Class 10 Mathematics · Chapter 13

13. Statistics

Step-by-step solutions to Exercise 13.1, 13.2 & 13.3. NCERT Mathematics Textbook for Class X (CBSE).

Exercise 13.1

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. (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) Which method did you use for finding the mean, and why?

Number of plants	Number of houses (f_i)	Class mark (x_i)	f_ix_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	Σf_i = 20		Σf_ix_i = 162

x̄ = Σf_ix_iΣf_i = 16220 = 8.1

Mean number of plants per house 8.1

We used the direct method, because the values of x_i and f_i are small.

2Consider 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; Number of workers: 12, 14, 8, 6, 10) Find the mean daily wages of the workers of the factory by using an appropriate method.

The values of x_i are large. So, we use the step-deviation method.

Daily wages (₹)	Number of workers (f_i)	x_i	u_i = x_i – 55020	f_iu_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	Σf_i = 50			Σf_iu_i = –12

Assumed mean a = 550, class size h = 20

x̄ = a + h × Σf_iu_iΣf_i = 550 + 20 × –1250 = 545.2

Mean daily wages ₹ 545.20

3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 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)

Daily allowance (₹)	Number of children (f_i)	x_i	f_ix_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	20f
21 – 23	5	22	110
23 – 25	4	24	96
Total	Σf_i = 44 + f		Σf_ix_i = 752 + 20f

x̄ = Σf_ix_iΣf_i ⇒ 18 = 752 + 20f44 + f

792 + 18f = 752 + 20f ⇒ 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. (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)

We use the step-deviation method.

Heartbeats per minute	Number of women (f_i)	x_i	u_i = x_i – 75.53	f_iu_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	Σf_i = 30			Σf_iu_i = 4

Assumed mean a = 75.5, class size h = 3

x̄ = a + h × Σf_iu_iΣf_i = 75.5 + 3 × 430 = 75.9

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. (Number of mangoes: 50–52, 53–55, 56–58, 59–61, 62–64; Number of boxes: 15, 110, 135, 115, 25) Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Number of mangoes	Number of boxes (f_i)	x_i	u_i = x_i – 573	f_iu_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	Σf_i = 400			Σf_iu_i = 25

Assumed mean a = 57, class size h = 3

x̄ = a + h × Σf_iu_iΣf_i = 57 + 3 × 25400 = 57.19

Mean number of mangoes in a box 57.19 (approx.)

We chose the step-deviation method, because the frequencies are large and the class marks are equally spaced.

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

We use the step-deviation method.

Daily expenditure (₹)	Households (f_i)	x_i	u_i = x_i – 22550	f_iu_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	Σf_i = 25			Σf_iu_i = –7

Assumed mean a = 225, class size h = 50

x̄ = a + h × Σf_iu_iΣf_i = 225 + 50 × –725 = 211

Mean daily expenditure on food ₹ 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: (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) Find the mean concentration of SO₂ in the air.

Concentration of SO₂ (ppm)	Frequency (f_i)	Class mark (x_i)	f_ix_i
0 – 0.04	4	0.02	0.08
0.04 – 0.08	9	0.06	0.54
0.08 – 0.12	9	0.10	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	Σf_i = 30		Σf_ix_i = 2.96

x̄ = Σf_ix_iΣf_i = 2.9630 = 0.099

Mean concentration of SO₂ 0.099 ppm (approx.)

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)

The class sizes are unequal. So, we use the assumed mean method.

Number of days	Number of students (f_i)	x_i	d_i = x_i – 17	f_id_i
0 – 6	11	3	–14	–154
6 – 10	10	8	–9	–90
10 – 14	7	12	–5	–35
14 – 20	4	17	0	0
20 – 28	4	24	7	28
28 – 38	3	33	16	48
38 – 40	1	39	22	22
Total	Σf_i = 40			Σf_id_i = –181

Assumed mean a = 17

x̄ = a + Σf_id_iΣf_i = 17 + –18140 = 12.475

Mean number of days a student was 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)

Literacy rate (%)	Number of cities (f_i)	x_i	u_i = x_i – 7010	f_iu_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	Σf_i = 35			Σf_iu_i = –2

Assumed mean a = 70, class size h = 10

x̄ = a + h × Σf_iu_iΣf_i = 70 + 10 × –235 = 69.43

Mean literacy rate 69.43 %

Exercise 13.2

1The 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.

Mode

The maximum class frequency is 23. So, the modal class is 35 – 45.

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 = 36.82

Mode 36.82 years (approx.)

Mean

Age (years)	Number of patients (f_i)	x_i	u_i = x_i – 3010	f_iu_i
5 – 15	6	10	–2	–12
15 – 25	11	20	–1	–11
25 – 35	21	30	0	0
35 – 45	23	40	1	23
45 – 55	14	50	2	28
55 – 65	5	60	3	15
Total	Σf_i = 80			Σf_iu_i = 43

Assumed mean a = 30, class size h = 10

x̄ = a + h × Σf_iu_iΣf_i = 30 + 10 × 4380 = 35.375

Mean 35.37 years (approx.)

Interpretation

Maximum number of patients admitted in the hospital are of the age 36.8 years (approx.).

The average age of a patient admitted to the hospital is 35.37 years.

2The 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.

The maximum class frequency is 61. So, the modal class is 60 – 80.

l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 60 + (61 – 52122 – 52 – 38) × 20

= 60 + 932 × 20 = 65.625

Modal lifetime of the components 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 ₹: 1000–1500: 24, 1500–2000: 40, 2000–2500: 33, 2500–3000: 28, 3000–3500: 30, 3500–4000: 22, 4000–4500: 16, 4500–5000: 7)

Mode

The maximum class frequency is 40. So, the modal class is 1500 – 2000.

l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 1500 + (40 – 2480 – 24 – 33) × 500

= 1500 + 1623 × 500 = 1847.83

Modal monthly expenditure ₹ 1847.83 (approx.)

Mean

Expenditure (₹)	Families (f_i)	x_i	u_i = x_i – 2750500	f_iu_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	Σf_i = 200			Σf_iu_i = –35

Assumed mean a = 2750, class size h = 500

x̄ = a + h × Σf_iu_iΣf_i = 2750 + 500 × –35200 = 2662.5

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: 15–20: 3, 20–25: 8, 25–30: 9, 30–35: 10, 35–40: 3, 40–45: 0, 45–50: 0, 50–55: 2)

Mode

The maximum class frequency is 10. So, the modal class is 30 – 35.

l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 30 + (10 – 920 – 9 – 3) × 5

= 30 + 18 × 5 = 30.625

Mode 30.6 (approx.)

Mean

Students per teacher	States / U.T. (f_i)	x_i	u_i = x_i – 32.55	f_iu_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	Σf_i = 35			Σf_iu_i = –23

Assumed mean a = 32.5, class size h = 5

x̄ = a + h × Σf_iu_iΣf_i = 32.5 + 5 × –2335 = 29.21

Mean 29.2 (approx.)

Interpretation

Most states/U.T. have a student-teacher ratio of about 30.6.

On an average, the ratio is about 29.2 students per teacher.

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: 3000–4000: 4, 4000–5000: 18, 5000–6000: 9, 6000–7000: 7, 7000–8000: 6, 8000–9000: 3, 9000–10000: 1, 10000–11000: 1)

The maximum class frequency is 18. So, the modal class is 4000 – 5000.

l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 4000 + (18 – 436 – 4 – 9) × 1000

= 4000 + 1423 × 1000 = 4608.70

Mode 4608.7 runs (approx.)

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)

The maximum class frequency is 20. So, the modal class is 40 – 50.

l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 40 + (20 – 1240 – 12 – 11) × 10

= 40 + 817 × 10 = 44.71

Mode 44.7 cars (approx.)

Exercise 13.3

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: 4, 85–105: 5, 105–125: 13, 125–145: 20, 145–165: 14, 165–185: 8, 185–205: 4)

Median

Monthly consumption (units)	Number of consumers	Cumulative frequency (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

The cumulative frequency just greater than 34 is 42. So, the median class is 125 – 145.

l = 125, cf = 22, f = 20, h = 20

Median = l + (n/2 – cff) × h = 125 + (34 – 2220) × 20

= 125 + 24020 = 137

Median 137 units

Mean

Consumption (units)	Consumers (f_i)	x_i	u_i = x_i – 13520	f_iu_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	Σf_i = 68			Σf_iu_i = 7

Assumed mean a = 135, class size h = 20

x̄ = a + h × Σf_iu_iΣf_i = 135 + 20 × 768 = 137.06

Mean 137.05 units (approx.)

Mode

The maximum class frequency is 20. So, the modal class is 125 – 145.

l = 125, h = 20, f₁ = 20, f₀ = 13, f₂ = 14

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 125 + (20 – 1340 – 13 – 14) × 20

= 125 + 713 × 20 = 135.77

Mode 135.76 units (approx.)

Comparison

The three measures are nearly equal: Mode (135.76) < Median (137) < Mean (137.05).

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

Class interval	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

n = 60 ⇒ 45 + x + y = 60 ⇒ x + y = 15 ... (1)

The median is 28.5, so the median class is 20 – 30.

l = 20, cf = 5 + x, f = 20, h = 10, n2 = 30

28.5 = 20 + (30 – (5 + x)20) × 10 ⇒ 8.5 = 25 – x2 ⇒ 25 – x = 17 ⇒ x = 8

From (1): y = 15 – 8 = 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 year. (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)

The given data is a cumulative frequency (less than type) distribution. We form the classes and frequencies first.

Age (years)	Number of policy holders	Cumulative frequency (cf)
15 – 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

The cumulative frequency just greater than 50 is 78. So, the median class is 35 – 40.

l = 35, cf = 45, f = 33, h = 5

Median = l + (n/2 – cff) × h = 35 + (50 – 4533) × 5

= 35 + 2533 = 35.76

Median age 35.76 years (approx.)

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: (Length in mm: 118–126: 3, 127–135: 5, 136–144: 9, 145–153: 12, 154–162: 5, 163–171: 4, 172–180: 2) Find the median length of the leaves.

The classes are converted to continuous classes: 117.5 – 126.5, 126.5 – 135.5, ..., 171.5 – 180.5.

Length (mm)	Number of leaves	Cumulative frequency (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

The cumulative frequency just greater than 20 is 29. So, the median class is 144.5 – 153.5.

l = 144.5, cf = 17, f = 12, h = 9

Median = l + (n/2 – cff) × h = 144.5 + (20 – 1712) × 9

= 144.5 + 2712 = 146.75

Median length of the leaves 146.75 mm

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

Life time (hours)	Number of lamps	Cumulative frequency (cf)
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 ⇒ n2 = 200

The cumulative frequency just greater than 200 is 216. So, the median class is 3000 – 3500.

l = 3000, cf = 130, f = 86, h = 500

Median = l + (n/2 – cff) × h = 3000 + (200 – 13086) × 500

= 3000 + 3500086 = 3406.98

Median life time of a lamp 3406.98 hours (approx.)

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: (Number of letters: 1–4: 6, 4–7: 30, 7–10: 40, 10–13: 16, 13–16: 4, 16–19: 4) 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.

Median

Number of letters	Number of surnames	Cumulative frequency (cf)
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 ⇒ n2 = 50

The cumulative frequency just greater than 50 is 76. So, the median class is 7 – 10.

l = 7, cf = 36, f = 40, h = 3

Median = l + (n/2 – cff) × h = 7 + (50 – 3640) × 3

= 7 + 4240 = 8.05

Median 8.05

Mean

Number of letters	Surnames (f_i)	x_i	u_i = x_i – 8.53	f_iu_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	Σf_i = 100			Σf_iu_i = –6

Assumed mean a = 8.5, class size h = 3

x̄ = a + h × Σf_iu_iΣf_i = 8.5 + 3 × –6100 = 8.32

Mean 8.32

Mode

The maximum class frequency is 40. So, the modal class is 7 – 10.

l = 7, h = 3, f₁ = 40, f₀ = 30, f₂ = 16

Mode = l + (f₁ – f₀2f₁ – f₀ – f₂) × h = 7 + (40 – 3080 – 30 – 16) × 3

= 7 + 1034 × 3 = 7.88

Modal size 7.88 (approx.)

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

Weight (kg)	Number of students	Cumulative frequency (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

The cumulative frequency just greater than 15 is 19. So, the median class is 55 – 60.

l = 55, cf = 13, f = 6, h = 5

Median = l + (n/2 – cff) × h = 55 + (15 – 136) × 5

= 55 + 106 = 56.67

Median weight 56.67 kg (approx.)

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