = 500
Σfᵢ = 200
Σfᵢuᵢ = (-72 – 80 – 33 + 0 + 30 + 44 + 48 + 28) = -185 + 150 = -35
Formula:
Mean (x̄) = a + (Σfᵢuᵢ / Σfᵢ) × h
Substitution:
Mean (x̄) = 2750 + [(-35) / 200] × 500
Mean (x̄) = 2750 – (35 × 5) / 2
Mean (x̄) = 2750 – 175 / 2
Mean (x̄) = 2750 – 87.5
Mean (x̄) = Rs 2662.50
Hence,
Modal monthly expenditure = Rs 1847.83
Mean monthly expenditure = Rs 2662.50
4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
| Number of students per teacher | Number of states / U.T. |
| 15 – 20 | 3 |
| 20 – 25 | 8 |
| 25 – 30 | 9 |
| 30 – 35 | 10 |
| 35 – 40 | 3 |
| 40 – 45 | 0 |
| 45 – 50 | 0 |
| 50 – 55 | 2 |
Soln: (i) To find the Mode:
From the given data, the maximum frequency is 10.
Modal class = 30 – 35
Here, Lower limit of modal class (l) = 30
Class size (h) = 35 – 30 = 5
Frequency of modal class (f₁) = 10
Frequency of class preceding modal class (f₀) = 9
Frequency of class succeeding modal class (f₂) = 3
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 30 + [(10 – 9) / (2 × 10 – 9 – 3)] × 5
Mode = 30 + [1 / (20 – 12)] × 5
Mode = 30 + [1 / 8] × 5
Mode = 30 + 5 / 8
Mode = 30 + 0.625
Mode = 30.63 (or 30.6)
(ii) To find the Mean:
Calculation Table:
| Number of students per teacher | Number of states / U.T. (fᵢ) | Class Mark (xᵢ) | uᵢ = (xᵢ – a)/h (a = 32.5, h = 5) | fᵢuᵢ |
| 15 – 20 | 3 | 17.5 | -3 | -9 |
| 20 – 25 | 8 | 22.5 | -2 | -16 |
| 25 – 30 | 9 | 27.5 | -1 | -9 |
| 30 – 35 | 10 | 32.5 (a) | 0 | 0 |
| 35 – 40 | 3 | 37.5 | 1 | 3 |
| 40 – 45 | 0 | 42.5 | 2 | 0 |
| 45 – 50 | 0 | 47.5 | 3 | 0 |
| 50 – 55 | 2 | 52.5 | 4 | 8 |
| Total | Σfᵢ = 35 | Σfᵢuᵢ = -23 |
Assumed Mean (a) = 32.5
Class size (h) = 5
Σfᵢ = 35
Σfᵢuᵢ = (-9 – 16 – 9 + 0 + 3 + 0 + 0 + 8) = -34 + 11 = -23
Formula:
Mean (x̄) = a + (Σfᵢuᵢ / Σfᵢ) × h
Substitution:
Mean (x̄) = 32.5 + [(-23) / 35] × 5
Mean (x̄) = 32.5 – 23 / 7
Mean (x̄) = 32.5 – 3.286
Mean (x̄) = 29.21 (approx.)
(iii) Interpretation:
On an average, the student-teacher ratio across states/U.T. is 29.2 (Mean).
Most states/U.T. have a student-teacher ratio of 30.6 (Mode).
Class 10 Maths Chapter 13.2 Statistics SEBA 2026-27
5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs scored | Number of batsmen |
| 3000 – 4000 | 4 |
| 4000 – 5000 | 18 |
| 5000 – 6000 | 9 |
| 6000 – 7000 | 7 |
| 7000 – 8000 | 6 |
| 8000 – 9000 | 3 |
| 9000 – 10000 | 1 |
| 10000 – 11000 | 1 |
Find the mode of the data.
Soln: From the given data, the maximum frequency is 18.
Modal class = 4000 – 5000
Here, Lower limit of modal class (l) = 4000
Class size (h) = 5000 – 4000 = 1000
Frequency of modal class (f₁) = 18
Frequency of class preceding modal class (f₀) = 4
Frequency of class succeeding modal class (f₂) = 9
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 4000 + [(18 – 4) / (2 × 18 – 4 – 9)] × 1000
Mode = 4000 + [14 / (36 – 13)] × 1000
Mode = 4000 + [14 / 23] × 1000
Mode = 4000 + 14000 / 23
Mode = 4000 + 608.695…
Mode = 4608.7 runs (approx.)
Hence, the mode of the given data is 4608.7 runs.
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
| Number of cars | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Soln: From the given data, the maximum frequency is 20.
Modal class = 40 – 50
Here, Lower limit of modal class (l) = 40
Class size (h) = 50 – 40 = 10
Frequency of modal class (f₁) = 20
Frequency of class preceding modal class (f₀) = 12
Frequency of class succeeding modal class (f₂) = 11
Formula:
Mode = l + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h
Substitution:
Mode = 40 + [(20 – 12) / (2 × 20 – 12 – 11)] × 10
Mode = 40 + [8 / (40 – 23)] × 10
Mode = 40 + [8 / 17] × 10
Mode = 40 + 80 / 17
Mode = 40 + 4.705…
Mode = 44.7 cars (approx.)
Hence, the mode of the data is 44.7 cars.
7. For the following distribution the upper limit of modal class is:
| Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
| Frequency | 11 | 15 | 20 | 21 | 9 |
(a) 10
(b) 15
(c) 20
(d) 25
Soln: Step 1: The maximum frequency is 21.
Step 2: The class corresponding to maximum frequency is 15 – 20.
Step 3: Upper limit of the modal class (15 – 20) is 20.
Correct Option: (c) 20
SEBA Class 10 Maths Ex 13.2 Question Answer
8. Which of the following is incorrect about modal class in grouped data?
(a) Class with highest frequency
(b) Used to calculate mode
(c) Has the smallest class interval
(d) May not always be unique
Soln: Explanation:
Modal class is defined as the class having the highest frequency, not necessarily having the smallest class interval.
Correct Option: (c) Has the smallest class interval
9. In the following questions a statement of assertion (A) is followed by a statement of reason (R).
Assertion (A) : Mode is the value that occures most frequently in a data set.
Reason (R) : Mode is the middle most observation of data when arranged in order.
Choose the correct option.
(a) Both assertion (A) and reason (R) are true and reason (R) is correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and reason (R) is not correct explanation of assertion (A).
(c) Assertion (A) is true, but reason (R) is false.
(d) Assertion (A) is false, but reason (R) is true.
Soln: Explanation:
- Assertion (A): Mode is the value that occurs most frequently in a data set. (True)
- Reason (R): Mode is the middle most observation of data when arranged in order. (False, because the middle most observation is the Median).
Correct Option: (c) Assertion (A) is true, but reason (R) is false
SEBA HSLC Class 10 Maths Chapter 13 Exercise 13.2 Solved PDF
10. Arrange the steps in order to calculate the mode of grouped data and then choose the correct option from the given alternatives.
(i) Identify modal class
(ii) Find l, h, f₁, f₀ and f₂
(iii) Apply mode formula
(iv) Locate class with highest frequency
(a) (i) → (iv) → (ii) → (iii)
(b) (iv) → (ii) → (iii) → (i)
(c) (i) → (ii) → (iii) → (iv)
(d) (iv) → (i) → (ii) → (iii)
Soln: Logical order of steps:
- Locate class with highest frequency → (iv)
- Identify modal class → (i)
- Find l, h, f₁, f₀ and f₂ → (ii)
- Apply mode formula → (iii)
Correct Sequence: (iv) → (i) → (ii) → (iii)
Correct Option: (d) (iv) → (i) → (ii) → (iii)
📊 HSLC Class 10 Maths Exercise 13.2 Notice: This page features the complete, step-by-step SEBA Class 10 Mathematics Chapter 13 (Statistics) Exercise 13.2 Official Textual Solutions updated for the 2026–27 academic session. Formatted according to the strict HSLC board marking criteria, every numerical problem—from locating the highest frequency modal class to correctly identifying f_0, f_1, and f_2 and comparing the mode with the mean to interpret central tendencies—is worked out with clear tabular layouts and step-by-step algebraic substitutions to ensure you score full marks in your 3-mark and 4-mark board exam questions.
💡 Master SEBA Class 10 Mode & Statistics:
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