Chapter 13 Statistics and Probability NCERT Exemplar Solutions Exercise 13.1 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 13 Statistics and Probability Exercise 13.1 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 13.1 Solutions
Multiple Choice Questions
1. In the formula for finding the mean of grouped data d_{i} 's = are deviation from a of
(a) lower limits of the classes
(b) upper limits of the classes
(c) midpoints of the classes
(d) frequencies of the class marks
Solution
(c) We know that, d_{i} = x_{ia}
i.e., d_{i} 's are the deviation from a of mid  points of the classes.
2. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Solution
(b) In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.
3. If x_{i} 's are the mid  points of the class intervals of grouped data f_{i} 's are the corresponding frequencies and x is the mean , then Î£(f_{i} x_{i}  x ) is equal to
(a) 0
(b) 1
(c) 1
(d) 2
Solution
4. In the formula x = a + h [(Î£(f_{i}u_{i})/(Î£(f_{i})], for finding the mean of grouped frequency distribution u_{i} is equal to
(a) (x_{i} + a)/h
(b) h(x_{i}  a)
(c) (x_{i}  a)/h
(d) (a  x_{i} )/h
Solution
5. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) All of these
Solution
(b) Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa.
6. For the following distribution,
Class 
0 – 5 
5 – 10 
10 – 15 
15 – 20 
20 – 25 
Frequency 
10 
15 
12 
20 
9 
the sum of lower limits of the median class and modal class is
(b) 25
(c) 30
Solution
(b) Here,
Class 
Frequency 
Cumulative frequency 
0 – 5 
10 
10 
5 – 10 
15 
25 
10 – 15 
12 
37 
15 – 20 
20 
57 
20 – 25 
9 
66 
Now, N/2 = 66/2 = 33, which lies in the interval 10  15 . Therefore, lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 1520. Therefore, lower limit of modal class is 15. Hence, required sum is 10 + 15 = 25.
7. For the following distribution,
Class 
0 – 5 
6 – 11 
12 – 27 
18 – 23 
24 – 29 
Frequency 
13 
10 
15 
8 
11 
(b) 17.5
(c) 18
Class 
Frequency 
Cumulative frequency 
−0.5 – 5.5 
13 
13 
5.5 – 11.5 
10 
23 
11.5 – 17.5 
15 
38 
17.5 – 23.5 
8 
46 
23.5 – 29.5 
11 
57 
8. For the following distribution,
Marks 
Number of students 
Below 10 
3 
Below 20 
12 
Below 30 
27 
Below 40 
57 
Below 50 
75 
Below 60 
80 
(b) 20  30
(c) 30  40
(c)
Marks 
Number of students 
Cumulative frequency 
Below 10 
3 = 3 
3 
10 – 20 
(12 – 3) = 9 
12 
20 – 30 
(27 – 12) = 15 
27 
30 – 40 
(57 – 27) = 30 
57 
40 – 50 
(75 – 57) = 18 
75 
50 – 60 
(80 – 75) = 5 
80 
Class 
65 – 85 
85 – 105 
105 – 125 
125 – 145 
145 – 165 
165 – 185 
185 – 205 
Frequency 
4 
5 
13 
20 
14 
7 
4 
(a) 0
(b) 19
(c) 20
(d) 38
(c)
Class 
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 
7 
63 
185 – 205 
4 
67 
Class 
13.8 – 14 
14 – 14.2 
14.2 – 14.4 
14.4 – 14.6 
14.6 – 14.8 
14.8 – 15 
Frequency 
2 
4 
5 
71 
48 
20 
(a) 11
(b) 71
(c) 82
(d) 130
Marks obtained 
Number of students 
More than or equal to 0 
63 
More than or equal to 10 
58 
More than or equal to 20 
55 
More than or equal to 30 
51 
More than or equal to 40 
48 
More than or equal to 50 
42 
(a) 3
(b) 4
(c) 3
(d) 4
Marks obtained 
Number of students 
0 – 10 
(63 – 58) = 5 
10 – 20 
(58 – 55) = 3 
20 – 30 
(55 – 51) = 4 
30 – 40 
(51 – 48) = 3 
40 – 50 
(48 – 42) = 6 
50….. 
42 = 42 
(b) 3/4
(c) 1/2
(d) 0
(b) 0.1
(c) 3
(d) 17/16
(b) 0.001
(c) 0.01
(d) 0.1
(b) P
(c) 1 – P
(d) 1  1/p
(b) less than 0
(d) anything but a whole number
(b) P(A) > 1
(c) 0 ≤ P(A) ≤ 1
(d) 1 ≤ P(A) ≤ 1
(b) 3/13
(c) 2/13
(d) 1/2
(a) 1/7
(b) 2/7
(c) 3/7
(d) 5/7
(a) 1/6
(b) 1/3
(c) 1/2
(d) 0
(b) 13
(c) 48
(d) 51
(b) 14
(c) 21
(d) 28
⇒ (Number of bad eggs)/(Total number of eggs) = 0.035
⇒ (Number of bad eggs)/400 = 0.035
∴ Number of bad eggs = 0.035 × 400 = 14
(b) 240
(c) 480
(d) 750
⇒ x = 0.08 × 6000
∴ x = 480
Hence, she bought 480 tickets.
(b) 3/5
(c) 4/5
(d) 1/3
∴ Required probability = 8/40 = 1/5
(b) 6/25
(c) 1/4
(d) 13/50
∴ Required probability = 25/100 = 1/4
(b) 6/23
(c) 8/23
(d) 17/23