SEBA Class 10 Mathematics Chapter 14 Statistics

By Dev Kirtonia / Class 10 Maths

SEBA Class 10 Mathematics Chapter 14 Statistics Solutions, SEBA Class 10 Maths Textbook Notes in English Medium, SEBA Class 10 Mathematics Chapter 14 Statistics Notes in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 10 Mathematics Chapter 14 Statistics Notes and select needs one.

SEBA Class 10 Mathematics Chapter 14 Statistics

Join Telegram channel

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 14 Statistics Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 14 Statistics Solutions for All Subject, You can practice these here.

Statistics

Chapter – 14

Exercise 14.1

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
Number of houses	1	2	3	4	5	6	7

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

Ans: To find the class mark (xi) for each interval, the following relation is used.

Class mark 𝒙₁=𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕+𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/ 2

Number of plants	Number of houses (fi)	xi	fixi
0 − 2	1	1	1 × 1 = 1
2 − 4	2	3	2 × 3 = 6
4 − 6	1	5	1 × 5 = 5
6 − 8	5	7	5 × 7 = 35
8 − 10	6	9	6 × 9 = 54
10 − 12	2	1	2 ×11 = 22
12 − 14	3	13	3 × 13 = 39
Total	20		162

From the table, it can be observed that

Therefore, the mean number of plants per house is 8.1.

2. Consider the following distribution of daily wages of 50 worker of a factory.

WhatsApp Group Join Now

Telegram Group Join Now

Instagram Join Now

Daily wages (in Rs)	100 − 120	120 − 140	140 −1 60	160 − 180	180 − 200
Number of workers	12	14	8	6	10

Ans: To find the class mark for each interval, the following relation is used.

X¹ = upper limit + Lower Limit// 2

Class size (h) of this data = 20

Taking 150 as assured mean (a) , d1, u1 and f1u1 can be calculate as follows

Daily wages (in Rs)	Number of workers (fi)	xi	di = xi − 150	𝒖𝒊 = 𝒅𝒊 /𝟐0	fiui
100 −120	12	110	− 40	-2	-24
120 − 140	14	130	− 20	-1	14
140 − 160	8	150	0	0	0
160 −180	6	170	20	1	6
180 − 200	10	190	40	2	20
Total	50				-12

From the table, it can be observed that

Therefore, the mean daily wage of the workers of the factory is Rs 145.20

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 Rs)	11 − 13	13 − 15	15 −17	17 − 19	19 − 21	21 − 23	23 − 25
Number of workers	7	6	9	13	f	5 4	4

Ans: To find the class mark (xi) for each interval, the following relation is used

X¹= 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 /2

Given that, mean pocket allowance, Taking 18 as assured mean (a), di and f¹d¹ are calculated as follows.

Daily pocket allowance (in Rs)	Number of children fi	Class mark xi	di = xi − 18	fidi
11 −13	7	12	-6	-42
13 − 15	6	14	-4	-24
15 − 17	9	16	-2	-18
17 −19	13	18	0	0
19 − 21	f	20	2	2f
21 − 23	5	22	4	20
23 − 25	4	24	6	24
Total				2f-40

Hence, the missing frequency, f, is 20.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	65 − 68	68 − 71	71 71 −74	74 − 77	77 − 80	80 − 83	83 − 8
Number of women	2	4	3	8	7	4	2

Ans: To find the class mark of each interval (xi), the following relation is used.

X₁ = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊 / 2

Class size , h , of this data = 3

Taking 75.5 as assumed mean (a) assumed (a) ui, f¹u¹ are calculated as follows.

Number of heart beats per minute	Number of women fi	xi	di = xi − 75.5	𝒖𝒊 = 𝒅𝒊/3	fiui
65 − 68	2	66.5	− 9	-3	-6
68 − 71	4	69.5	− 6	-2	-8
71 − 74	3	69.5	− 3	-1	-3
74 − 77	8	75.5	0	0	0
77 − 80	7	78.5	3	3	7
80 − 83	4	81.5 6	6	6	8
83 − 86	2	84.5	9	9	6
Total	30				4

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a 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?

Ans:

Number of mangoes	Number of boxes fi
50 − 52	15
53 − 55	110
56 − 58	135
59 − 61	115
62 − 64	25

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit.

and 1/2 has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/ 𝟐

Class size (h) of this data = 3

Taking 57 as assumed mean (a), di, ui, f₁u₁are calculated as follows

Class interval	fi	xi	di = xi − 57	𝒖𝒊 = 𝒅i/3
49.5 − 52.5	15	51	-6	-2	-30
52.5 − 55.5	110	54	-3	-1	-110
55.5 − 58.5	135	57	0	0	0
58.5 − 61.5	115	60	3	1	115
61.5 − 64.5	25	63	6	2	50
Total 4	400

Mean number of mangoes kept in a packing box is 57.19. Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all.

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)	100 − 150	150 − 200	200 − 250	250 − 300	300 − 350
Number of households	4	5	12	2	3

Find the mean daily expenditure on food by a suitable method.

Ans: To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 / 𝟐

Class size = 50

Taking 225 as assumed mean (a), di, ui, f₁u₁are calculated as follows.

Daily expenditure (in Rs)	f₁	xi	di = xi − 225	𝒖𝒊 = 𝒅𝒊 50 f	f₁u₁
100 − 150	4	125	-100	-2	-8
100 − 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	25

Therefore, mean daily expenditure on food is Rs 211.

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:

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

To find the class marks for each interval, the following relation is used.

Find the mean concentration if the SO² in the air .

Ans:

Concentration of SO² (in ppm)	Frequency f	Class mark xi	di = xi − 0.14	𝒖𝒊 = 𝒅𝒊 0.04	fiui
0.00 − 0.04	4	0.02	-0.12	− 3	− 12
0.04 − 0.08	9	0.06	-0.08	− 2	− 18
0.08 − 0.12	9	0.10	-0.04	− 1	− 9
0.12 − 0.16	2	0.14	0	0	0
0.16 − 0.20	4	0.18	1	1	4
0.20 − 0.24	2	0.22	2	2	4
Total	30				−31

Therefore, the mean concentration of SO2 in the air is 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	20 − 28	38 − 40
Number of students	11	10	7	4	4	3	1

Ans: To find the class mark of each interval, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊/2

Taking 17 as assumed mean (a), di and f₁d₁are calculated as follows.

Number of days	Number of students	xi	di = xi − 17	fid
0 − 6	11	3	− 14	− 154
6 − 10	10	8	− 9	− 90
10 − 14	7	12	− 5	− 35
14 − 20	4	4	0	0
20 − 28	4	24	7	28
28 − 38	3	33	16	48
38 − 40	1	39	22	22
Total	40			− 181

From the table, we obtain

Therefore, the mean number of days is 12.48 days for which a student was absent.

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
Number of cities	3	10	11	8	3

Ans: To find the class marks, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) for this data = 10

Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.

Literacy rate (in %)	Number of cities fi	x₁	di = x₁ − 70	𝒖𝒊 = 𝒅𝒊 1	f₁ u₁
45 − 55	3	50	-20	-2	-6
55 − 65	10	60	-10	-10	-10
65 − 75	11	70	-0	0	0
75 − 85	8	80	10	1	8
85 − 95	3	90	20	2	6
Total	35				− 2

From the table, we obtain

Therefore, the mean literacy rate is 69.43%.

Exercise 14.2

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.

Ans: To find the class marks (xi), the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 30 as assumed mean (a), di and f₁d₁are calculated as follows.

Age (in years)	Number of patients fi	Class mark xi	di = xi − 30	fidi
5 − 15	6	10	-20	-120
15 − 25	11	20	– 10	-110
25 − 35	21	30	0	0
35 − 45	23	40	10	230
45 − 55	14	50	20	280
55 − 65	5	60	30	150
Total	80			430

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval

35 − 45.

Modal class = 35 − 45

Lower limit (l) of modal class = 35

Frequency (f¹) of modal class = 23

Class size (h) = 10

Frequency (f) of class preceding the modal class = 21

Frequency (f²⁾ of class succeeding the modal class = 14

2.Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.

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.

Ans: From the data given above, it can be observed that the maximum class frequency is61, belonging to class interval 60 − 80.

Therefore, modal class = 60 − 80

Lower class limit (l) of modal class = 60

Frequency (f₁) of modal class = 61

Frequency (f) of class preceding the modal class = 52

Frequency (f₂) of class succeeding the modal class = 38

Class size (h) = 20

Therefore, the modal lifetime of electrical components is 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 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	16

Ans:

Therefore, modal monthly expenditure was Rs 1847.83.

To find the class mark, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) of the given data = 500

Taking 2750 as assumed mean (a), di, ui, and are calculated as follows.

Expenditure (in Rs)	Number of families fi	xi	di = xi − 275	𝒖𝒊 = 𝒅𝒊 𝟓𝟎	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	200				−35

Therefore, the mean monthly expenditure was 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

Ans: It can be observed from the given data that the maximum class frequency is 10

belonging to class interval 30 − 35.

Therefore, modal class = 30 − 35

Class size (h) = 5

Lower limit (l) of modal class = 30

Frequency (f¹) of modal class = 10

Frequency (f⁰) of class preceding modal class = 9

Frequency (f²) of class succeeding modal class = 3

Mode = 30.6

It represents that most of the states/U.T have a teacher-student ratio as 30.6.

To find the class marks, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/2

Taking 32.5 as assumed mean (a), di, ui, and f_iu_iare calculated as follows.

Number of students per teacher	Number of states/U.T (fi)	xi	di = xi − 32.5	𝒖𝒊 = 𝒅𝒊/5	f₁u₁
15 − 20	3	17.5	− 15	-3	− 9
20 − 25	8	22.5	− 10	-2	−16
25 − 30	9	27.5	− 5	-1	− 9
30 − 35	10	32.5	0	0	0
35 − 40	3	37.5	5	1	3
40 − 45	0	42.5	10	2	0
45 − 50	0	47.5	15	3	0
50 − 55	2	52.5	20	4	8
Total	35

Therefore, the mean of the data is 29.2.

It represents that on an average, teacher−student ratio was 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.

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.

Ans:

Therefore, 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:

Ans:

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

From the given data, it can be observed that the maximum class frequency is 20,

belonging to 40 − 50 class intervals. Therefore, modal class = 40 − 50

Lower limit (l) of modal class = 40

Frequency (f₁) of modal class = 20

Frequency (f) of class preceding modal class = 12

Frequency (f₂) of class succeeding modal class = 11

Class size = 10

Therefore, the mode of this data is 44.7 cars.

Exercise 14.3

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)	Number of consumers
65 − 85	4
85 − 105	5
105 − 125	13
125 − 145	20
145 − 165	14
165 − 185	8
185 − 205	4

Ans:

Monthly consumption (in units)	Number of consumers (f i)	xi class mark	di= xi− 135	𝒖𝒊 = 𝒅𝒊 𝟐𝟎	fiui
65 − 85	4	75	− 60	− 3	−12
85 − 105	5	95	− 40	− 2	− 10
105 − 125 1	13	115	− 40	− 1	− 13
125 − 145	20	135	0	0	0
145 − 165	14	155	20	1	14
165 − 185	8	175	40	2	16
185 − 205	4	195	60	3	12
Total	68				12

From the table, it can be observed that the maximum class frequency is 20, belonging

to class interval 125 − 145.

Modal class = 125 − 145

Lower limit (l) of modal class = 125

Class size (h) = 20

Frequency (f¹) of modal class = 20

Frequency (f⁰) of class preceding modal class = 13

Frequency (f²) of class succeeding the modal class = 14

To find the median of the given data, cumulative frequency is calculated as follows.

Monthly consumption (in units)	Number of consumers	Cumulative frequency
65 − 85	4	4
85 − 105	5	4 + 5 = 9
105 − 125	13	9 + 13 = 22
125 − 145	20	22 + 20 = 42
145 − 165	14	42 + 14 = 56
165 − 185	8	56 + 8 = 64
185 − 205	4	64 + 4 = 68

From the table, we obtain n = 68 Cumulative frequency (cf) just greater than n/2

( i.e,68/2=34) is 42 belonging to intervals 125 − 145.

Therefore, median class = 125 − 145

Lower limit (l) of median class = 125

Class size (h) = 20

Frequency (f) of median class = 20

Cumulative frequency (cf) of class preceding median class = 22

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 Respectively.

The three measures are approximately the same in this case.

2. If the median of the distribution is given below is 28.5, find the values of x and y.

Classinterval	0-10	10-20	20-30	30-40	40-50	50-60	total
frequency	5	x	20	15	y	5	60

Ans:

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 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

Ans: Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

Age (in years)	Number of policy holders (fi)	Cumulative frequency (cf
18 − 20	2	2
20 − 25	6 − 2 = 4	6
25 − 30	24 − 6 = 18	24
30 − 35	45 − 24 = 21	45
35 − 40	78 − 45 = 33	78
40 − 45	89 − 78 = 11	89
45 − 50	92 − 89 = 3	92
50 − 55	98 − 92 = 6	98
55 − 60	100 − 98 = 2	100
Total (n)

From the table, it can be observed that n = 100.

Cumulative frequency (cf) just greater than is 78, belonging to interval 35 − 40.

Therefore, median class = 35 − 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Therefore, the median age is 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:

Length (in mm)	Number or leaves f
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.

Ans:

Length (in mm)	Number or leaves f	Cumulative frequency
117.5 − 126.5	3	3
126.5 − 135.5	5	3 + 5 = 8
135.5 − 144.5	9	8 + 9 = 17
144.5 − 153.5	12	17 + 12 = 29
153.5 − 162.5	5	29 + 5 = 34
162.5 − 171.5	4	34 + 4 = 38
171.5 − 180.5	2	38 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater than

is 29, belonging to class interval 144.5 − 153.5.

Median class = 144.5 − 153.5

Lower limit (l) of median class = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

Therefore, median length of leaves is 146.75 mm.

5. Find the following table gives the distribution of the life time of 400 neon lamps:

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

Ans: The cumulative frequencies with their respective class intervals are as follows.

Life time	Number of lamps (fi)	Cumulative frequency
1500 − 2000	14	14
2000 − 2500	56	14 + 56 = 70
2500 − 3000	60	70 + 60 = 130
3000 − 3500	86	130 + 86 = 216
3500 − 4000	74	216 + 74 = 290
4000 − 4500	62	290 + 62 = 352
4500 − 5000	48	352 + 48 = 400
Total (n)	400

It can be observed that the cumulative frequency just greater than n/2 (i.e 400/2 = 200)

Median class = 3000 − 3500

Lower limit (l) of median class = 3000

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

= 3406.976

Therefore, median life time of lamps is 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:

Number of letters	1 − 4	4 − 7	7 − 10	10 − 13	13 − 16	16 − 19
Number of surnames	6	30	40	6	4	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.

Ans: The cumulative frequencies with their respective class intervals are as follows.

Number of letters	Frequency (fi)	Cumulative frequency
1 − 4	6	6
4 − 7	30	30 + 6 = 36
7 − 10	40	36 + 40 = 76
10 − 13	16	76 + 16 = 92
13 − 16	4	92 + 4 = 96
16 − 19	4	96 + 4 = 100
Total (n)	100

It can be observed that the cumulative frequency just greater than n/2( i .e , 100/2 = 50) is 76, belonging to class interval 7 − 10.

Median class = 7 − 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36

Frequency (f) of median class = 40

Class size (h) = 3

= 8.05

To find the class marks of the given class intervals, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows.

(ii) Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows.

Ans:

Number of letters	Number of surnames fi	fi xi	di = xi− 11.5	𝒖𝒊 = 𝒅𝒊 𝟑	fiui
1 − 4	6	2.5	− 9	− 3	− 18
4 − 7	30	5.5	− 6	− 2	− 60
7 − 10	40	8.5	− 3	− 1	− 40
10 − 13	16	11.5	0	0	0
13 − 16	4	14.5	3	1	4
16 − 19	4	17.5	6	2	8
Total	100

From the table, we obtain

∑fiui = −106

∑fi = 100

Mean ,

= 11.5 − 3.18 = 8.32

The data in the given table can be written as

Number of letters	Frequency (fi)
1 − 4	6
4 − 7	30
7 − 10	40
13 − 16	16
13 − 16	4
16 − 19	4
	100

From the table, it can be observed that the maximum class frequency is 40

belonging to class interval 7 − 10. Modal class = 7 − 10

Lower limit (l) of modal class = 7

Class size (h) = 3

Frequency (f¹) of modal class = 40

Frequency (f⁰) of class preceding the modal class = 30

Frequency (f²) of class succeeding the modal class = 16

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.

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
Number of students	2	3	8	6	6	3	2

The cumulative frequencies with their respective class intervals are as follows.

Ans:

Weight (in kg)	Frequency (fi)	Cumulative frequency
40 − 45	2	2
45 − 50	3	2 + 3 = 5
50 − 55	8	5 + 8 = 13
55 − 60	6	13 + 6 = 19
60 − 65	6	19 + 6 = 25
65 − 70	3	25 + 3 = 28
70 − 75	2	28 + 2 = 30
Total (n)	2

Cumulative frequency just greater than is 19, belonging to class

interval 55 − 60.

Median class = 55 − 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) of median class = 13

Class size (h) = 5

= 56.67

Therefore, the median weight is 56.67 kg.

Exercise 14.4

1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs)	100−120	120−140	140−160	160−180	180−200
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its given.

Ans: The frequency distribution table of less than type is as follows.

Daily income (in Rs) (upper class limits)	Cumulative frequency
Less than 120	12
Less than 140	12 + 14 = 26
Less than 160	26 + 8 = 34
Less than 180	34 + 6 = 40
Less than 200	40 + 10 = 50

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
ess than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verifies the result by using the formula.

Ans: The given cumulative frequency distributions of less than type are:

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
ess than 52	35

Taking upper class limits on x-axis and their respective cumulative frequencies on y – axis, its ogive can be drawn as follows.

Here, n = 35

So, n/2 = 17.5

Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore,

the median of this data is 46.5.

It can be observed that the difference between two consecutive upper class limits is

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield in (kg/ha)	50 – 55	55 -60	60 – 65	65 – 70	70 – 75	75 -80
Number of firms	2	8	12	24	38	16

Change the distribution to a more than type distribution , and drawn its given.

Ans:

production yield	Number of Frequency (f)	Cumulative frequency more than type
50 – 55	2	50 or more than 50	100 = 100
55 -60	8	55 or more than 55	(100 – 2) = 98
60 – 65	12	60 or more than 60	(98 – 8 ) = 90
65 – 70	34	65 or more than 65	(90 – 12) = 78
70 – 75	38	70 or more than 70	(78 – 24 ) = 54
75 -80	16	75 or more than 75	(54 – 38) = 16
Total	N = 100

Now , we will draw the given by plotting the pointing the the point ( 50, 100), ( 55, 98), (60,90), (65,78), (70,54) and (75,16)

Dev Kirtonia

Hi, I’m Dev Kirtonia, Founder & CEO of Dev Library. A website that provides all SCERT, NCERT 3 to 12, and BA, B.com, B.Sc, and Computer Science with Post Graduate Notes & Suggestions, Novel, eBooks, Biography, Quotes, Study Materials, and more.

SEBA Class 10 Mathematics Chapter 14 Statistics

Statistics

Related Posts

Leave a Comment Cancel Reply