NCERT Solutions for Class 10 Maths Chapter 13 Statistics

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 14
Chapter Name : Statistics

Exercise 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.

Which method did you use for nding the mean, and why?

Answer. To nd the class mark (x ) for each interval, the following relation is used.
i

Class mark (x )
Upper class limit + Lower class limit

i
2

xi and fXi can be calculated as follows

From the table it can be observed that
∑ fi = 20

∑ fr xi = 162

∑ ff

Mean x =
i
¯
¯¯

∑ fi

162
= = 8.1
20

Therefore mean number of plants per house is 8.1 Here, direct method has been used as the
values of class marks (x ) and f are small
i i

Page : 270 , Block Name : Exercise 14.1

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

Answer. To nd the class mark for each interval, the following relation is used,
Upper class limit + Lower class limit
xi =
2

Class size (h ) of this data =20
Taking 150 as assured mean (a) d, u and f can be calculated as follows.

From the table it can be observed that
∑ fi = 50

∑ fi ui = −12
∑ f i ui
Mean x = a + (
¯
¯¯
)h
∑ fi

−12
= 150 + ( ) 20
50

24
= 150 −
5

= 150 − 4.8

= 145.2

Therefore the mean daily wage of the workers of the factory is Rs 145.20

Page : 270 , Block Name : Exercise 14.1

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.

Answer. To nd the class mark (x ) for each interval, the following relation is used.
i

Upper class limit + Lower class limit
xi =
2

Given that mean pocket allowance x = Rs 18 ¯
¯¯

Taking 18 as assured mean (a), d and f d are calculated as follows.
i i

From the table, we obtain

∑ fi = 44 + f

∑ fi ul = 2f − 40
∑ fi di
¯
¯¯
x = a +
∑ fl

2f −40
18 = 18 + ( )
44+f

2f −40
0 = ( )
44+f

2f − 40 = 0

2f = 40

f = 20

Hence the missing frequency f is 20

Page : 270 , Block Name : Exercise 14.1

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

Answer. To nd the class mark of each interval (x,), the following relation is used.
Upper class limit + Lower class limit
xi =
2

Class size h of this data =3
Taking 75.5 as assumed mean (a), di, u , f u are calculated as follows.
it i i

From the table we obtain
∑ f i = 30

∑ fμ, = 4
∑ f ui
Mean x = a + (
¯
¯¯

∑ fi
) × h

4
= 75.5 + ( ) × 3
30

= 75.5 + 0.4 = 75.9

Therefore ,mean hear beats per minute for these women are 75.9 beats per minute.

Page : 271 , Block Name : Exercise 14.1

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 mangoes kept in a packing box. Which method of nding the mean did
you choose?

Answer.

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,) can be obtained by using the following elation.
Upper class limit + Lower class limit
xi =
2

Class size (h) of this data = 3
Taking 57 as assumed mean , (a), d , u f i it μi are calculated as follows.

It can be observed that
∑ fi = 400

∑ fi ui = 25
∑ f i ui
Mean x = a + (
¯
¯¯
) × h
∑ fi

25
= 57 + ( ) × 3
400

3
= 57 + = 57 + 0.1875
16

= 57.1875

= 57.19

Mean number of mangoes kept in a packing box is 57.19 Step deviation method is used here as
the values of f d are big and also there is a common multiple between all d
i, i i

Page : 271 , Block Name : Exercise 14.1

Q6 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.

Answer. To nd the class mark (X ) for each interval, the following relation is used,
i

Upper class limit + Lower class limit
x1 =
2

Class size =50
Taking 225 as assumed mean (a), d , u , f i i μi are calculated as follows.

From the table , we obtain
Σfi = 25

∑ fi ui = −7
∑ f i ui
Mean x = a + (
¯
¯¯
) × h
∑ fi

−7
= 225 + ( ) × (50)
25

= 225 − 14

= 211

Therefore, mean daily expenditure on food is Rs 211

Page : 271 , Block Name : Exercise 14.1

Q7 To nd out the concentration of SO in the air (in parts per million, i.e., ppm), the data was
2

collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO in the air. 2

Answer. To nd the class marks for each interval the following relation is used
Upper class limit + Lower class limit
xi =
2

Class sizeof this data =0.04
Taking 0.14 as assumed mean (a), d u f it ir μi are calculated as follows.

From the table we obtain
∑ fi = 30

∑ fi ui = −31
∑ f i ui
Mean x = a + (
¯
¯¯
) × h
∑ fi

−31
= 0.14 + ( ) (0.04)
30

= 0.14 − 0.04133

= 0.09867

= 0.099ppm

Therefore mean concentration of SO in the air is 0.099 ppm
2

Page : 271 , Block Name : Exercise 14.1

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

Answer. To nd the class mark of each interval the following relation is used
Upper class limit + Lower class limit
xi =
2

Taking 17 as assumed mean (a), d and f are calculated as follows.
i di

From the table we obtain
Σfi = 40

Σfi di = −181
∑ fj di
Mean x = a + (
¯
¯¯

∑ fi
)

−181
= 17 + ( )
40

= 17 − 4.525

= 12.475

= 12.48

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

Page : 272 , Block Name : Exercise 14.1

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

Answer. To nd the class marks the following relation is used
Upper class limit + Lower class limit
xi =
2

Class size (h) for this data =10
Taking 70 as assumed mean (a), d , u i it and f′ ui are calculated as follows

From the table we obtain
Σfi = 35

∑ fi ui = −2
∑ f i ui
Mean x = a + (
¯
¯¯

∑ ff
) × h

−2
= 70 + ( ) × (10)
35

20
= 70 −
35
4
= 70 −
7

= 70 − 0.57

= 69.43

Therefore ,mean literacy rate is 69.43%

Page : 272 , Block Name : Exercise 14.1

Exercise 14.2

Q1 The following table shows the ages of the patients admitted in a hospital during a year

Find the mode and the mean of the given data above .Compare and interpret the two measures
of central tendency .

Answer. To nd the class marks (x ) the following relation is used
i

Upper class limit + Lower class limit
xi =
2

Taking 30 as assumed mean d and f d are calculated as follows.
i f i

From the table we obtain
∑ fi = 80

∑ fi di =430

∑ fl d j
¯
¯¯
Mean, x = a +
∑ fi

430
= 30 + ( )
80

= 30 + 5.375

= 35.375

= 35.38

Mean of this data 35.38
It represents that on an average the age of a patient admitted to hospital was 35.38 years

It can be observed that the maximum class frequency is 23 belonging to class interval 35-45
Modal class = 35-45
Lower limit (l) of modal class =35
Frequency (f ) of modal class =23
1

Class size (h) =10
Frequency (f ) of class preceding the modal class =21
0

Frequency (f ) of class succeeding the modal class =14
2

Mode=
f1 − f0
l + ( ) × h
2f1 − f0 − f2

23 − 21
= 35 + ( ) × 10
2(23) − 21 − 14

2
= 35 + [ ] × 10
46 − 35

20
= 35 +
11

= 35 + 1.81

= 36.8

Mode is 36.8 It represents that the age of maximum number of patients admitted in hospital was
36.8 years.

Page : 275 , Block Name : Exercise 14.2

Q2 The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :

Answer. From the data given above, it can be observed that the maximum class frequency is 61,
belonging to class interval 60 -80.
Therefore, modal class =60 - 80
Lower class limit (l) of modal class =60
Frequency (f ) of modal class = 61
1

Frequency (f ) of class preceding the modal class = 52
0

Frequency (f ) of class succeeding the modal class 38
2

Class size (h) = 20
f1 − f0
Mode = l + ( ) × h
2f1 − f0 − f2

61 − 52
= 60 + ( ) (20)
2(61) − 52 − 38

9
= 60 + ( ) (20)
122−90

9×20
= 60 + ( )
32

90
= 60 + = 60 + 5.625
16

= 65.625

Therefore ,modal lifetime of electrical components is 65.625 hours.

Page : 275 , Block Name : Exercise 14.2

Q3 The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, nd the mean
monthly expenditure.

Answer. It can be observed from the given data that the maximum class frequency s 40,
belonging to 1500 — 2000 inte vals.
Therefore, modal class =1500 — 2000
Lower limit (l) of modal class = 1500
Frequency (f )of modal class = 40
1

Frequency (f ) of class preceding modal class=24
0

Frequency (f ) of class succeeding modal class = 33
2

Class size (h) =500
f1 − f0
Mode = l + ( ) × h
2f1 − f0 − f2

40 − 24
= 1500 + ( ) × 500
2(40) − 24 − 33

16
= 1500 + ( ) × 500
80 − 57

8000
= 1500 +
23

= 1847.826 = 1847.83

Therefore , modal monthly expenditure was Rs 1847.83
To nd the class mark , the following relation is used
Upper class limit + Lower class limit
Class mark = 2

Class Size(h) of the given data=500
Talking 2750 as assumed mean (a), d , u , and f are calculated as follows.
i i μ

From the table we obtain

∑ fi = 200

∑ fi uf = −35

∑ f i ul
¯
¯¯
x( mean ) = a + ( ) × h
∑ fi

−35
¯
¯¯
x = 2750 + ( ) × 500
200

= 2750 − 87.5

= 2662.5

Therefore, mean monthly expenditure was Rs 2662.50

Page : 275 , Block Name : Exercise 14.2

Q4 The following distribution gives the state -wise teacher student ratio in higher secondary
schools of india .Find the mode and mean of this data .Interpret the two measures.

Answer. It can be observed from the given data that the maximum class frequency s 10
belonging to class interval 30 — 35.
Therefore, modal class = 30 — 35
Class size (h) = 5
Lower limit (l) of modal class =30
Frequency (f ) of modal class = 10
1

Frequency (f ) of class preceding modal class = 9
0

Frequency (f ) of class succeeding modal class = 3
2

f1 − f0
Mode = l + ( ) × h
2f1 − f0 − f2

10 − 9
= 30 + ( ) × (5)
2(10) − 9 − 3

5
1
= 30 + ( )
20 − 12

5
= 30 + = 30.625
8

Mode = 30.6

It represents that most of the states/LJ.T have a teacher-student ratio as 30.6. To nd the class
marks, the following relation is used.
Upper class limit + Lower class limit
Class mark = 2

Taking 32.5 as assumed mean d , u , and f μ are calculated as follows.
i i ′ i

∑ fi u i
¯
¯¯
Mean, x = a + ( )h
∑ fi

−23
= 32.5 + ( ) × 5
35

23
= 32.5 − = 32.5 − 3.28
7

= 29.22

Therefore ,mean of the data is 29.2 It represents that on an average , teacher-student ratio was
29.2

Page : 276 , Block Name : Exercise 14.2

Q5 The given distribution shows the number of runs scored by some top batsmen of the world in
one-day international cricket matches.

Find the mode of the data

Answer. From the given data it can be observed that the maximum class frequency is 18,
belonging to class interval 4000-5000
Therefore ,modal class =4000-5000
Lower limit (l) of modal class = 4000
Frequency (f ) of modal class 18
1

Frequency (f ) of class preceding modal class =4
0

Frequency (f ) of class succeeding modal class = 9
2

Class size = 1000
f1 − f0
Mod e = l + ( ) × h
2f1 − f0 − f2

18 − 4
= 4000 + ( ) × 1000
2(18) − 4 − 9

14000
= 4000 + ( )
23

= 4000 + 608.695

= 4608.695

Therefore mode of the given data is 4608.7 runs

Page : 276 , Block Name : Exercise 14.2

Q6 A student noted the number of cars passing through a spot on a road for 100 periods each of
3 minutes and summarised it in the table given below. Find the mode of the data:

Answer. From the given data, it can be observed that the maximum class frequency is 20
belonging to 40 — 50 class intervals.
Therefore, modal class =40 - 50
Lower limit of modal class = 40
Frequency (f ) of modal class =20
1

Frequency (f ) of class preceding modal class=12
0

Frequency (f ) of class succeeding modal class =11
2

Class size = 10
f1 − f0
Mode = l + ( ) × h
2f1 − f0 − f2

20 − 12
= 40 + [ ] × 10
2(20) − 12 − 11

80
= 40 + ( )
40 − 23

80
=40 +
17

=40 + 4.7

=44.7

Therefore , mode of this data is 44.7 cars

Page : 276 , Block Name : Exercise 14.2

Exercise 14.3

Q1 The following frequency distribution gives the monthly consumption of electricity of 68
consumers of a locality. Find the median, mean and mode of the data and compare them

Answer. To nd the class marks the following relation is used
Upper class limit + + Lower class limit
=
2

Taking 135 as assumed mean (a), d u i in f μi are calculated according to step deviation method as
follows

From the table ,we obtain
∑ fi u i = 7

∑ fi = 68

Class size (h) = 20

∑ f ui
¯
¯¯
Mean, x = a + ( ) × h
∑ fi

7
= 135 + × 20
68

140
= 135 +
68

= 137.058

From the table it can be observed that the maximum class frequency is 20,
belonging to class interval 125 - 145.
Modal class =125- 145
Lower limit ( l )of modal class = 125

Class size (h) 20
Frequency (f ) of modal class =20
1

Frequency (f ) of class preceding modal class = 13
0

Frequency(f ) of class succeeding the modal class = 14
2

f1 − f0
Mode = l + ( ) × h
2f1 − f0 − f2

20 − 13
= 125 + [ ] × 20
2(20) − 13 − 14

7
= 125 + × 20
13

140
= 125 + = 135.76
13

To nd the median of the given data, cumulative frequency is calculated as follows.

From the table we obtain
n=68
Cumulative frequency (cf) just greater than
interval 125 - 145.
Therefore, median class = 125 — 145
Lower limit (I) of median class =125
Class size (h) =20
Frequency (cf) of median class =20
Cumulative frequency (cf) of class preceding median class = 22

n
− cf
2
Median = l + ( ) × h
f

34 − 22
= 125 + ( ) × 20
20

= 125 + 12

= 137

Therefore median mode mean of the given data is 137, 135.76 and 137.05 respectively.
The three measures are approximately the same .

Page : 287 , Block Name : Exercise 14.3

Q2 If the median of the distribution given below is 28.5, nd the values of x and y.

Answer. The cumulative frequency for Hoe given data is calculated as follows.

From the table it can be observed that n=60
45 + x + y = 60

x + y = 15(1)

Median of the data is given as 28.5 which lies in interval 20-30.
Therefore, median class =20 — 30
Lower limit (l) of median class=20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
n
−cf
2
Median = l + ( ) × h
f

60
−(5+x)
2
28.5 = 20 + [ ] × 10
20

25−x
8.5 = ( )
2

17 = 25 − x

x = 8

From equation (1)
8 + y = 15

y = 7

Hence the value of x and y are 8 and 7 respectively

Page : 287 , Block Name : Exercise 14.3

Q3 A life insurance agent found the following data for distribution of ages of 100 policy holders.
Calculate the median age, if policies are given only to persons having age 18 years onwards but
less than 60 year.

Answer. Here, class width is not the same. There is no requirement of adjusting the frequencies
according to class intervals. The given frequency table is of less than type represented with
upper class limits. The policies were given only to persons with age 18 years onwards but less
than 60 years. Therefore, class intervals with their respective cumulative frequency can be
de ned as below

From the table, it can be observed that n = 100
Cumulative frequency (cf) just greater than n

2
( i.e.,
100

2
= 50) is 78 belonging to interval 35-
40
Therefore, median class =35 — 40
Lower limit (l) of median class= 35
Class size (h) =5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
n
− cf
2
Median = l + ( ) × h
f

50 − 45
= 35 + ( ) × 5
33

25
= 35 +
33

= 35.76

Therefore ,median age is 35.76 years.

Page : 287 , Block Name : Exercise 14.3

Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the
data obtained is represented in the following table:

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for nding the median, since the
formula assumes continuous classes. The classes then change to 117.5 ,126.5, 126.5 135.5...
171.5 180 5)

Answer. The given data does not have continuous class intervals. It can be observed that the
difference between two class intervals is 1. Therefore, = 0.5 has to be added and subtracted
1

2

to upper class limits and lower class limits respectively.
Continuous class intervals with respective cumulative frequencies can be represented
as follows.

From the table it can be observed that the cumulative frequency just greater than
n

2
( i.e.,
40

2
= 20) is 29 belonging to class interval 144.5 - 153.5
Median class = 144.5 - 153.5
Lower limit (l) of median class =144 5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
n
−cf

Median = l + ( 2

f
) × h

20−17
= 144.5 + ( ) × 9
12

9
= 144.5 + = 146.75
4

Therefore median length of leaves is 146.75 mm

Page : 288 , Block Name : Exercise 14.3

Q5 The following table gives the distribution of the life time of 400 neon lamps :

Find the median lifetime of a lamp.

Answer. The cumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than n

2
( i.e.,
400

2
= 200) is 216,
belonging to class interval 3000

-3500.
Median class = 3000 — 3500
Lower limit (l) of median class =3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
n
−cf

Median = l + ( 2

f
) × h

200 − 130
= 3000 + ( ) × 500
86

70 × 500
= 3000 +
86

= 3406.976

Therefore, median lifetime of lamps is 3406.98 hours.

Page : 289 , Block Name : Exercise 14.3

Q6 100 surnames were randomly picked up from a local telephone directory and the frequency
distribution of the number of letters in the English alphabets in the surnames was obtained as
follows:

Determine the median number of letters in the surnames. Find the mean number of letters in
the surnames? Also, nd the modal size of the surnames.

Answer. The cumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than is 76,
n 100
( i.e., = 50)
2 2

belonging to class interval 7-10.
Median class = 7 - 10
Lower limit (l) of median class= 7
Cumulative frequency (cf) of class preceding median class =36
Frequency (f) of median class = 40
Class size (h) = 3
n
−cf

Median = l + ( 2

f
) × h

50 − 36
= 7 + ( ) × 3
40

14 × 3
= 7 +
40

= 8.05

To nd the class marks of the given class intervals the following relation is used
Upper class limit + Lower class limit
Class mark = 2

Taking 11.5 as assumed mean d u it ir and f, ui are calculated according to step deviation method
as follows .

From the given table
∑ fi μi = −106

Σfi = 100
∑ f i ui
¯
¯¯
x = a + ( )h
∑ fi

−106
= 11.5 + ( ) × 3
100

= 11.5 − 3.18 = 8.32

The data in the given table can be written as

From the table, it can be observed that the maximum class frequency is 40
belonging to class interval 7 — 10
Modal class =7 — 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f ) of modal class == 40
1

Frequency (f ) Of class preceding the modal class = 30
0

Frequency (f ) of class succeeding the modal class =16
2

f1 − f0
Mode =l + ( ) × h
2f1 − f0 − f2

40 − 30
= 7 + [ ] × 3
2(40) − 30 − 16

10
= 7 + × 3
34

30
= 7 + = 7.88
34

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively
while modal size of surnames is 7.88.

Page : 289 , Block Name : Exercise 14.3

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of
the students.

Answer. The cumulative frequencies with their respective class intervals are as follows.

Cumulative frequency just greater than interval 55 — 60.
Median class = 55 — 60
Lower limit (l) of median class=55
Frequency (f) of median class = 6
is 19, belonging to class
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
n
−cf

Median = l + ( 2

f
) × h

15−13
= 55 + ( ) × 5
6

10
= 55 +
6

= 56.67

Therefore the median weight is 56.67 kg

Page : 289 , Block Name : Exercise 14.3

Exercise 14.4

Q1 The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw
its ogive.

Answer. The frequency distribution table of less than type is as follows.

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis,
its ogive can be drawn as follows.

Page : 293 , Block Name : Exercise 14.4

Q2 During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph
verify the result by using the formula.

Answer. The given cumulative frequency distributions of less than type are
Taking upper class limits on x-axis and their respective cumulative frequencies on y- axis, its
ogive can be drawn as follows.

Here n= 35
So , = 17.5
n

2

Mark the point A whose ordinate is 17.5 and its co-ordinate is 46.5 .Therefore median of this
data is 46.5

It can be observed that the difference between two consecutive upper class limits is
2. The class marks with their respective frequencies are obtained as below.

The cumulative frequency just greater than n

2
( i.e.,
35

2
= 17.5) is 28, belonging to class
interval 46-48
Median class = 46 — 48
Lower class limit (l) of median class 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) =2
n
− cf
2
Median = l + ( ) × h
f

17.5 − 14
= 46 + ( ) × 2
14

3.5
= 46 +
7

= 46.5

Therefore, median of this data is 46.5. Hence, the value of median is veri ed.

Page : 293 , Block Name : Exercise 14.4

Q3 The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution and draw ogive.

Answer. The cumulative frequency distribution of more than type can be obtained as follows.

Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its
ogive can be obtained as follows.

Page : 293 , Block Name : Exercise 14.4