NCERT Solutions for Class 10 Maths Chapter 14 Statistics

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Updated at 2021-05-07


NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.1

Q1 ) A survey was conducted by a group of students as a part of their environment awareness program, 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.

Number of Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of Houses 1 2 1 5 6 2 3

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



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


In order to find the mean value, we will use direct method because the numerical value of fi and xi are small.

The midpoint of the given interval is found by formula: Midpoint (xi) = upper limit + lower limit2

No. of plants (Class interval) No. of houses (Frequency (fi) ) Mid-point (xi) fixi
0-2 1 1 1
2-4 2 3 6
4-6 1 5 5
6-8 5 7 35
8-10 6 9 54
10-12 2 11 22
12-14 3 13 39
? fi = 20 ? fixi = 162

The formula to find the mean is: ?? x? = ? fixi?fi

= 16220 = 8.1

? The mean number of plants per house is 8.1

Q2 ) Consider the following distribution of daily wages of 50 workers 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.



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The midpoint of the given interval is found by formula: Midpoint (xi) = upper limit + lower limit2

In this case, the value of mid-point ( xi) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So, ui = xi ? Ah 
? ui = xi ? 15020

Daily wages (Class interval) Number of workers (frequency (fi) ) Mid-point ( xi ) ui = xi ? 15020 fiui
100-120 12 110 -2 -24
120-140 14 130 -1 -14
140-160 8 150 0 0
160-180 6 170 1 6
180-200 10 190 2 20
Total ?fi = 50 ?fiui = ?12

So, the formula to find out the mean is :  x? = A +h? fixi?fi
= 150 + 20 × ?1250
= 150 ? 4.8
= 145.20
? Mean daily wage of the workers = 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 Allowence(in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-35
Number of children 7 6 9 13 f 5 4



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Given, x? = 18

Class interval Number of children (frequency (fi) ) Mid-point (xi) fiui
11-13 7 12 84
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
Total ?fi = 44 + f ?fixi = 752 + 20f

Thus,  x? = ? fixi?fi

? 18 = 752 + 20f44 + f

? 18(44 + f) = (752 + 20f)

? 792 + 18f = 752 + 20f

? 792 ? 752 = 20f ? 18f 
? 40 = 2f

? f = 20

So, the missing frequency, f = 20.

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 heart beats per minute for these women, choosing a suitable method.

Number of heartbeats 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



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

From the given data, let us assume the mean as A = 75.5
Class size (h) = 3

So, ui = xi ? Ah 
? ui = xi ? 75.5h

Class Interval Number of women ( fi ) Mid-point ( xi ) ui = xi ? 75.5h fiui
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
?fi =30 ?fiui =4

x? = A +h? fixi?fi

=75.5 + 3 × 430 
= 75.5 + 410 
= 75.9

? The mean heart beats per minute for these women is 75.9

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?



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 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 (fi) Mid-point (xi) di = xi ? A fidi
49.5-52.5 15 51 -6 90
52.5-55.5 110 54 -3 -330
55.5-58.5 135 57 0 0
58.5-61.5 115 60 3 345
61.5-64.5 25 63 6 150
?fi = 400 ?fidi = 75

The formula to find out the Mean is:

x? = A +h? fidi?fi = 57 + 3×75400

= 57 + 0.1875 = 57.19

? The mean number of mangoes kept in a packing box is 57.19

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.

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



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Let us assume the mean (A) = 225, Class size (h) = 50

Class Interval Number of households (fi) Mid-point (xi) di = xi ? A ui = di50 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
?fi = 25 ?fiui = ?7

x? = A +h? fiui?fi

= 225 + 50×?725

= 225 ? 14 = 211

? The mean daily expenditure on food is 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
Find the mean concentration of SO2 in the air.



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Calculate mean by direct method:

Concentration of SO2 (in ppm) Frequency (fi) Mid-point (xi) fixi
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 ?fi = 30 ?fixi = 2.96

The formula to find out the mean is

x? = ? fixi?fi

= 2.9630

= 0.099 ppm

? The mean concentration of SO2 in air is 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



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Class interval Frequency (fi) Mid-point (xi) fixi
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
?fi = 40 ?fixi = 499

The mean formula is,

x? = ? fixi?fi

= 49940

= 12.48 days

? The mean number of days a student was absent = 12.48.

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 85-98
Number of cities 3 10 11 8 3



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

In this case, the value of mid-point (xi) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, ui = xi ? Ah

=> ui = xi ? 7010

Class Interval Frequency (fi) (xi) di = xi ? A ui = dih 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
?fi = 35 ?fiui = ?2

So,

x? = A +h? fixi?fi

= 70 + ?235 × 10

= 69.42

? The mean literacy part is 69.42

NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.2

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

Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

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



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 35-45 has the maximum frequency, therefore, this is the modal class.

Here l = 35 , h = 10 , f1 = 23 , f0 = 21 , f2 =14

Now, let us substitute these values in the formula

Mode = l + (f1 ? f02f1 ? f0 ? f2) × h

= 35 + 23 ? 212 × 23 ? 21 ? 14 × 10

= 35 + 246 ? 21 ? 14 × 10

= 35 + 211 × 10

= 35 + 1.8 = 36.8

Now, to calculate the Mean,

Class Interval Frequency ( fi ) Mid-point ( xi ) fixi
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 23 40 920
45-55 14 50 700
55-65 5 60 300
?fi = 80 ?fixi = 2830

The formula for mean is :

x? = ? fixi?fi

= 283080 = 35.37 years

? The mean of the given data = 35.37 years

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

Lifetime (in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 60-80 has the maximum frequency, therefore, this is the modal class.

Here l = 60 , h = 20 , f1 = 61 , f0 = 52 , and f2 = 38

Now, let us substitute these values in the formula

Mode = l + (f1 ? f02f1 ? f0 ? f2) × h

= 60 + (61 ? 522 × 61 ? 52 ? 38) × 20

= 60 + (61 ? 52122 ? 52 ? 38) × 20

= 60 + 932 × 20

= 60 + 5.625 = 65.625

? The modal lifetimes of the components is 65.625 hours.

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, find the mean monthly expenditure :

Expenditure Number of families
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 1500 – 2000 has the maximum frequency. Therefore, it is the modal class.

Here l = 1500,
h = 500,
f1 =40 ,
f0 =24 and
f2 =33

Now, let us substitute these values in the formula

Mode = l + (f1 ? f02f1 ? f0 ? f2) × h

= 1500 + (40 ? 242 × 40 ? 24 ? 33) × 500

= 1500 + 40 ? 2480 ? 24 ? 33 × 500

= 1500 + 1623 × 500

= 1500 + 347.83 = 1847.83

? Modal monthly expenditure = Rs. 1847.83

Let the assumed mean be A = 2750 and h = 500.

Class Interval fi xi di = xi ? A ui = dih fiui
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28
?fi = 200 ?fiui = ?35

The formula to calculate the mean,

x? = A +h? fixi?fi

= 2750 + (?35200) × 500

= 2750 ? 87.50 = 2662.50

So, the mean monthly expenditure of the families is Rs 2662.50

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

No of Students per teacher Number of states / U.T
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30,
h = 5, f1 = 10,
f0 = 9 and
f2 = 3

Now, let us substitute these values in the formula

Mode = l + (f1 ? f02f1 ? f0 ? f2) × h

= 30 + (10 ? 92 × 10 ? 9 ? 3) × 5

= 30 + 10 ? 920 ? 9 ? 3 × 5

= 30 + 18 × 5 = 30 + 0.625

= 30.625 = 30.6

Calculation of mean :

Class Interval Frequency (fi) Mid-point (xi) fixi
15-20 3 17.5 52.5
20-25 8 22.5 180.0
25-30 9 27.5 247.5
30-35 10 32.5 325.0
35-40 3 37.5 112.5
40-45 0 42.5 0
45-50 0 47.5 0
50-55 2 52.5 105.5
?fi = 35 ?fixi = 1022.5

x? = ? fixi?fi

= 1022.535 = 29.2

? mean = 29.2

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

Run Scored Number of Batsman
3000-4000 4
4000-5000 18
5000-6000 9
6000-7000 7
7000-8000 6
8000-9000 3
9000-10000 1
10000-11000 1
Find the mode of the data.



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 4000-5000 has the maximum frequency ,
therefore, this is the modal class.

Here l = 4000,
h = 1000,
f1 = 18,
f0 = 4 and
f2 = 9

Now, let us substitute these values in the formula

Mode =l + (f1 ? f02f1 ? f0 ? f2) × h

= 4000 + (18 ? 42 × 18 ? 4 ? 9) × 1000

= 4000 + 18 ? 436 ? 4 ? 9 × 1000

= 4000 + 1423 × 1000

= 4000 + 608.695 =4608.7

? The mode of the given data is 4608.7 runs.

6. 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:

Number of cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


The class 40-50 has the maximum frequency, therefore, this is the modal class.

Here l = 40, h =10, f1 = 20, f0 = 12 and f2 = 11

Now, let us substitute these values in the formula
Mode = l + (f1 ? f02f1 ? f0 ? f2) × h

= 40 + (20 ? 1240 ? 12 ? 11) × 10

= 40 + 817 × 10

= 40+4.7 = 44.7

? Mode = 44.7 cars

NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.3

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

Monthly consumption(in units) No. of customers
65-85 4
85-105 5
105-125 13
125-145 20
145-165 14
165-185 8
185-205 4



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :

Class Interval Frequency Cumulative frequency
65-85 4 4
85-105 5 9
105-125 13 22
125-145 20 42
145-165 14 56
165-185 8 64
185-205 4 68
N=68

We have n = 68
? n2 = 682 = 34

The cumulative frequency just greater than n2 , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that n2 = 34,
l = 125 ,
c.f. = 22 ,
f = 20 , h = 20.

Median = l + (n2 ? cff) × h,

Median = 125 + (34 ? 2220) × 20 
= 125 + 12 = 137 units

To calculate the mode :

Modal class = 125?145,
f1=20,
f0=13,
f2=14, and
 h=20

Mode formula :

Mode = l+(f1?f0)(2f1?f0?f2)×h

Mode = 125+(20?13)(40?13?14)×20

=125+14013

=125+10.77=135.77

Therefore, mode = 135.77

Calculation of mean :

Let the assumed mean, A = 135, class interval, h = 20

so ui = xi ? Ah = xi ? 13520

Class Interval fi xi di = xi ? A ui = dih fiui
65-85 4 75 -60 -3 -12
85-105 5 95 -40 -2 -10
105-125 13 115 -20 -1 -13
125-145 20 135 0 0 0
145-165 14 155 20 1 14
165-185 8 175 40 2 16
185-205 4 195 60 3 12
?fi = 68 ?fiui = 7

x? = A +h? fixi?fi

= 135 + 20(768)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

Q2 ) If the median of a distribution given below is 28.5 then, find the value of an x &y.

Class Interval Frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60



NCERT Solutions for Class 10 Maths Chapter 14 Statistics


Answer :


Given that median is 28.5 and n = ?fi = 60

Here n = 60
? n2 = 30

Since the median is given to be 28.5, thus the median class is 20-30.