CBSE Class 10 - Maths - CH 14 - Statistics (Ex 14.1)

Statistics

(NCERT Ex 14.1)

Q1: 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.


Answer:

We use the following relation to find the class mark (x_i) for each interval:

Class mark (x_i) =

Calculating x_i and f_ix_i as follows:

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, it is observed that

Σ f_i = 20

Σ f_ix_i = 162

Mean , x =

= = 8.1

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (x_i) and f_i are small.


Q2: 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

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

Answer:



To find the class mark (x_i) for each interval, using the following relation:

Class mark (x_i) =

Class size (h) = 20

Taking 150 as assumed mean (a), d_i, u_i, and f_iu_i can be calculated as follows:

Daily wages(in Rs)	Number of workers(fi)	xi	di = xi - 150	ui =	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

Σ f_i = 50

Σ f_iu_i = -12

Mean , x = a +()h

             = 150 + ()20

             = 150 -

             = 150 - 4.8

             = 145.2

∴ the mean daily wage of the workers of the factory = Rs 145.20.

Q3: 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 allownace(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

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Given that, mean pocket allowance, x = Rs 18

Taking 18 as assumed mean (a), d_i and f_id_i are calculated as:

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	Σ fi = 44 + f			2f - 40

From the table:

Σ f_i = 44 + f

Σ f_id_i = 2f - 40

x = a +

18 = 18 + ()

0 =

2f - 40 = 0

2f = 40

f = 20

∴ the missing frequency, f, is 20.

Q4: 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 - 86
Number of woman	2	4	3	8	7	4	2

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Class size, h, of this data = 3

Considering 75.5 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows.

Number of heart beats per minute	Number of woman(fi)	xi	di = xi - 75.5	ui =	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

From the table:

Σ f_i = 30

Σ f_iu_i = 4

Mean , x = a +()☓ h

              = 75.5 + 4/30 ☓ 3

              = 75.5 + 0.4 = 75.9

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

Q5: 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?

Answer:

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

Since the class intervals are not continuous, there is a gap of 1 between two class intervals. Therefore adding    to the upper class limit and subtract   from the lower class limit of each interval.

Obtaining Class mark (x_i) by using the following relation.

x_i =

Class size (h) of this data = 3

Taking 57 as assumed mean (a), d_i, u_i, f_iu_i are calculated as:

Class interval	fi	xi	di = xi - 57	ui =	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

From the table:

Σ f_i = 400

Σ f_iu_i = 25

Mean , x = a +()☓ h

              = 57 + ()☓ 3

              = 57 + 3/16= 57 + 0.1875

              = 57.1875

              ≈ 57.19

Mean number of mangoes kept in a packing box = 57.19.

Using step deviation method as the values of f_i, d_i are big and also, there is a common multiple between all d_i.

Q6: 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.

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Class size = 50

Taking 225 as assumed mean (a), d_i, u_i, f_iu_i are calculated as:

Daily expenditure(in Rs)	fi	xi	di = xi - 225	ui =	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

From the table:

Σ f_i = 25

Σ f_iu_i = -7

Mean , x = a +()☓ h

              = 225 + ()☓ 50

              = 225 - 14

              = 211

Thus the mean daily expenditure on food is Rs 211.

Q7: 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

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), calculating d_i, u_i, f_iu_i  as:

concentration of SO2 (in ppm)	Frequency(fi)	Class mark(xi)	di = xi - 0.14	ui =	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

From the table:

Σ f_i = 30

Σ f_iu_i = -31

Mean , x = a +()☓ h

              = 0.14 + ()☓ 0.04

              = 0.14 - 0.04133

              = 0.09867

              ≈ 0.099 ppm

Therefore, mean concentration of SO₂ in the air = 0.099 ppm.

Q8: 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

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Taking 17 as assumed mean (a), calculating d_i and f_id_i as:

Number of days	Number of students(fi)	xi	di = xi - 17	fidi
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	40			- 181

From the table:

Σ f_i = 40

Σ f_id_i = -181

Mean, x = a +

            = 17 + ()

            = 17 - 4.525

            = 12.475

            ≈12.48

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

Q9: 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

Answer:

To find the class mark (x_i) for each interval, using the following relation:

x_i =

Class size(h) for this data = 10

Taking 70 as assumed mean (a), calculating d_i, u_i, f_iu_{i :}

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

From the table:

Σ f_i = 35

Σ f_iu_i = -2

Mean , x = a +()☓ h

              = 70 + ()☓ 10

              = 70 -

              = 70 -

              = 70 - 0.57

              = 69.43

Therefore, mean literacy rate is 69.43%.