NCERT Solutions for Class 10 Maths Chapter 13: Statistics (NCERT 2026–27)

These Class 10 Maths Chapter 13 solutions cover Statistics from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 13.1, 13.2 and 13.3 is solved step by step, showing the full working tables for the mean, mode and median of grouped data so you can revise quickly and score full marks.

Class: 10 Subject: Mathematics Chapter: 13 – Statistics Exercises: 13.1, 13.2, 13.3 Topics: Mean, Mode, Median (grouped data) Session: 2026–27
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Chapter 13 Overview

Chapter 13 of Class 10 Maths, Statistics, extends the three measures of central tendency — mean, median and mode — from ungrouped data to grouped frequency distributions. You learn three ways to find the mean (the direct method, the assumed-mean method and the step-deviation method), how to locate the modal class and apply the mode formula, and how to build a cumulative frequency table to find the median class and the median. The chapter also introduces the empirical relationship 3 Median = Mode + 2 Mean and explains when each measure is the most suitable. The Class 10 Maths Chapter 13 solutions below work through every exercise question with complete tables.

Key Concepts & Definitions

Class mark (xi): the mid-point of a class interval = (upper limit + lower limit) ÷ 2. It represents all observations in that class.

Mean: the average of all observations. For grouped data it is computed using class marks weighted by frequency.

Modal class: the class interval that has the maximum frequency. The mode is a value lying inside this class.

Cumulative frequency (cf): the running total of frequencies up to and including a class.

Median class: the class whose cumulative frequency is greater than (and nearest to) n÷2, where n is the total frequency.

Continuous classes: before using the mode and median formulae, gaps between classes (e.g. 118–126, 127–135) must be removed to make the classes continuous (117.5–126.5, 126.5–135.5, …).

Important Formulas (Chapter 13)

Mean – Direct method: x̄ = (Σfixi) ÷ (Σfi).

Mean – Assumed-mean method: x̄ = a + (Σfidi) ÷ (Σfi), where di = xi − a.

Mean – Step-deviation method: x̄ = a + h × (Σfiui) ÷ (Σfi), where ui = (xi − a) ÷ h.

Mode: Mode = l + [ (f1 − f0) ÷ (2f1 − f0 − f2) ] × h, where l = lower limit of modal class, h = class size, f1 = modal-class frequency, f0 = frequency before it, f2 = frequency after it.

Median: Median = l + [ (n÷2 − cf) ÷ f ] × h, where l = lower limit of median class, cf = cf of the class before it, f = frequency of median class, h = class size.

Empirical relation: 3 Median = Mode + 2 Mean.

Exercise 13.1 — Mean of Grouped Data

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number of plants: 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14 with number of houses 1, 2, 1, 5, 6, 2, 3. Which method did you use for finding the mean, and why?

SOLUTION We use the direct method because the values of xi and fi are small.
Classfixifixi
0–2111
2–4236
4–6155
6–85735
8–106954
10–1221122
12–1431339
Total20162
x̄ = Σfixi ÷ Σfi = 162 ÷ 20 = 8.1 plants per house.

2. Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in ₹): 500–520, 520–540, 540–560, 560–580, 580–600 with number of workers 12, 14, 8, 6, 10. Find the mean daily wages of the workers of the factory by using an appropriate method.

SOLUTION Values are large, so we use the step-deviation method with a = 550 and h = 20.
Classfixidi=xi−550ui=di/20fiui
500–52012510−40−2−24
520–54014530−20−1−14
540–5608550000
560–58065702016
580–6001059040220
Total50−12
x̄ = a + h × (Σfiui ÷ Σfi) = 550 + 20 × (−12 ÷ 50) = 550 − 4.8 = ₹545.20.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Daily pocket allowance (in ₹): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25 with number of children 7, 6, 9, 13, f, 5, 4.

SOLUTION
Classfixifixi
11–1371284
13–1561484
15–17916144
17–191318234
19–21f2020f
21–23522110
23–2542496
Total44 + f752 + 20f
Mean = (752 + 20f) ÷ (44 + f) = 18. 752 + 20f = 18(44 + f) = 792 + 18f ⇒ 2f = 40 ⇒ f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats per minute: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86 with number of women 2, 4, 3, 8, 7, 4, 2.

SOLUTION Use the step-deviation method with a = 75.5 and h = 3.
Classfixiui=(xi−75.5)/3fiui
65–68266.5−3−6
68–71469.5−2−8
71–74372.5−1−3
74–77875.500
77–80778.517
80–83481.528
83–86284.536
Total304
x̄ = 75.5 + 3 × (4 ÷ 30) = 75.5 + 0.4 = 75.9 heartbeats per minute.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? Number of mangoes: 50–52, 53–55, 56–58, 59–61, 62–64 with number of boxes 15, 110, 135, 115, 25.

SOLUTION The classes are inclusive, so make them continuous (subtract 0.5 from lower and add 0.5 to upper limits): 49.5–52.5, 52.5–55.5, … The class marks are unchanged (51, 54, 57, 60, 63). Use the step-deviation method with a = 57 and h = 3 because the frequencies are large.
Classfixiui=(xi−57)/3fiui
50–521551−2−30
53–5511054−1−110
56–581355700
59–61115601115
62–642563250
Total40025
x̄ = 57 + 3 × (25 ÷ 400) = 57 + 0.1875 ≈ 57.19 mangoes per box.

6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method. Daily expenditure (in ₹): 100–150, 150–200, 200–250, 250–300, 300–350 with number of households 4, 5, 12, 2, 2.

SOLUTION Use the step-deviation method with a = 225 and h = 50.
Classfixiui=(xi−225)/50fiui
100–1504125−2−8
150–2005175−1−5
200–2501222500
250–300227512
300–350232524
Total25−7
x̄ = 225 + 50 × (−7 ÷ 25) = 225 − 14 = ₹211 (per household per day).

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below. Find the mean concentration of SO2 in the air. Concentration of SO2 (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 with frequency 4, 9, 9, 2, 4, 2.

SOLUTION Use the direct method (small class marks).
Classfixifixi
0.00–0.0440.020.08
0.04–0.0890.060.54
0.08–0.1290.100.90
0.12–0.1620.140.28
0.16–0.2040.180.72
0.20–0.2420.220.44
Total302.96
x̄ = 2.96 ÷ 30 ≈ 0.099 ppm.

8. 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. Number of days: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40 with number of students 11, 10, 7, 4, 4, 3, 1.

SOLUTION Class sizes are unequal, so we use the direct method with class marks.
Classfixifixi
0–611333
6–1010880
10–1471284
14–2041768
20–2842496
28–3833399
38–4013939
Total40499
x̄ = 499 ÷ 40 = 12.475 days (about 12.48 days).

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Literacy rate (in %): 45–55, 55–65, 65–75, 75–85, 85–95 with number of cities 3, 10, 11, 8, 3.

SOLUTION Use the step-deviation method with a = 70 and h = 10.
Classfixiui=(xi−70)/10fiui
45–55350−2−6
55–651060−1−10
65–75117000
75–8588018
85–9539026
Total35−2
x̄ = 70 + 10 × (−2 ÷ 35) = 70 − 0.571 ≈ 69.43%.

Exercise 13.2 — Mode of Grouped Data

1. The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65 with number of patients 6, 11, 21, 23, 14, 5.

SOLUTION Mode. Maximum frequency 23 is in 35–45, so the modal class is 35–45: l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10. Mode = 35 + [(23 − 21) ÷ (2×23 − 21 − 14)] × 10 = 35 + (2 ÷ 11) × 10 = 35 + 1.81 ≈ 36.8 years. Mean. Using the direct method with class marks 10, 20, 30, 40, 50, 60: Σfixi = 60 + 220 + 630 + 920 + 700 + 300 = 2830, Σfi = 80. x̄ = 2830 ÷ 80 = 35.37 years. Interpretation: the mode (36.8) shows the age that the maximum number of patients have, while the mean (35.37) is the average age of all the patients admitted.

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components. Lifetimes (in hours): 0–20, 20–40, 40–60, 60–80, 80–100, 100–120 with frequency 10, 35, 52, 61, 38, 29.

SOLUTION Maximum frequency 61 is in 60–80, so the modal class is 60–80: l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20. Mode = 60 + [(61 − 52) ÷ (2×61 − 52 − 38)] × 20 = 60 + (9 ÷ 32) × 20 = 60 + 5.625 = 65.625 hours.

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 ₹): 1000–1500, 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000 with number of families 24, 40, 33, 28, 30, 22, 16, 7.

SOLUTION Mode. Maximum frequency 40 is in 1500–2000, so the modal class is 1500–2000: l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500. Mode = 1500 + [(40 − 24) ÷ (2×40 − 24 − 33)] × 500 = 1500 + (16 ÷ 23) × 500 = 1500 + 347.83 ≈ ₹1847.83. Mean. Use the step-deviation method with a = 2750, h = 500.
Classfixiuifiui
1000–1500241250−3−72
1500–2000401750−2−80
2000–2500332250−1−33
2500–300028275000
3000–3500303250130
3500–4000223750244
4000–4500164250348
4500–500074750428
Total200−35
x̄ = 2750 + 500 × (−35 ÷ 200) = 2750 − 87.5 = ₹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: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55 with number of states/U.T. 3, 8, 9, 10, 3, 0, 0, 2.

SOLUTION Mode. Maximum frequency 10 is in 30–35, so the modal class is 30–35: l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5. Mode = 30 + [(10 − 9) ÷ (2×10 − 9 − 3)] × 5 = 30 + (1 ÷ 8) × 5 = 30 + 0.625 = 30.6 (approx). Mean. Step-deviation with a = 32.5, h = 5: fiui values = −9, −16, −9, 0, 3, 0, 0, 12 ⇒ Σfiui = −19, Σfi = 35. x̄ = 32.5 + 5 × (−19 ÷ 35) = 32.5 − 2.71 = 29.2 (approx). Interpretation: most states have a student-teacher ratio of about 30.6, while the average ratio across all states is about 29.2.

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data. Runs scored: 3000–4000, 4000–5000, 5000–6000, 6000–7000, 7000–8000, 8000–9000, 9000–10000, 10000–11000 with number of batsmen 4, 18, 9, 7, 6, 3, 1, 1.

SOLUTION Maximum frequency 18 is in 4000–5000, so the modal class is 4000–5000: l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000. Mode = 4000 + [(18 − 4) ÷ (2×18 − 4 − 9)] × 1000 = 4000 + (14 ÷ 23) × 1000 = 4000 + 608.7 ≈ 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 with frequency 7, 14, 13, 12, 20, 11, 15, 8.

SOLUTION Maximum frequency 20 is in 40–50, so the modal class is 40–50: l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10. Mode = 40 + [(20 − 12) ÷ (2×20 − 12 − 11)] × 10 = 40 + (8 ÷ 17) × 10 = 40 + 4.7 ≈ 44.7 cars.

Exercise 13.3 — Median of Grouped Data

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units): 65–85, 85–105, 105–125, 125–145, 145–165, 165–185, 185–205 with number of consumers 4, 5, 13, 20, 14, 8, 4.

SOLUTION
Classficfxifixi
65–854475300
85–1055995475
105–12513221151495
125–14520421352700
145–16514561552170
165–1858641751400
185–205468195780
Total689320
Median. n = 68, n÷2 = 34. The cf just exceeding 34 is 42, so the median class is 125–145: l = 125, cf = 22, f = 20, h = 20. Median = 125 + [(34 − 22) ÷ 20] × 20 = 125 + 12 = 137 units. Mean. x̄ = 9320 ÷ 68 = 137.06 units. Mode. Modal class 125–145: l = 125, f1 = 20, f0 = 13, f2 = 14, h = 20. Mode = 125 + [(20 − 13) ÷ (40 − 13 − 14)] × 20 = 125 + (7 ÷ 13) × 20 = 125 + 10.77 ≈ 135.77 units. Comparison: the three measures (median 137, mean 137.06, mode 135.77) are very close, showing the distribution is nearly symmetric.

2. If the median of the distribution given below is 28.5, find the values of x and y. Class interval: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60 with frequency 5, x, 20, 15, y, 5 and Total 60.

SOLUTION
Classficf
0–1055
10–20x5 + x
20–302025 + x
30–401540 + x
40–50y40 + x + y
50–60545 + x + y
Total60
Total: 45 + x + y = 60 ⇒ x + y = 15  (1). Median = 28.5 lies in the class 20–30, so the median class is 20–30: l = 20, cf = 5 + x, f = 20, h = 10, n÷2 = 30. 28.5 = 20 + [(30 − (5 + x)) ÷ 20] × 10 ⇒ 8.5 = (25 − x) ÷ 2 ⇒ 17 = 25 − x ⇒ x = 8. From (1): 8 + y = 15 ⇒ y = 7.

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Age (in years) given as “Below” cumulative figures: Below 20 = 2, Below 25 = 6, Below 30 = 24, Below 35 = 45, Below 40 = 78, Below 45 = 89, Below 50 = 92, Below 55 = 98, Below 60 = 100.

SOLUTION Convert the “less than” data into class intervals and frequencies (lowest class 18–20).
Classficf
18–2022
20–2546
25–301824
30–352145
35–403378
40–451189
45–50392
50–55698
55–602100
Total100
n = 100, n÷2 = 50. The cf just exceeding 50 is 78, so the median class is 35–40: l = 35, cf = 45, f = 33, h = 5. Median = 35 + [(50 − 45) ÷ 33] × 5 = 35 + 25 ÷ 33 = 35 + 0.76 ≈ 35.76 years.

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves. Length (in mm): 118–126, 127–135, 136–144, 145–153, 154–162, 163–171, 172–180 with number of leaves 3, 5, 9, 12, 5, 4, 2. (Hint: convert to continuous classes 117.5–126.5, …)

SOLUTION Make classes continuous by subtracting/adding 0.5.
Classficf
117.5–126.533
126.5–135.558
135.5–144.5917
144.5–153.51229
153.5–162.5534
162.5–171.5438
171.5–180.5240
Total40
n = 40, n÷2 = 20. The cf just exceeding 20 is 29, so the median class is 144.5–153.5: l = 144.5, cf = 17, f = 12, h = 9. Median = 144.5 + [(20 − 17) ÷ 12] × 9 = 144.5 + 27 ÷ 12 = 144.5 + 2.25 = 146.75 mm.

5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp. Life time (in hours): 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000 with number of lamps 14, 56, 60, 86, 74, 62, 48.

SOLUTION
Classficf
1500–20001414
2000–25005670
2500–300060130
3000–350086216
3500–400074290
4000–450062352
4500–500048400
Total400
n = 400, n÷2 = 200. The cf just exceeding 200 is 216, so the median class is 3000–3500: l = 3000, cf = 130, f = 86, h = 500. Median = 3000 + [(200 − 130) ÷ 86] × 500 = 3000 + (70 × 500) ÷ 86 = 3000 + 406.98 ≈ 3406.98 hours.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. Number of letters: 1–4, 4–7, 7–10, 10–13, 13–16, 16–19 with number of surnames 6, 30, 40, 16, 4, 4.

SOLUTION
Classficfxifixi
1–4662.515
4–730365.5165
7–1040768.5340
10–13169211.5184
13–1649614.558
16–19410017.570
Total100832
Median. n = 100, n÷2 = 50. The cf just exceeding 50 is 76, so the median class is 7–10: l = 7, cf = 36, f = 40, h = 3. Median = 7 + [(50 − 36) ÷ 40] × 3 = 7 + 42 ÷ 40 = 7 + 1.05 = 8.05 letters. Mean. x̄ = 832 ÷ 100 = 8.32 letters. Mode. Modal class 7–10: l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3. Mode = 7 + [(40 − 30) ÷ (80 − 30 − 16)] × 3 = 7 + (10 ÷ 34) × 3 = 7 + 0.88 ≈ 7.88 letters.

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg): 40–45, 45–50, 50–55, 55–60, 60–65, 65–70, 70–75 with number of students 2, 3, 8, 6, 6, 3, 2.

SOLUTION
Classficf
40–4522
45–5035
50–55813
55–60619
60–65625
65–70328
70–75230
Total30
n = 30, n÷2 = 15. The cf just exceeding 15 is 19, so the median class is 55–60: l = 55, cf = 13, f = 6, h = 5. Median = 55 + [(15 − 13) ÷ 6] × 5 = 55 + 10 ÷ 6 = 55 + 1.67 ≈ 56.67 kg.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to make inclusive classes (e.g. 50–52, 53–55) continuous before applying the mode or median formula.
  • Using the wrong f0 and f2 in the mode formula — they are the frequencies of the classes just before and just after the modal class.
  • Picking the median class from the frequency instead of the cumulative frequency that first exceeds n÷2.
  • Taking cf of the median class itself; the formula uses the cf of the class preceding the median class.
  • Choosing a step-deviation assumed mean ‘a’ that is not a class mark, which makes ui non-integer and error-prone.
  • Mixing units — always state the answer with the correct unit (years, hours, kg, ₹, etc.).

Practice MCQs & Assertion–Reason

1. The class mark of the interval 25–35 is:

(a) 25    (b) 30    (c) 35    (d) 10

2. In the step-deviation method ui equals:

(a) xi − a    (b) (xi − a)/h    (c) a − xi    (d) h(xi − a)

3. The modal class is the class with the:

(a) least frequency    (b) maximum frequency    (c) least class mark    (d) cf nearest to n/2

4. To find the median class we locate the cumulative frequency just greater than:

(a) n    (b) n/4    (c) n/2    (d) 2n

5. The empirical relationship between mean, median and mode is:

(a) Mode = 3 Median − 2 Mean    (b) Mean = 3 Mode − 2 Median    (c) Median = 3 Mode − 2 Mean    (d) Mode = 2 Median + 3 Mean

6. While computing the mean by the direct method, xi represents the:

(a) frequency    (b) lower limit    (c) class mark    (d) cumulative frequency

7. If n = 50, then n/2 used to find the median class is:

(a) 50    (b) 25    (c) 12.5    (d) 100

8. The cumulative frequency of a class is found by:

(a) dividing frequencies    (b) adding frequencies of all preceding classes and itself    (c) multiplying frequencies    (d) subtracting the mean

9. Before applying the median formula to classes 1–10, 11–20, … we must:

(a) take class marks    (b) make the classes continuous    (c) double the frequencies    (d) find the mode first

10. The measure of central tendency that is least affected by extreme values is the:

(a) mean    (b) range    (c) median    (d) sum

Answer key: 1-(b), 2-(b), 3-(b), 4-(c), 5-(a), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The class mark of the interval 60–70 is 65.

Reason: The class mark is the average of the upper and lower class limits.

A-R 2. Assertion: The mode of a grouped distribution always lies in the class with the maximum frequency.

Reason: The modal class is defined as the class having the maximum frequency.

A-R 3. Assertion: The median can be found without forming a cumulative frequency table.

Reason: The median class is located using the cumulative frequency just greater than n/2.

A-R 4. Assertion: For inclusive classes such as 50–52, 53–55, the median formula can be applied directly.

Reason: The median formula assumes the class intervals are continuous.

A-R 5. Assertion: The mean takes every observation into account.

Reason: The mean is the sum of all observations divided by the number of observations.

Answer key: 1-(A), 2-(A), 3-(D), 4-(D), 5-(A).

Quick Revision Summary

  • Class mark xi = (upper limit + lower limit) ÷ 2 represents each class.
  • Mean of grouped data: direct (Σfixi/Σfi), assumed-mean (a + Σfidi/Σfi) or step-deviation (a + h·Σfiui/Σfi).
  • Mode = l + [(f1 − f0)/(2f1 − f0 − f2)] × h, using the modal class (maximum frequency).
  • Cumulative frequency is the running total of frequencies; the median class is where cf first exceeds n÷2.
  • Median = l + [(n÷2 − cf)/f] × h.
  • Empirical relation: 3 Median = Mode + 2 Mean.
  • Always make inclusive classes continuous before finding mode and median.

How to score full marks in this chapter

Lay out a neat working table every time — columns for xi, fi, and either fixi/ui (for mean) or cf (for median). Pick a class mark in the middle of the data as the assumed mean ‘a’ to keep ui small. Clearly identify the modal class or median class and write down l, f, cf, h before substituting. Round only at the final step and always attach the correct unit so you do not lose easy marks.

Frequently Asked Questions

What is Class 10 Maths Chapter 13 Statistics about?

Chapter 13, Statistics, teaches how to find the three measures of central tendency — mean, median and mode — for grouped data, using class marks, cumulative frequency, the modal class and the median class, along with the empirical relationship 3 Median = Mode + 2 Mean.

How many exercises are there in Class 10 Maths Chapter 13?

There are three exercises: Exercise 13.1 on the mean of grouped data, Exercise 13.2 on the mode of grouped data, and Exercise 13.3 on the median of grouped data — all solved step by step on this page.

Which method is best for finding the mean of grouped data?

Use the direct method when the class marks and frequencies are small. When the values are large, the assumed-mean or step-deviation method is faster; the step-deviation method is best when all deviations share a common factor (the class size).

Are these Class 10 Maths Chapter 13 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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