NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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Q1.A survey was conducted by a group of students as a part of their environmental awareness programm 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 ?

Ans.

No. of Plants

Class Mark (x_i)

No. of Houses (f_i)

f_ix_i

0 – 2

2 – 4

4 – 6

6 – 8

8 – 10

10 – 12

12 – 14

∑f_i= 20

∑f_ix_i= 162

We have

$\overline{X}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{162}{20}=8.1$

Hence mean number of the plants is 8.1

The direct method is proper to use here because the values of observations and their frequencies are small.

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

Q2.Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans.

Class Interval	Class Mark (x_i)	Frequency(f_i)	u_i=(x_i-A)/h	f_ix_i
100 – 120	110	12	-2	-24
120 – 140	130	14	-1	-14
140 – 160	150 = A	8	0	0
160 – 180	170	6	1	6
180 – 200	190	10	2	20
		∑f_i= 50		∑f_iu_i= -12

Since the class interval is of the size,h= 20 and sum of frequencies is 50, both of them are large therefore applying the step deviation method

$\overline{X}=A + \frac{\sum f_{i}u_{i}}{\sum f_{i}}$

$\overline{X}=A +\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h$

$=150+\frac{-12}{50}\times 20=150-\frac{240}{50}=150-4.8=145.20$

Hence the mean daily wages of the workers of the factory is 145.20

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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.

Ans.

Daily pocket allowance (in Rs)

Class mark(x_i)

Frequency(f_i)

d_i= x_i-A

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

∑f_i= 44 + f

∑f_id_i= 2f – 40

Applying the shortcut method of mean

$\overline{X}=A+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$

Where A = 18, ∑f_id_i= 2f – 40 , ∑f_i= 44 + f and $\overline{X}=18$

$18=18+\frac{2f-40}{44+f}$

$\frac{2f-40}{44+f}=0$

2f – 40 = 0⇒ f = 20

Hence missing frequency is 20

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

Q4.Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Ans.

No of heart beats per minute	Class mark(x_i)	Frequency(f_i)	f_ix_i
65 – 68	66.5	2	133
68 – 71	69.5	4	278
71 – 74	72.5	3	217.5
74 – 77	75.5	8	604
77 – 80	78.5	7	549.5
80 – 83	81.5	4	326
83 – 86	84.5	2	169
		∑f_i= 30	∑f_ix_i= 2277

Applying the direct method for getting the mean of the data

$\overline{X}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

$\overline{X}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{2277}{30}=75.9$

Hence mean of the heartbeats per minute of the women is 75.9

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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.

Find the mean number of the mangoes kept in a packing box. Which method of finding the mean did you choose.

Ans.

No of mangoes

Class mark(x_i)

Frequency (f_i)

u_i=(x_i-A)/h

f_iu_i

50 – 52

53 – 55

56 – 58

59 – 61

62 – 64

57= A

110

135

115

-2

-1

-30

-110

115

∑f_i= 400

∑f_iu_i= 25

Here we have to apply step deviation because class intervals are not continuous

$\overline{X}=A+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h=57+\frac{25}{400}\times 3=57 +0.19 = 57.19$

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

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

Q6.The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by suitable method.

Ans..

Daily expenditure(in Rs)	Class marks(x_i)	Frequency (f_i)	u_i= (x_i-A)/h	f_iu_i
100 – 150	125	4	-2	-8
150 – 200	175	5	-1	-5
200 – 250	225	12	0	0
250 – 300	275	2	1	2
300 – 350	325	2	2	4
		∑f_i=25		∑f_iu_i= -7

$\overline{X}=A+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h=225+\frac{-7}{25}\times 50=225-\frac{350}{25}=225-14=211$

Hence the mean daily expenditure on food is 211

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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.

Ans.

Concentration of SO2(PPM)	Class Mark(x_i)	Frequency(f_i)	u_i=(x_i -A)/h	f_ix_i
o.oo – 0.04	0.02	4	-2	-8
0.04 – 0.08	0.06	9	-1	-9
0.08 – 0.12	0.10	9	0	0
0.12 – 0.16	0.14	2	1	2
0.16 – 0.20	0.18	4	2	8
0.20 – 0.24	0.22	2 ∑f_i=30	3	6 ∑f_iu_i=-1

Applying the step deviation method for getting the mean

$\overline{X}=A+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h$

$=0.10+\frac{-1}{30}\times 0.04$

$=0.10-\frac{0.04}{30}= 0.10-0.0013=0.0987$

Hence the average concentration of SO2 is 0.0987

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

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.

Ans.

No. of days	Class Mark(x_i)	Frequency (f_i)	f_ix_i
0 – 6	3	11	33
6 – 10	8	10	80
10 – 14	12	7	84
14 -20	17	4	68
20 – 28	24	4	96
28 – 38	33	3	99
38 – 40	39	1	39
		∑f_i= 40	∑f_ix_i= 499

Applying direct method of getting the mean

$\overline{X}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

$=\frac{499}{40}$

=12 .48

The mean number of days a student was absent is 12.48

NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics

Q9.The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Ans.

Litracy rate (%)	Class mark(x_i)	Frequency(f_i)	u_i= (x_i– A)/h	f_ix_i
45 – 55	50	3	-2	-6
55 – 65	60	10	-1	-10
65 – 75	70 = A	11	0	0
75 – 85	80	8	1	8
85 – 95	90	3 ∑f_i= 35	2	6 ∑f_iu_i=-2

Applying the step deviation method

$\overline{X}=A +\frac{\sum f_{i}x_{i}}{\sum f}\times h$

Where A = 70, h = 10

$=70 +\frac{-2}{35}\times 10$

$=70 -\frac{20}{35}$

= 70 – 0.57 = 69.43 %

Hence the mean litracy rate is 69.43 %

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NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics