NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.1.

Students find NCERT Solutions for Class 10 Maths Chapter 14 Statistics quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 14 are provided here for you for free.

• Statistics Class 10 Ex 14.2
• Statistics Class 10 Ex 14.3
• Statistics Class 10 Ex 14.4

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

NCERT Solutions for Class 10 Maths

Question 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.

Which method did you use for finding the mean, and why?
Solution:
Let us find class marks (x_i) for each interval by using the relation.

Now we may compute x_i and f_ix_i as following

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	11	2 —11 = 22
12 – 14	3	13	3 — 13 = 39
Total	20		162

From the table we may observe that

So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (x_i) and f_i are small.

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

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

Solution:
Let us find class mark for each interval by using the relation.

Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, u_i and f_iu_i as following.

Daily wages (in Rs)	Number of workers (fi)	xi	di = xi – 150		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 we may observe that

So mean daily wages of the workers of the factory is Rs.145.20

Question 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

Solution:
We may find class mark (x_i) for each interval by using the relation.

Given that mean pocket allowance \(\overline { x } \) = Rs.18
Now taking 18 as assured mean (a) we may calculate d_i and f_id_i as following.

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	2 f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total				2f – 40

From the table we may obtain

Hence the missing frequency f is 20.

Question 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	71-74	74 – 77	77 – 80	80 – 83	83 – 86
Number of women	2	4	3	8	7	4	2

Solution:
We may find class mark of each interval (x_i) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following.

Number of heart beats per minute	Number of women fi	xi	di = xi -75.5		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	30				4

Now we may observe from table that

So mean heart beats per minute for these women are 75.9 beats per minute.

Question 5.
In 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?
Solution:

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

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add 1/2 to upper class limit and subtract 1/2 from lower class limit of each interval.
And class mark (x_i) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following –

Class interval	fi	xi	di = xi – 57		fiui
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	400				25

Now we may observe that

Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of f_i, d_i are big and also there is a common multiple between all d_i.

Question 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.
Solution:
We may calculate cla

mark (x_i) for each interval by using the relation

Class size = 50

Now taking 225 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following

Daily expenditure (in Rs)	fi	xi	di = xi – 225		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	25				-7

Now we may observe that –

Question 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 pmm)	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 SO₂ in the air.

Solution:

Concentration of SO2 (in ppm)	Frequency	Class mark xi	di = xi – 0.14		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	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

Question 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	28 – 38	38 –  40
Number of students	11	10	7	4	4	3	1

Solution:
We may find class mark of each interval by using the relation

Now taking 16 as assumed mean (a) we may calculate d_i and f_id_i as following

Number of days	Number of students fi	xi	di= xi – 16	fidi
0 – 6	11	3	-13	-143
6 -10	10	8	-8	-80
10 – 14	7	12	-4	-28
14 – 20	4	16	0	0
20 – 28	4	24	8	32
28 – 38	3	33	17	51
38 – 40	1	39	23	23
Total	40			-145

Now we may observe that

So, mean number of days is 12.38 days, for which a student was absent.

Question 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

Solution:
We may find class marks by using the relation

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate d_i, u_i, and f_iu_i as following

Literacy rate  (in %)	Number of cities  fi	xi	di= xi – 70	ui	fiui
45 – 55	3	50	-20	-2	-6
55 – 65	10	60	-10	-1	-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

Now we may observe that

So, mean literacy rate is 69.43%.

 

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