Chapter 13 Statistics and Probability NCERT Exemplar Solutions Exercise 13.3 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 13 Statistics and Probability Exercise 13.3 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 13.3 Solutions
1. Find the mean of the distribution
Class 
1 – 3 
3 – 5 
5 – 7 
7 – 10 
Frequency 
9 
22 
27 
17 
Solution
Class 
Class marks 
Frequency 
f_{i} x_{i} 
1 – 3 
2 
9 
18 
3 – 5 
4 
22 
88 
5 – 7 
6 
27 
162 
7 – 10 
8.5 
17 
144.5 


Î£f_{i} = 75 
Î£f_{i} x_{i} = 412.5 
Therefore, mean (x) = (Î£f_{i} x_{i})/(Î£f_{i}) = 412.5/75 = 5.5
Hence, mean of the given distribution is 5.5.
Marks 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Number of students 
2 
4 
7 
6 
1 
Marks 
Class marks 
Frequency 
f_{i} x_{i} 
10 – 20 
15 
2 
30 
20 – 30 
25 
4 
100 
30 – 40 
35 
7 
245 
40 – 50 
45 
6 
270 
50 – 60 
55 
1 
55 
Î£f_{i} = 20 
Î£f_{i} x_{i} = 700 
Therefore, mean (x) = (Î£f_{i} x_{i})/(Î£f_{i}) = 700/20 = 35
Hence, the mean of scores of 20 students in mathematics test is 35.
3. Calculate the mean of the following data
Class 
4 – 7 
8 – 11 
12 – 15 
16 – 19 
Frequency 
5 
4 
9 
10 
Class 
Class marks 
Frequency 
f_{i} x_{i} 
3.5  7.5 
5.5 
5 
27.5 
7.5 – 11.5 
9.5 
4 
38 
11.5 – 15.5 
13.5 
9 
121.5 
15.5 – 19.5 
17.5 
10 
175 


Î£f_{i} = 28 
Î£f_{i} x_{i} = 362 
Therefore, mean (x) = (Î£f_{i} x_{i})/(Î£f_{i})= 362/28 = 12.93
Hence, mean of the given data is 12.93.
Number of pages written per day 
16 – 18 
19 – 21 
22 – 24 
25 – 27 
28 – 30 
Number of days 
1 
3 
4 
9 
13 
Since,
Class  marks 
Mid  value 
Number of days 
f_{i} x_{i} 
15.5 – 18.5 
17 
1 
17 
18.5 – 21.5 
20 
3 
60 
21.5 – 24.5 
23 
4 
92 
24.5 – 27.5 
26 
9 
234 
27.5 – 30.5 
29 
13 
377 
Total 
Î£f_{i} = 30 
Î£f_{i} x_{i} = 780 
Income (in ₹) 
1 – 200 
201 – 400 
400 – 600 
601 – 800 
Number of employees 
14 
15 
14 
7 
∴ Assumed mean, a = 300.5
Class width, h = 200
and total observations, N = 50
By step deviation method,
= 300.5 + 200 × (1/50) × 14
= 300.5 + 56 = 356.5
Number of seats 
100 – 104 
104 – 108 
108 – 112 
112 – 116 
116 – 120 
Freqency 
15 
20 
32 
18 
15 
Determine the mean number of seats occupied over the flights.
Solution
∴ Assumed mean, a = 110,
Class width, h = 4
and total observation, N = 100
By assumed mean method,
Mean ( x ) = a + (Î£f_{i} d_{i} )/(Î£f_{i} )
= 110 + (8/100) = 110  0.08 = 109.92
Weight (in kg) 
100 – 110 
110 – 120 
120 – 130 
130 – 140 
140 – 150 
Number of wrestlers 
4 
14 
21 
8 
3 
∴ Assumed mean (a) = 125
Class width (h) = 10 and total observation (N) = 50
By assumed mean method,
= 125 + (80)/50
= 125  16 = 123.4 kg
Mileage(kmL^{1} ) 
10 – 12 
12 – 14 
14 – 16 
16 – 18 
Number of cars 
7 
12 
18 
13 
Do you agree with this claim?
Here, Î£f_{i} = 50
and Î£f_{i} x_{i} = 724
= 724/50 = 14.48
Weight (in kg) 
40 – 45 
45 – 50 
50 – 55 
55 – 60 
60 – 65 
65 – 70 
70 – 75 
75 – 80 
Number of persons 
4 
4 
13 
5 
6 
5 
2 
1 
Weight (in kg) 
Cumulative frequency 
Less than 45 
4 
Less than 50 
4 + 4 = 8 
Less than 55 
8 + 13 = 21 
Less than 60 
21 +5 = 26 
Less than 65 
26 + 6 = 32 
Less than 70 
32 + 5 = 37 
Less than 75 
37 + 2 = 39 
Less than 80 
39 + 1 = 40 
10. The following table shows the cumulative frequency distribution of marks of 800 students in an examination.
Marks 
Number of students 
Below 10 
10 
Below 20 
50 
Below 30 
130 
Below 40 
270 
Below 50 
440 
Below 60 
570 
Below 70 
670 
Below 80 
740 
Below 90 
780 
Below 100 
800 
Marks (Out of 90) 
Number of candidates 
More than or equal to 80 
4 
More than or equal to 70 
6 
More than or equal to 60 
11 
More than or equal to 50 
17 
More than or equal to 40 
23 
More than or equal to 30 
27 
More than or equal to 20 
30 
More than or equal to 10 
32 
More than or equal to 0 
34 
Class interval 
Number of candidates 
0 – 10 
34 – 32 = 2 
10 – 20 
32 – 30 = 2 
20 – 30 
30 – 27 = 3 
30 – 40 
27 – 23 = 4 
40 – 50 
23 – 17 = 6 
50 – 60 
17 – 11 = 6 
60 – 70 
11 – 6 = 5 
70 – 80 
6 – 4 = 2 
80 – 90 
4 
12. Find the unknown entries o, b, c, d, e and f in the following distribution of heights of students in a class
Height (in cm) 
Frequency 
Cumulative frequency 
150 – 155 
12 
A 
155 – 160 
b 
25 
160 – 165 
10 
C 
165 – 170 
d 
43 
170 – 175 
e 
48 
175 – 180 
2 
f 
Total 
50 

On comparing last two tables, we get
∴ 12 + b = 25
⇒ b = 25  12 = 13
22 + b = c
⇒ c = 22 + 13 = 35
22 + b + d = 43
⇒ 22 + 13 + d = 43
⇒ d = 43  35 = 8
and 22 + b + d + e = 48
⇒ 22 + 13 + 8 + e = 48
⇒ e = 48  43 = 5
and 24 + b + d + e = f
⇒ 24 + 13 + 8 + 5 = f
∴ f = 50
Age (in year) 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
Number of patients 
60 
42 
55 
70 
53 
20 
Marks 
Below 20 
Below 40 
Below 60 
Below 80 
Below 100 
Number of students 
17 
22 
29 
37 
50 
Form the frequency distribution table for the data.
Solution
Marks 
Number of students 
0 – 20 
17 
20 – 40 
22 – 17 = 5 
40 – 60 
29 – 22 = 7 
60 – 80 
37 – 29 = 8 
80 – 100 
50 – 37 = 13 
Weekly Income(in ₹) 
Number of families 
0 – 1000 
250 
1000 – 2000 
190 
2000 – 3000 
100 
3000 – 4000 
40 
4000 – 5000 
15 
5000 – 6000 
5 
Total 
600 
It is given that, n = 600
∴ n/2 = 600/2 = 300
Since, cumulative frequency 440 lies in the interval 1000  2000.
Here, (lower median class)l = 1000,
f = 190, cf = 250, (class width) h = 1000
and (total observation ) n = 600
= 1000 + 263.15 = 1263.15
Speed (in km/h) 
85 – 100 
100 – 115 
115 – 130 
130 – 145 
Number of players 
11 
9 
8 
5 
It is given that, n = 33
∴ n/2 = 33/2 = 16.5
So, the median class is 100  115,
frequency (f) = 9,
cumulative frequency (cf) = 11
and class width (h) = 15
= 109.17
Hence, the median bowling speed is 109.17 km/h.
Income (in ₹) 
Number of families 
0 – 5000 
8 
5000 – 10000 
26 
10000 – 15000 
41 
15000 – 20000 
16 
20000 – 25000 
3 
25000 – 30000 
3 
30000 – 35000 
2 
35000 – 40000 
1 
Calculate the model income.
Hence, the modal income is ₹ 11875.
Weight (in g) 
Number of packets 
200 – 201 
12 
201 – 202 
26 
202 – 203 
20 
203 – 204 
9 
204 – 205 
2 
205 – 206 
1 
Determine the model weight.
Solution
Hence, the modal weight is 201.7 g.
∴ Number of possible outcomes = 6
Now, required probability = 6/36 = 1/6
= 36  Number of possible outcomes for same number on both dice
= 36  6 = 30
∴ Required probability = 3036 = 5/ 6
(ii) a prime number ?
(iii) 1 ?
Number of possible ways = 6
∴ Required probability = 6/36 = 1/6
So, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5) .
∴ Required probability = 15/36 = 5/12
It is not possible, so its probability is zero.
(ii) 12
(iii) 7
Number of possible ways = 4
∴ Required probability = 4/36 = 1/9
Number of possible ways = 4
∴ Required probability = 4/36 = 1/9
S = {(HH), (TT), (HT), (TH)}
∴ n(S) = 4
Let E = Event of getting at most one head
∴ n(E) = 3
Hence, required probability = n(E)/n(S) = 3/4
(ii) at least 2 heads
Let, E = Event of getting the numbers whose difference is 2
= {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
∴ n(E) = 8
∴ P(E) = n(E)/n(S) = 8/36 = 2/9
(ii) green ball
(iii) not a blue ball
(ii) a king
(ii) 10 of hearts
(ii) greater than 7
(iii) Less than 7
E = Card value 7 may be of a spade, a diamond, a club or a heart
∴ P(E_{1}) = n(E_{1})/n(S) = 4/40 = 1/10
∴ n(E_{2}) = 3 × 4 = 12
P(E_{2} ) = n(E_{2})/n(S) = 12/40 = 3/10
= Event of getting a card whose value is 1, 2, 3, 4, 5 or 6
∴ n(E_{3}) = 6 × 4 = 24
∴ P(E_{3}) = n(E_{3})/n(S) = 24/40 = 3/5
(ii) not divisible by 7?
n(S)= 99
= Event of choosing an integer which is multiple of 7
∴ n(E_{1}) = 14
∴ P(E_{1}) = n(E_{1})/n(S) = 14/99
(ii) a square number
[in an AP, l = a + (n1)d, here l = 100, a = 2 and d = 2
∴ n(E_{1} ) = 50
∴ Required probability = n(E_{1} )/n(S) = 50/100 = 1/2
= {(2)^{2} ,(3)^{2} , (4)^{2} , (5)^{2} , (6)^{2} , (7)^{2} , (8)^{2} , (9)^{2} , (10)^{2} }
∴ n(E_{2} ) = 9
Hence, required probability = n(E_{2} )/n(S) = 9/100
are, n(S) = 26
Let E = Event of choosing a english alphabet, which is a consonant
∴ n(E) = 21
∴ P(E) = n(E)/n(S) = 690/1000 = 69/100 = 0.69
From the chart it is clear that, there are 11 slips which are marked other than ₹ 1.
Let E_{1} = Event of selecting not defective bulb = Event of selecting good bulbs
n(E_{1}) = 18
∴ P(E_{1}) = n(E_{1})/n(S) = 18/24 = 3/4
Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton, n(S) = 23.
In them, 18 are good bulbs and 5 are defective bulbs.
(ii) square
(iii) square of blue colour
(iv) triangle of red
(i) P (lost piece is a triangle ) = 8/18 = 4/9
(iii) P (square of blue colour) = 6/18 = 1/3
(iv) P(triangle of red colour) = 5/18
P(E_{1}) = n(E_{1})/n(S) = 1/8
= Sweta gets heads one or two times
= (HTT), (THT), (TTH), (HHT) , (HTH), (THH)}
∴ n(E_{3}) = 6
∴ P(E_{3}) = n(E_{3})/n(S) = 6/8 = 3/4
(ii) Let E = Event of getting a sum 7
∴ n(E) = 12
∴ P(E) = n(E)/n(S) = 12/36 = 1/3
(i) Let E_{1} = Event that Varnika will buy a mobile phone
∴ P(E_{1}) = n(E_{1} )/n(S) = 42/48 = 7/8
(ii) Let E_{2} = Event that trader will buy only when it has no major defects
∴ P(E_{2}) = n(E_{2})/n(S) = 45/48 = 15/16
(ii) white
∴ Number of red balls = x = 4
Number of white balls = 2x = 2 ×4 = 8
∴ n(E_{1}) = Number of white balls + Number of blue balls
⇒ n(E_{1}) = 8 + 12 = 20
∴ Required probability = n(E_{1})/n(S) = 20/24 = 5/6
So, required probability = n(E_{2})/n(S) = 8/24 = 1/3
(i) the first player wins a prize?
If the selected card has a perfect square greater than 500, then player wins a prize.
∴ n(E_{2}) = 9  1 = 8
So, required probability = n(E_{2})/n(S') = 8/999