NCERT Solutions For Class 10 Math Chapter – 14 Exercise – 14.1

NCERT SOLUTIONS FOR CLASS 10 MATH CHAPTER – 14 EXERCISE – 14.1

The NCERT Solutions for Class 10 Maths Chapter 14 Statistics are undoubtedly an essential study materials for the students studying in CBSE Class 10 NCERT Solutions provided here along with the downnloads PDF can help the students prepare effectively for their first term exams The chapter is a continuation of what was taught in chapter probability in Class 9 and further explains the different concept related to it.

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 her house.

No of plants

No of houses

0 – 2

2-4

4-6

6-8

8-10

10-12

12-14

Solution:-

Number of Plants

Class mark(x_i)

Number of houses (f_i)

f_ix_i

0-2

2-4

4-6

6-8

8-10

10-12

12-14

Total

Fi = 20

Fixi = 162

Mean = (x) = fixi/fi

= 162/80

= 8.1

Q2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily Wage

No of workers

100-120

120-140

140-160

160-180

180-200

Solution:-

Class intervals

equency

Class marks

U_i = x_i – a/h

Fiui

100-120

120-140

140-160

160-180

180-200

Fi= 50

110

130

150=A

170

190

-2

-1

-24

-14

Fiui = 120

X = a + h(fiui/fi)

= 150 + 20(-12/50)

= 726/5

= 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 Allowances

No of children

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Solution:-

Daily pocket allowance

Class Mark

No of children

D_i = x_i – 18

f_id_i

11-13

13-15

15-17

17-19

19-21

21-23

23-25

18=a

-6

-4

-2

-42

-24

-18

Total

F_i = 44 + f

F_id_{i =}2f – 40

Mean = x = f_ix_i/f_i

= (752 + 20f)/(44+f)

18 = (752 + 20f)/(44+f)

F = 20

So , the missing the frequency f = 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 s suitable method.

Number of heart beat 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:-

Class Intervals	Number of women	Mid point	U_i = (x_i – 75.5)/h	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	3	8
83-86	2	84.5	3	6

Sum f_i = 30 Sum f_iu_i = 4

Mean = x = A + hf_iu_i/f_i

= 75.5 + 3 x (4/30)

= 75.5 + 0.4

= 75.9

Therefore , the mean heart beats per minute for these women is 75.9

Q5. In a retail market fruit vendors were selling mangoes kept in packaging 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:

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval	Number of boxes (f_i)	Mid-point (x_i)	d_i = x_i – A	f_id_i
49.5-52.5	15	51	-6	90
52.5-55.5	110	54	-3	-330
55.5-58.5	135	57 = A	0	0
58.5-61.5	115	60	3	345
61.5-64.5	25	63	6	150
	Sum f_i = 400			Sum f_id_i = 75

The formula to find out the Mean is:

Mean = x̄ = A +h ∑f_id_i /∑f_i

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily expenditure(in c)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

Let is assume the mean (A) = 225

Class size (h) = 50

Class Interval	Number of households (f_i)	Mid-point (x_i)	d_i = x_i – A	u_i= d_i/50	f_iu_i
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
	Sum f_i = 25				Sum f_iu_i = -7

Mean = x̄ = A +h∑f_iu_i /∑f_i

= 225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on food is 211

To find out the concentration of SO₂in 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 SO₂ in the air.

Solution:

To find out the mean, first find the midpoint of the given frequencies as follows:

Concentration of SO₂(in ppm)	Frequency (f_i)	Mid-point (x_i)	f_ix_i
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.20	0.40
Total	Sum f_i = 30		Sum (f_ix_i) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑f_ix_i /∑f_i

= 2.96/30

= 0.099 ppm

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

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:

Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

Class interval	Frequency (f_i)	Mid-point (x_i)	f_ix_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
	Sum f_i = 40		Sum f_ix_i = 499

The mean formula is,

Mean = x̄ = ∑f_ix_i /∑f_i

= 499/40

= 12.48 days

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

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	85-98
Number of cities	3	10	11	8	3

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

In this case, the value of mid-point (x_i) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.

So, u_i= (x_i-A)/h = u_i= (x_i-70)/10

Substitute and find the values as follows:

Class Interval	Frequency (f_i)	(x_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
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
	Sum f_i = 35				Sum f_iu_i = -2

So, Mean = x̄ = A+(∑f_iu_i /∑f_i)×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42