SEBA Class 10 Maths Chapter 13 Statistics

SEBA Class 10 Maths Chapter 13 Statistics Solutions English Medium As Per New Syllabus. to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Maths Chapter 13 Statistics Notes and select need one. SEBA Class 10 Maths Chapter 13 Statistics Question Answers Download PDF. SEBA Class 10 General Mathematics Solutions.

SEBA Class 10 Maths Chapter 13 Statistics

Also, you can read the SEBA book online in these sections Solutions by Expert Teachers as per Board of Secondary Education, Assam (SEBA) Book guidelines. SEBA Class 10 Maths Chapter 13 Statistics are part of All Subject Solutions. Here we have given SEBA Solutions For Class 10 Maths for All Chapters, You can practice these here.

NEW VIDEO CLASS

Class 10 Mathematics Video Classes

Step-by-step easy concept explanation aur full questions Answer ke liye watch karein ya app download karein:

Watch on YouTube

Get Full Class on App

Statistics

Chapter: 13

Exercise 13.1

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	1	5	6	2	3

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 (f₁)	x₁	f₁xx₁
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	11	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 workers of a factory.

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), d_i, u_i and f_iu_i can be calculated as follows

Daily wages (in Rs)	Number of workers (f_i)	x_i	d_i = x_i − 150	𝒖_𝒊 = 𝒅_𝒊 /20=x₁-150/20	f_iu_i
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

Ans: To find the class mark (x_i) 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 f₁d₁are calculated as follows.

Daily pocket allowance (in Rs)	Number of children f_i	Class mark x_i	d_i = x_i − 18	f_id_i
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. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	65 − 68	68 − 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 (x_i), the following relation is used.

X₁ = Upper limit + Lower limit/2

Class size, h, of this data = 3

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

Number of heart beats per minute	Number of women f_i	x_i	d_i = x_i − 75.5	𝒖_𝒊 = 𝒅_𝒊/3	f_iu_i
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

Therefore, mean heart 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: 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 (x_i) can be obtained by using the following relation.

x_i = Upper limit + Lower limit/2

Class size (h) of this data = 3

Taking 57 as assumed mean (a), d_i, u_i, f₁u₁are calculated as follows

Class interval	f_i	x_i	d_i = x_i − 57	𝒖_𝒊 = x_i-57_i/3	f_iu_i
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	2

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

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

x_i = Upper limit + Lower limit/2

Class size = 50

Taking 225 as assumed mean (a), d_i, u_i, f₁u₁are calculated as follows.

Daily expenditure (in Rs)	f₁	x_i	d_i = x_i − 225	𝒖_𝒊 = x_𝒊-225/50	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				-7

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

Find the mean concentration of the SO₂in the air.

Ans:

Concentration of SO₂ (in ppm)	Frequency f	Class mark x_i	d_i = x_i − 0.14	𝒖_𝒊 = x_𝒊 -0.01/0.04	f_iu_i
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:

Number of days	Number of students f_i	Class-mark 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	∑f_i = 40		∑f_ix_i = 499

Here, we use direct method to find the required mean.

Using the direct method, we have

Therefore, the required mean number of days a student was absent is 12.475.

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.

x_i = Upper limit + Lower limit/2

Class size (h) for this data = 10

Taking 70 as assumed mean (a), d_i, u_i, and fiui are calculated as follows.

Literacy rate (in %)	Number of cities f_i	x₁	d_i = x₁ − 70	𝒖_𝒊 = x_𝒊-70/10	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%.

10. Mean of 10 observations is 2.5. If one observation, namely 2.5 is deleted, determine the new mean.

Ans: Total of 10 observation

= 2.5 × 10

= 25

When 2.5 is deleted

Total of 9 observation

= 25 − 2.5

= 22.5

∴ New mean = 22.5/9

9 observation =2.5

11. If the mean of marks of five students is 33 and that of the marks of first four of them is 32.5, then find the marks obtained by the 5th student.

Ans: Mean of five students = 33

∴ Total marks obtained by 5 students = 5 × 33 = 165

Mean of four students = 32.5

∴ Total marks obtained by 4 students = 4 × 32.5 = 130

∴ Marks obtained by 5th student = 165 − 130 = 35

12. Mean of 25 observations was found to be 78.4. But later on it was discovered that 96 was misread as 69. Find the correct mean.

Ans: Mean of 25 observation = 78.4

∴ Total of 25 observation = 25 × 78.4 = 1960

Corrected value = 96 − 69

= 27

Corrected Total = 1960 + 27 = 1987

Mean of 25 observation when corrected = 1987/25 = 79.48

13. In the following frequency distribution, the frequency of the class interval (40–50) is missing. It is known that the mean of the distribution is 52. Find the missing frequency.