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Chapter 13: Statistics (Statistics)

Step-by-step NCERT solutions for Class 10 Maths Chapter 13, Statistics (2026-27 reprint): Exercise 13.1 mean of grouped data (direct and step-deviation methods), Exercise 13.2 mode, and Exercise 13.3 median, with every frequency table worked in full. All 22 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-13-statistics

The class mark is $x_{i} = \frac{lower limit + upper limit}{2}$ . Mean: direct $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , or step-deviation $\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h$ with $u_{i} = \frac{x _{i} - a}{h}$ (best when the class marks are large and the classes are of equal width). Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ . Median $= l + \frac{\frac{n}{2} - c f}{f} \times h$ . For mode and median the classes must be continuous.

Exercise 13.1

The number of plants in 20 houses is given below. Find the mean number of plants per house. Which method did you use, and why?

Solution

Number of plants	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
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	20		162

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{162}{20} = 8.1$

We used the direct method because the class marks and frequencies are small numbers, so $f_{i} x_{i}$ is easy to work out.

8.1 plants per house (direct method, since the numbers are small).

Find the mean daily wages of the 50 workers using an appropriate method.

Solution

Take the assumed mean $a = 550$ and the class width $h = 20$ (step-deviation method).

Daily wages (₹)	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 550}{20}$	$f_{i} u_{i}$
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	50			−12

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 550 + \frac{- 12}{50} \times 20 = 545.2$

₹545.20

The mean daily pocket allowance of the children is ₹18. Find the missing frequency f.

Solution

Allowance (₹)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
11 – 13	7	12	84
13 – 15	6	14	84
15 – 17	9	16	144
17 – 19	13	18	234
19 – 21	$f$	20	$20 f$
21 – 23	5	22	110
23 – 25	4	24	96
Total	$44 + f$		$752 + 20 f$

$\frac{752 + 20 f}{44 + f} = 18 \Rightarrow 752 + 20 f = 792 + 18 f \Rightarrow 2 f = 40$

f = 20

Find the mean heartbeats per minute of the 30 women, choosing a suitable method.

Solution

Step-deviation with $a = 75.5$ , $h = 3$ :

Heartbeats per minute	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 75.5}{3}$	$f_{i} u_{i}$
65 – 68	2	66.5	−3	−6
68 – 71	4	69.5	−2	−8
71 – 74	3	72.5	−1	−3
74 – 77	8	75.5	0	0
77 – 80	7	78.5	1	7
80 – 83	4	81.5	2	8
83 – 86	2	84.5	3	6
Total	30			4

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 75.5 + \frac{4}{30} \times 3 = 75.9$

75.9 heartbeats per minute

Find the mean number of mangoes in a packing box. Which method did you choose?

Solution

The class marks are 51, 54, … and the frequencies are large, so we use the step-deviation method with $a = 57$ , $h = 3$ . (The mean does not change if we first make the classes continuous, 49.5 – 52.5 and so on, because the class marks stay the same.)

Number of mangoes	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 57}{3}$	$f_{i} u_{i}$
50 – 52	15	51	−2	−30
53 – 55	110	54	−1	−110
56 – 58	135	57	0	0
59 – 61	115	60	1	115
62 – 64	25	63	2	50
Total	400			25

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 57 + \frac{25}{400} \times 3 = 57.188$

About 57.19 mangoes per box (step-deviation method).

Find the mean daily expenditure on food of the 25 households.

Solution

Step-deviation with $a = 225$ , $h = 50$ :

Daily expenditure (₹)	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 225}{50}$	$f_{i} u_{i}$
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	25			−7

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 225 + \frac{- 7}{25} \times 50 = 211$

₹211

Find the mean concentration of SO₂ in the air of 30 localities.

Solution

SO₂ (ppm)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0 – 0.04	4	0.02	0.08
0.04 – 0.08	9	0.06	0.54
0.08 – 0.12	9	0.1	0.9
0.12 – 0.16	2	0.14	0.28
0.16 – 0.2	4	0.18	0.72
0.2 – 0.24	2	0.22	0.44
Total	30		2.96

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2.96}{30} = 0.099$

≈ 0.099 ppm

Find the mean number of days a student was absent (40 students).

Solution

The class widths are unequal, so we use the direct method.

Days absent	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
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	40		499

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{499}{40} = 12.475$

≈ 12.48 days

Find the mean literacy rate of the 35 cities.

Solution

Step-deviation with $a = 70$ , $h = 10$ :

Literacy rate (%)	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 70}{10}$	$f_{i} u_{i}$
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	35			−2

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 70 + \frac{- 2}{35} \times 10 = 69.429$

≈ 69.43%

Exercise 13.2

Find the mode and the mean of the ages of patients admitted to a hospital. Compare and interpret the two measures.

Solution

Mode

The highest frequency is 23, so the modal class is 35 – 45: $l = 35$ , $h = 10$ , $f_{1} = 23$ , $f_{0} = 21$ , $f_{2} = 14$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{2}{11} \times 10 = 36.818$

Mean

Age (years)	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
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	80		2830

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2830}{80} = 35.375$

Mode ≈ 36.8 years, mean = 35.375 years. The largest number of patients admitted are about 36.8 years old, while the average age of a patient is about 35.4 years.

Determine the modal lifetime of the 225 electrical components.

Solution

The highest frequency is 61, so the modal class is 60 – 80: $l = 60$ , $h = 20$ , $f_{1} = 61$ , $f_{0} = 52$ , $f_{2} = 38$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 60 + \frac{9}{32} \times 20 = 65.625$

65.625 hours

Find the modal monthly expenditure and the mean monthly expenditure of the 200 families.

Solution

Mode

The highest frequency is 40, so the modal class is 1500 – 2000: $l = 1500$ , $h = 500$ , $f_{1} = 40$ , $f_{0} = 24$ , $f_{2} = 33$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 1500 + \frac{16}{23} \times 500 = 1847.826$

Mean (step-deviation, $a = 2750$ , $h = 500$ )

Expenditure (₹)	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 2750}{500}$	$f_{i} u_{i}$
1000 – 1500	24	1250	−3	−72
1500 – 2000	40	1750	−2	−80
2000 – 2500	33	2250	−1	−33
2500 – 3000	28	2750	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	200			−35

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 2750 + \frac{- 35}{200} \times 500 = 2662.5$

Modal expenditure ≈ ₹1847.83; mean expenditure ₹2662.50

Find the mode and mean of the state-wise teacher–student ratio, and interpret them.

Solution

Mode

The highest frequency is 10, so the modal class is 30 – 35: $l = 30$ , $h = 5$ , $f_{1} = 10$ , $f_{0} = 9$ , $f_{2} = 3$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 30 + \frac{1}{8} \times 5 = 30.625$

Mean (step-deviation, $a = 32.5$ , $h = 5$ )

Students per teacher	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 32.5}{5}$	$f_{i} u_{i}$
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	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	35			−23

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 32.5 + \frac{- 23}{35} \times 5 = 29.214$

Mode ≈ 30.6, mean ≈ 29.2. Most states/UTs have about 30.6 students per teacher, while on average the ratio is about 29.2.

Find the mode of the runs scored by top batsmen in one-day internationals.

Solution

The highest frequency is 18, so the modal class is 4000 – 5000: $l = 4000$ , $h = 1000$ , $f_{1} = 18$ , $f_{0} = 4$ , $f_{2} = 9$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 4000 + \frac{14}{23} \times 1000 = 4608.696$

≈ 4608.7 runs

Find the mode of the number of cars passing a spot in 100 periods of 3 minutes.

Solution

The highest frequency is 20, so the modal class is 40 – 50: $l = 40$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = 12$ , $f_{2} = 11$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{8}{17} \times 10 = 44.706$

≈ 44.7 cars

Exercise 13.3

Find the median, mean and mode of the monthly electricity consumption of 68 consumers and compare them.

Solution

Median

Units	Frequency	Cumulative frequency
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

$n = 68$ , so $\frac{n}{2} = 34$ . The first cumulative frequency above 34 is 42, so the median class is 125 – 145: $l = 125$ , $c f = 22$ , $f = 20$ , $h = 20$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 125 + \frac{34 - 22}{20} \times 20 = 137$

Mean (step-deviation, $a = 135$ , $h = 20$ )

Units	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 135}{20}$	$f_{i} u_{i}$
65 – 85	4	75	−3	−12
85 – 105	5	95	−2	−10
105 – 125	13	115	−1	−13
125 – 145	20	135	0	0
145 – 165	14	155	1	14
165 – 185	8	175	2	16
185 – 205	4	195	3	12
Total	68			7

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 135 + \frac{7}{68} \times 20 = 137.059$

Mode

The highest frequency is 20, so the modal class is 125 – 145: $l = 125$ , $h = 20$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 14$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 125 + \frac{7}{13} \times 20 = 135.769$

Median = 137 units, mean ≈ 137.06 units, mode ≈ 135.77 units (the answer key truncates these to 137.05 and 135.76). The three measures are almost the same here.

The median of the distribution (total frequency 60) is 28.5. Find x and y.

Solution

Class	Frequency	Cumulative frequency
0 – 10	5	5
10 – 20	$x$	$5 + x$
20 – 30	20	$25 + x$
30 – 40	15	$40 + x$
40 – 50	$y$	$40 + x + y$
50 – 60	5	$45 + x + y$

The total is 60, so $45 + x + y = 60$ , i.e. $x + y = 15$ .

The median 28.5 lies in 20 – 30, so $l = 20$ , $c f = 5 + x$ , $f = 20$ , $h = 10$ :

$28.5 = 20 + \frac{30 - ( 5 + x )}{20} \times 10 = 20 + \frac{25 - x}{2} \Rightarrow 17 = 25 - x \Rightarrow x = 8$

Then $y = 15 - 8 = 7$ .

x = 8, y = 7

Calculate the median age of 100 policy holders, given as a "less than" table (policies are given only to people aged 18 to under 60).

Solution

Convert the cumulative table to class frequencies. The youngest holder is 18, so the first class is 18 – 20; each later frequency is the difference of consecutive cumulative frequencies (6 − 2 = 4, 24 − 6 = 18, …).

Age (years)	Frequency	Cumulative frequency
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

$n = 100$ , so $\frac{n}{2} = 50$ . The first cumulative frequency above 50 is 78, so the median class is 35 – 40: $l = 35$ , $c f = 45$ , $f = 33$ , $h = 5$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 35 + \frac{50 - 45}{33} \times 5 = 35.758$

Median age ≈ 35.76 years

Find the median length of the 40 leaves.

Solution

Make the classes continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit:

Length (mm)	Frequency	Cumulative frequency
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

$n = 40$ , so $\frac{n}{2} = 20$ . The first cumulative frequency above 20 is 29, so the median class is 144.5 – 153.5: $l = 144.5$ , $c f = 17$ , $f = 12$ , $h = 9$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 144.5 + \frac{20 - 17}{12} \times 9 = 146.75$

Median length = 146.75 mm

Find the median life time of the 400 neon lamps.

Solution

Life time (hours)	Frequency	Cumulative frequency
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400

$n = 400$ , so $\frac{n}{2} = 200$ . The first cumulative frequency above 200 is 216, so the median class is 3000 – 3500: $l = 3000$ , $c f = 130$ , $f = 86$ , $h = 500$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 3000 + \frac{200 - 130}{86} \times 500 = 3406.977$

Median life ≈ 3406.98 hours

For the number of letters in 100 surnames, find the median, the mean and the modal size.

Solution

Median

Number of letters	Frequency	Cumulative frequency
1 – 4	6	6
4 – 7	30	36
7 – 10	40	76
10 – 13	16	92
13 – 16	4	96
16 – 19	4	100

$n = 100$ , so $\frac{n}{2} = 50$ . The first cumulative frequency above 50 is 76, so the median class is 7 – 10: $l = 7$ , $c f = 36$ , $f = 40$ , $h = 3$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 7 + \frac{50 - 36}{40} \times 3 = 8.05$

Mean (step-deviation, $a = 8.5$ , $h = 3$ )

Number of letters	$f_{i}$	$x_{i}$	$u_{i} = \frac{x _{i} - 8.5}{3}$	$f_{i} u_{i}$
1 – 4	6	2.5	−2	−12
4 – 7	30	5.5	−1	−30
7 – 10	40	8.5	0	0
10 – 13	16	11.5	1	16
13 – 16	4	14.5	2	8
16 – 19	4	17.5	3	12
Total	100			−6

$\overset{x}{ˉ} = a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 8.5 + \frac{- 6}{100} \times 3 = 8.32$

Mode

The highest frequency is 40, so the modal class is 7 – 10: $l = 7$ , $h = 3$ , $f_{1} = 40$ , $f_{0} = 30$ , $f_{2} = 16$ .

$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 7 + \frac{10}{34} \times 3 = 7.882$

Median = 8.05, mean = 8.32, modal size ≈ 7.88 letters

Find the median weight of the 30 students.

Solution

Weight (kg)	Frequency	Cumulative frequency
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

$n = 30$ , so $\frac{n}{2} = 15$ . The first cumulative frequency above 15 is 19, so the median class is 55 – 60: $l = 55$ , $c f = 13$ , $f = 6$ , $h = 5$ .

$Median = l + \frac{\frac{n}{2} - c f}{f} \times h = 55 + \frac{15 - 13}{6} \times 5 = 56.667$

Median weight ≈ 56.67 kg

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