Class 10 Maths NCERT Solutions Chapter 13

Key Concepts & Formulae at a Glance

Mean — Direct Method:x̄ = Σfixi ÷ Σfi, using each class's mid-point (class mark) as xi.
Mean — Assumed Mean / Step-Deviation:x̄ = a + Σfidi÷Σfi (di=xi−a), or x̄ = a + h(Σfiui÷Σfi) (ui=xi−ah) — same answer, easier arithmetic for large numbers.
Mode:Mode = l + [f₁−f₀2f₁−f₀−f₂]×h, using the modal class (highest frequency); f₀,f₂ are the precedingsucceeding class frequencies.
Median:Median = l + [(n2−cf)/f]×h, using the first class whose cumulative frequency reaches n2.
Cumulative Frequency:A running total of frequencies up to and including each class — needed to locate the median class.
Choosing a Measure:Mean uses every value and suits comparisons; median suits data with extreme values; mode identifies the most common value. Empirical relation: 3×Median = Mode + 2×Mean.

Exercise 13.1

9 Questions

1A 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. Which method did you use for finding the mean, and why?

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

Class interval	fi	xi	fixi
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	Σfi = 20		Σfixi = 162

Mean = Σfixi ÷ Σfi = 16220 = 8.1
Used the Direct Method — the class marks (1,3,5,...,13) and frequencies are all small, single-digit numbers, so there's no real arithmetic advantage to shifting or scaling them first.

Mean number of plants per house = 8.1.

2Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages (in ₹)	500–520	520–540	540–560	560–580	580–600
Number of workers	12	14	8	6	10

Class interval	fi	xi	di=xi−550	ui=di20	fiui
500–520	12	510	−40	−2	−24
520–540	14	530	−20	−1	−14
540–560	8	550	0	0	0
560–580	6	570	20	1	6
580–600	10	590	40	2	20
Total	Σfi = 50				Σfiui = −12

Used the step-deviation method (xi are large, in the 500s) with a = 550, h = 20.
Mean = a + h(Σfiui÷Σfi) = 550 + 20×(−1250) = 550 − 4.8

Mean daily wages = ₹545.20.

3The 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
Number of children	7	6	9	13	f	5	4

Class interval	fi	xi	fixi
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	Σfi = 44+f		Σfixi = 752+20f

Mean = Σfixi ÷ Σfi = 18 → 752+20f44+f = 18
752+20f = 792+18f → 2f = 40 → f = 20

Missing frequency f = 20.

4Thirty 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
Number of women	2	4	3	8	7	4	2

Class interval	fi	xi	di=xi−75.5	ui=di3	fiui
65–68	2	66.5	−9	−3	−6
68–71	4	69.5	−6	−2	−8
71–74	3	72.5	−3	−1	−3
74–77	8	75.5	0	0	0
77–80	7	78.5	3	1	7
80–83	4	81.5	6	2	8
83–86	2	84.5	9	3	6
Total	Σfi = 30				Σfiui = 4

Used step-deviation with a = 75.5, h = 3.
Mean = a + h(Σfiui÷Σfi) = 75.5 + 3×(430) = 75.5 + 0.4

Mean heartbeats per minute = 75.9.

5In 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
Number of boxes	15	110	135	115	25

These classes are inclusive (50–52, 53–55, ...), with a gap between them, so first make them continuous by extending each boundary by 0.5: 49.5–52.5, 52.5–55.5, ..., 61.5–64.5. This doesn't change any class mark, since the width just shifts symmetrically.

Class interval	fi	xi	fixi
49.5–52.5	15	51	765
52.5–55.5	110	54	5940
55.5–58.5	135	57	7695
58.5–61.5	115	60	6900
61.5–64.5	25	63	1575
Total	Σfi = 400		Σfixi = 22875

Used the Direct Method — the class marks (51 to 63) are close together and manageable, so shifting to an assumed mean isn't needed.
Mean = 22875400 = 57.1875

Mean number of mangoes per box ≈ 57.19.

6The 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
Number of households	4	5	12	2	2

Class interval	fi	xi	di=xi−225	ui=di50	fiui
100–150	4	125	−100	−2	−8
150–200	5	175	−50	−1	−5
200–250	12	225	0	0	0
250–300	2	275	50	1	2
300–350	2	325	100	2	4
Total	Σfi = 25				Σfiui = −7

Used step-deviation with a = 225, h = 50.
Mean = 225 + 50×(−725) = 225 − 14

Mean daily expenditure = ₹211.

7To 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)	Frequency
0.00–0.04	4
0.04–0.08	9
0.08–0.12	9
0.12–0.16	2
0.16–0.20	4
0.20–0.24	2

Class interval	fi	xi	fixi
0.00–0.04	4	0.02	0.08
0.04–0.08	9	0.06	0.54
0.08–0.12	9	0.10	0.90
0.12–0.16	2	0.14	0.28
0.16–0.20	4	0.18	0.72
0.20–0.24	2	0.22	0.44
Total	Σfi = 30		Σfixi = 2.96

Mean = 2.9630 = 0.0987 (approx.)

Mean concentration of SO₂ ≈ 0.099 ppm.

8A 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
Number of students	11	10	7	4	4	3	1

Class interval	fi	xi	fixi
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	Σfi = 40		Σfixi = 499

Used the Direct Method — the class sizes are unequal here (widths of 6, 4, 4, 6, 8, 10, 2), so the step-deviation shortcut (which needs a single common class size h) doesn't cleanly apply.
Mean = 49940 = 12.475

Mean number of days absent ≈ 12.48.

9The 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
Number of cities	3	10	11	8	3

Class interval	fi	xi	fixi
45–55	3	50	150
55–65	10	60	600
65–75	11	70	770
75–85	8	80	640
85–95	3	90	270
Total	Σfi = 35		Σfixi = 2430

Mean = 243035 = 69.4286 (approx.)

Mean literacy rate ≈ 69.43%.

Exercise 13.2

6 Questions

1The 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
Number of patients	6	11	21	23	14	5

Modal class = 35–45 (highest frequency, 23). l=35, h=10, f₁=23, f₀=21, f₂=14.
Mode = l + [f₁−f₀2f₁−f₀−f₂]×h = 35 + [23−2146−21−14]×10 = 35 + (211)×10 = 35 + 1.818

Class interval	fi	xi	fixi
5–15	6	10	60
15–25	11	20	220
25–35	21	30	630
35–45	23	40	920
45–55	14	50	700
55–65	5	60	300
Total	Σfi = 80		Σfixi = 2830

Mean = 283080 = 35.375

Mode ≈ 36.82 years, Mean = 35.375 years. The two are close, so both suggest a typical patient age in the mid-30s; the mode identifies the single most common age-band peak, while the mean reflects the overall average across all admitted patients.

2The 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
Frequency	10	35	52	61	38	29

Modal class = 60–80 (highest frequency, 61). l=60, h=20, f₁=61, f₀=52, f₂=38.
Mode = 60 + [61−52122−52−38]×20 = 60 + (932)×20 = 60 + 5.625

Modal lifetime = 65.625 hours.

3The 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 ₹)	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

Modal class = 1500–2000 (highest frequency, 40). l=1500, h=500, f₁=40, f₀=24, f₂=33.
Mode = 1500 + [40−2480−24−33]×500 = 1500 + (1623)×500 = 1500 + 347.83

Class interval	fi	xi	di=xi−2750	ui=di500	fiui
1000–1500	24	1250	−1500	−3	−72
1500–2000	40	1750	−1000	−2	−80
2000–2500	33	2250	−500	−1	−33
2500–3000	28	2750	0	0	0
3000–3500	30	3250	500	1	30
3500–4000	22	3750	1000	2	44
4000–4500	16	4250	1500	3	48
4500–5000	7	4750	2000	4	28
Total	Σfi = 200				Σfiui = −35

Mean = 2750 + 500×(−35200) = 2750 − 87.5

Modal monthly expenditure ≈ ₹1847.83. Mean monthly expenditure = ₹2662.50.

4The 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 statesU.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

Modal class = 30–35 (highest frequency, 10). l=30, h=5, f₁=10, f₀=9, f₂=3.
Mode = 30 + [10−920−9−3]×5 = 30 + (18)×5 = 30 + 0.625

Class interval	fi	xi	fixi
15–20	3	17.5	52.5
20–25	8	22.5	180
25–30	9	27.5	247.5
30–35	10	32.5	325
35–40	3	37.5	112.5
40–45	0	42.5	0
45–50	0	47.5	0
50–55	2	52.5	105
Total	Σfi = 35		Σfixi = 1022.5

Mean = 1022.535 = 29.214 (approx.)

Mode ≈ 30.6 students per teacher, Mean ≈ 29.2 students per teacher. Most states have a ratio close to 30.6, but a couple of states with very high ratios (50–55) pull the overall average down slightly below the mode.

5The 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	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

Modal class = 4000–5000 (highest frequency, 18). l=4000, h=1000, f₁=18, f₀=4, f₂=9.
Mode = 4000 + [18−436−4−9]×1000 = 4000 + (1423)×1000 = 4000 + 608.7

Mode ≈ 4608.7 runs.

6A 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

Modal class = 40–50 (highest frequency, 20). l=40, h=10, f₁=20, f₀=12, f₂=11.
Mode = 40 + [20−1240−12−11]×10 = 40 + (817)×10 = 40 + 4.706

Mode ≈ 44.7 cars.

Exercise 13.3

7 Questions

1The 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
Number of consumers	4	5	13	20	14	8	4

Class interval	fi	cf
65–85	4	4
85–105	5	9
105–125	13	22
125–145	20	42
145–165	14	56
165–185	8	64
185–205	4	68

n = 68, n2 = 34. Median class = 125–145 (first class with cf ≥ 34, cf=42). l=125, cf(preceding)=22, f=20, h=20.
Median = 125 + [34−2220]×20 = 125 + 12 = 137.

Modal class = 125–145 (highest frequency, 20). l=125, h=20, f₁=20, f₀=13, f₂=14.
Mode = 125 + [20−1340−13−14]×20 = 125 + (713)×20 = 125 + 10.77.

Mean (direct method, xi = 75,95,115,135,155,175,195): Σfixi = 9310, Σfi = 68, Mean = 931068 = 137.06.

Median = 137, Mean ≈ 137.06, Mode ≈ 135.77 units — all three measures are close together, showing the data is fairly symmetric around its centre.

2If 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	Total
Frequency	5	x	20	15	y	5	60

Σfi = 60 → 5+x+20+15+y+5 = 60 → x+y = 15 — (1)
Since the median (28.5) lies between 20 and 30, the median class is 20–30: l=20, cf(preceding) = 5+x, f=20, h=10, n=60, n2=30.
28.5 = 20 + [30−5−x20]×10 → 8.5 = 25−x2 → 25−x = 17 → x = 8.
From (1): 8+y = 15 → y = 7.

x = 8, y = 7.

3A 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 years.

Age (in years)	Number of policy holders
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

Since policies start at 18, the classes are 18–20, 20–25, 25–30, ..., 55–60. Converting the "less than" cumulative data into class frequencies by subtraction:

Class interval	fi	cf
18–20	2	2
20–25	4	6
25–30	18	24
30–35	21	45
35–40	33	78
40–45	11	89
45–50	3	92
50–55	6	98
55–60	2	100

n = 100, n2 = 50. Median class = 35–40 (first class with cf ≥ 50, cf=78). l=35, cf(preceding)=45, f=33, h=5.
Median = 35 + [50−4533]×5 = 35 + 2533

Median age ≈ 35.76 years.

4The 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. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, ..., 171.5-180.5.)

Length (in mm)	118–126	127–135	136–144	145–153	154–162	163–171	172–180
Number of leaves	3	5	9	12	5	4	2

Class interval	fi	cf
117.5–126.5	3	3
126.5–135.5	5	8
135.5–144.5	9	17
144.5–153.5	12	29
153.5–162.5	5	34
162.5–171.5	4	38
171.5–180.5	2	40

n = 40, n2 = 20. Median class = 144.5–153.5 (first class with cf ≥ 20, cf=29). l=144.5, cf(preceding)=17, f=12, h=9.
Median = 144.5 + [20−1712]×9 = 144.5 + 2.25

Median length of leaves = 146.75 mm.

5The 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)	Number of lamps
1500–2000	14
2000–2500	56
2500–3000	60
3000–3500	86
3500–4000	74
4000–4500	62
4500–5000	48

Cumulative frequencies: 14, 70, 130, 216, 290, 352, 400.
n = 400, n2 = 200. Median class = 3000–3500 (first class with cf ≥ 200, cf=216). l=3000, cf(preceding)=130, f=86, h=500.
Median = 3000 + [200−13086]×500 = 3000 + 3500086

Median life time ≈ 3406.98 hours.

6100 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
Number of surnames	6	30	40	16	4	4

Class interval	fi	xi	fixi	cf
1–4	6	2.5	15	6
4–7	30	5.5	165	36
7–10	40	8.5	340	76
10–13	16	11.5	184	92
13–16	4	14.5	58	96
16–19	4	17.5	70	100
Total	Σfi = 100		Σfixi = 832

Median: n=100, n2=50. Median class = 7–10 (first cf ≥ 50, cf=76). l=7, cf(preceding)=36, f=40, h=3.
Median = 7 + [50−3640]×3 = 7 + 1.05 = 8.05

Mean: Mean = 832100 = 8.32

Mode: Modal class = 7–10 (highest frequency, 40). f₁=40, f₀=30, f₂=16.
Mode = 7 + [40−3080−30−16]×3 = 7 + (1034)×3 = 7 + 0.88

Median ≈ 8.05 letters, Mean = 8.32 letters, Mode ≈ 7.88 letters.

7The 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
Number of students	2	3	8	6	6	3	2

Class interval	fi	cf
40–45	2	2
45–50	3	5
50–55	8	13
55–60	6	19
60–65	6	25
65–70	3	28
70–75	2	30

n = 30, n2 = 15. Median class = 55–60 (first class with cf ≥ 15, cf=19). l=55, cf(preceding)=13, f=6, h=5.
Median = 55 + [15−136]×5 = 55 + 106

Median weight ≈ 56.67 kg.

Frequently Asked Questions

The assumed mean and step-deviation methods are just algebraic rearrangements of the direct method's formula — subtracting a fixed value 'a' from every xi (and optionally dividing by the class size h) only shifts and scales the numbers to make the arithmetic easier. Since the same shift and scale is undone at the end by adding 'a' back (and multiplying by h), the final answer is mathematically identical, whichever method you pick.

The modal class is simply the class interval with the highest frequency. But with grouped data you only know how many observations fall somewhere inside that class, not their exact individual values, so you can't point to a single "most repeated number" the way you can with ungrouped data. Instead, the mode formula estimates where inside that class the peak of the distribution most likely sits, using the frequencies of the classes on either side of it.

The median is the value at the halfway point of the data, so you first need to know how many observations lie below each class boundary — that's exactly what a running (cumulative) total gives you. Once you find the class where the cumulative frequency first reaches or passes n2, you know the median falls inside that class, and the median formula pins down where.

For a moderately skewed distribution, 3×Median ≈ Mode + 2×Mean. It's mainly used as a quick cross-check or to estimate one measure when only the other two are known, rather than as an exact formula — the three measures generally only coincide exactly for a perfectly symmetric distribution.

Yes — a printable Formula Card covering all 7 units, including Statistics, is available for ₹99.

Continue to the Next Chapter

Chapter 14 (Probability) is the final chapter of the Class 10 Maths syllabus.

All 14 Chapters Chapter 14, Probability →

Chapter 13Statistics