Chapter 13 Statistics and Probability NCERT Exemplar Solutions Exercise 13.4 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 13 Statistics and Probability Exercise 13.4 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 13.4 Solutions
1. Find the mean marks of students for the following distribution.
Marks 
Number of students 
0 and above 
80 
10 and above 
77 
20 and above 
72 
30 and above 
65 
40 and above 
55 
50 and above 
43 
60 and above 
28 
70 and above 
16 
80 and above 
10 
90 and above 
8 
100 and above 
0 
Solution
Marks 
Class marks (x_{i}) 
Number of students 
f_{i} 
f_{i} x_{i} 
0 – 10 
5 
80 
3 
15 
10 – 20 
15 
77 
5 
75 
20 – 30 
25 
72 
7 
175 
30 – 40 
35 
65 
10 
350 
40 – 50 
45 
55 
12 
540 
50 – 60 
55 
43 
15 
825 
60 – 70 
65 
28 
12 
780 
70 – 80 
75 
16 
6 
450 
80 – 90 
85 
10 
2 
170 
90 – 100 
95 
8 
8 
760 
100 – 110 
105 
0 
0 
0 




Î£f_{i} x_{i} = 4140 
Mean = (Î£f_{i} x_{i})/ N = 4140/80 = 51.75
2.
Marks 
Number of students 
Below 10 
5 
Below 20 
9 
Below 30 
17 
Below 40 
29 
Below 50 
45 
Below 60 
60 
Below 70 
70 
Below 80 
78 
Below 90 
83 
Below 100 
85 
Here, (assumed mean) a = 45
and (class width) h = 10
By step deviation method,
Mean (x) = a + (Î£f_{i} u_{i})/(Î£f_{i}) × h
= 45 + 3.41
Age equal and above (in years) 
0 
10 
20 
30 
40 
50 
60 
70 
Number of persons 
100 
90 
75 
50 
25 
15 
5 
0 
Here, (assumed mean) a = 35
and (class width) h = 10
By step deviation method,
Mean (x) = a + (Î£f_{i} u_{i})/(Î£f_{i}) × h
= 35 + (40/100) × 10
= 35  4 = 31 .
Hence, the required mean age is 31 yr.
Weight (in g) 
Number of packets 
200 – 201 
13 
201 – 202 
27 
202 – 203 
18 
203 – 204 
10 
204 – 205 
1 
205 – 206 
1 
Find the mean weight of packets.
Solution
and (class width) h = 1
By assumed mean method,
= 203.5 + 108/70
= 203.5  1.54
Less than type 

Weight (in g) 
Number of packets 
Less than 200 
0 
Less than 201 
0 + 13 = 13 
Less than 202 
27 + 13 = 40 
Less than 203 
18 + 40 = 58 
Less than 204 
10 + 58 = 68 
Less than 205 
1 + 68 = 69 
Less than 206 
1 + 69 = 70 
Now,
Median weight = 201.8 g
Salary (in ₹ thousand) 
Number of persons 
5 – 10 
49 
10 – 15 
133 
15 – 20 
63 
20 – 25 
15 
25 – 30 
6 
30 – 35 
7 
35 – 40 
4 
40 – 45 
2 
45 – 50 
1 
Cumulative frequency (cf) = 49 and class width (h) = 5
= ₹ 13.421 (in thousand)
= 13.421 × 1000
= ₹ 13421
= ₹ 12.727 (in thousand)
Class 
0 – 20 
20 – 40 
40 – 60 
60 – 80 
80 – 100 
Frequency 
17 
f_{1} 
32 
f_{2} 
19 
Given that, sum of all frequencies = 120
⇒ Î£f_{i} = 68 + f_{1} + f_{2} = 120
⇒ f_{1} + f_{2} = 52
Here, (assumed mean) a = 50
and (class width) h = 20
By step deviation method,
⇒ 4 + f_{2}  f_{1} = 0
⇒  f_{2} + f_{1} = 4
On adding Eqs. (i) and (ii), we get
⇒ f_{1} = 28
Put the value of f_{1} in Eq. (i), we get
⇒ f_{2} = 24
Hence, f_{1} = 28 and f_{2} = 24.
Marks 
Frequency 
20 – 30 
Î¡ 
30 – 40 
15 
40 – 50 
25 
50 – 60 
20 
60 – 70 
q 
70 – 80 
8 
80 – 90 
10 
Solution
Height (in cm) 
Number of children 
124 – 128 
5 
128 – 132 
8 
132 – 136 
17 
136 – 140 
24 
140 – 144 
16 
144 – 148 
12 
148 – 152 
6 
152 – 156 
4 
156 – 160 
3 
160 – 164 
1 
Height (in cm) 
Number of children 
Less than 124 
0 
Less than 128 
5 
Less than 132 
13 
Less than 136 
30 
Less than 140 
54 
Less than 144 
70 
Less than 148 
82 
Less than 152 
88 
Less than 156 
92 
Less than 160 
95 
Less than 164 
96 
Size of agricultural holdings(in hec) 
Number of families 
0 – 5 
10 
5 – 10 
15 
10 – 15 
30 
15 – 20 
80 
20 – 25 
40 
25 – 30 
20 
30 – 35 
5 
Size of agricultural holdings(in hec) 
Number of families (f_{i}) 
Cumulative frequency 
0 – 5 
10 
10 
5 – 10 
15 
25 
10 – 15 
30 
55 
15 – 20 
80 
135 
20 – 25 
40 
175 
25 – 30 
20 
195 
30 – 35 
5 
200 
Now, N/2 = 200/2 = 100, which lies in the interval 15  20 .
Lower limit, l = 15, h = 5, f = 80 and cf = 55
(ii) In a given table 80 is the highest frequency.
Rainfall(in cm) 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Number of days 
22 
10 
8 
15 
5 
6 
Calculate the median rainfall using ogives (or move than type and of less than type)
Solution
∵ Total number of days(n) = 66
Now, n/2 = 33
Duration (in s) 
Number of calls 
95 – 125 
14 
125 – 155 
22 
155 – 185 
28 
185 – 215 
21 
215 – 245 
15 
Solution
Here, (assumed mean) a = 170,
and (class width)h = 30
By step deviation method,
Hence, average duration is 170.3 s .
∴ n/2 = 100/2 = 50
14. 50 students enter for a school javelin throw competition. The distance (in metre) thrown are recorded below
Distance (in m) 
0 – 20 
20 – 40 
40 – 60 
60 – 80 
80 – 100 
Number of students 
6 
11 
17 
12 
4 
Now, n/2 = 50/2 = 25
which lies is the interval 40  60 .
∴ l = 40, h = 20, cf = 17 and f = 17
= 40 + 9.41
= 49.41
(iv) Yes, median distance calculated by parts (ii) and (iii) are same.