RBSE Class 10 Maths Chapter 17 – Statistics Important questions and solutions are given here. All these important questions available have stepwise solutions so that students can understand them easily. The RBSE Class 10 important questions and solutions provided at BYJU’S will help the students in acquiring problem-solving and analytical skills.
Chapter 17 of RBSE Class 10 has 7 exercises which cover the measures of central tendency, namely mean, median and mode. After practicing these questions, students will be able to understand how to find the mean or median or mode for ungrouped and grouped data. Several situations are given here so that they will be guided with an approach on how to solve them using appropriate methods.
RBSE Maths Chapter 17: Exercise 17.1 Textbook Important Questions and Solutions
Question 1: The monthly salaries (in Rs.) of subordinate employees of a school are 1720, 1750, 1760 and 1710, then find the arithmetic mean.
Solution:
Arithmetic mean = Sum of observations/Number of observations
= Sum of monthly salaries of employees/Number of employees
= (1720 + 1750 + 1760 + 1710)/4
= 6940/4
= 1735
Therefore, the arithmetic mean of monthly salaries of subordinate employees is Rs. 1735.
Question 2: If the arithmetic mean of the marks 3, 4, 8, 5, x, 3, 2, 1 is 4, then find the value of x.
Solution:
Arithmetic mean = Sum of observations/Number of observations
= Sum of marks/Number of students
4 = (3 + 4 + 8 + 5 + x + 3 + 2 + 1)/8
4 × 8 = 26 + x
32 = 26 + x
⇒ x = 32 – 26
⇒ x = 6
Question 3: Calculate the arithmetic mean of the following marks obtained by 10 students in English in the monthly test.
Roll No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Marks obtained | 30 | 28 | 32 | 12 | 18 | 20 | 25 | 15 | 26 | 14 |
Solution:
Arithmetic mean = Sum of observations/Number of observations
= Sum of marks/Number of students
= (30 + 28 + 32 + 12 + 18 + 20 + 25 + 15 + 15 + 26 + 14)/10
= 235/10
= 23.5
Hence, the required arithmetic mean is 23.5.
Question 4: The average weight of 25 students of section A of a class is 51 kg, whereas the average weight of 35 students of section B is 54 kg. Find the average weight of 60 students of this class.
Solution:
Given,
Average weight of 25 students of section A = 51 kg
Thus, the total weight of 25 students = 51 × 25 = 1275 kg
Average weight of 35 students of section B = 54 kg
Thus, the total weight of 35 students = 54 × 35 = 1890 kg
Total weight of the students of the class = 1275 + 1890 = 3165
Total number of students in the class = 25 + 35 = 60
Average weight of 60 students = 3165/60 = 52.75
Hence, the average weight of students of the class is 52.75 kg.
Question 5: The mean of 5 numbers is 18. If one number is excluded, then mean becomes 16, then find the excluded number.
Solution:
Given,
Average of 5 numbers = 18
Sum of 5 numbers = 18 × 5 = 90
Let x be the number excluded.
Average of 4 numbers = 16 (given)
Sum of 4 numbers = 16 × 4 = 64
Thus, x + 64 = 90
x = 90 – 64
x = 26
Therefore, 26 is the excluded number.
Question 6: The mean of 13 numbers is 24. If 3 is added to each number, then find their new mean.
Solution:
Given,
Mean of 13 numbers = 24
Sum of 13 numbers = 24 × 13 = 312
3 is added to each of these 13 numbers.
Thus, the new sum of the numbers = 312 + (3 × 13)
= 312 + 39
= 351
Mean of new numbers = 351/13 = 27
Question 7: The monthly salary of 5 employees of a school is Rs. 3000. On the retirement of one employee, the average monthly salary of remaining employees is Rs. 3200. What was the salary of a retired employee at the time of retirement?
Solution:
Given,
Average salary of 5 employees = Rs. 3000
Sum of salary of 5 employees = Rs. 3000 × 5 = Rs. 15000
Average of 4 employees after one employee got retired = Rs. 3200
Sum of salary of 4 employees = Rs. 3200 × 4 = Rs. 12800
Salary of the 5th employee = Rs. 15000 – Rs. 12800 = Rs. 2200
Therefore, the salary of an employee at the time of retirement is Rs. 2200.
RBSE Maths Chapter 17: Exercise 17.2 Textbook Important Questions and Solutions
Question 8: Find the mean of the following frequency distribution.
x | 2 | 5 | 7 | 9 | 11 |
f | 1 | 5 | 4 | 7 | 3 |
Solution:
Table for calculating sum is:
x_{i} | f_{i} | f_{i}x_{i} |
2 | 1 | 2 |
5 | 5 | 25 |
7 | 4 | 28 |
9 | 7 | 63 |
11 | 3 | 33 |
∑f_{i} = 20 | ∑f_{i}x_{i} = 151 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 151/20
= 7.55
Therefore, the arithmetic mean of the given data is 7.55.
Question 9: Find the mean of the following frequency distribution.
x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
f | 30 | 60 | 20 | 40 | 10 | 50 |
Solution:
Table for calculating sum is:
x_{i} | f_{i} | f_{i}x_{i} |
0.1 | 30 | 3 |
0.2 | 60 | 12 |
0.3 | 20 | 6 |
0.4 | 40 | 16 |
0.5 | 10 | 5 |
0.6 | 50 | 30 |
∑f_{i} = 210 | ∑f_{i}x_{i} = 72 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 72/210
= 0.343
Therefore, the arithmetic mean of the given data is 0.343.
Question 10: In hundred families, the number of children are:
Number of children | 1 | 2 | 3 | 4 | 5 | 6 |
Number of families | 45 | 25 | 19 | 8 | 2 | 1 |
Find their arithmetic mean.
Solution:
Table of calculating the sum is:
Number of children
x_{i} |
Number of families
f_{i} |
f_{i}x_{i} |
1 | 45 | 45 |
2 | 25 | 50 |
3 | 19 | 57 |
4 | 8 | 32 |
5 | 2 | 10 |
6 | 1 | 6 |
∑f_{i} = 100 | ∑f_{i}x_{i} = 200 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 200/100
= 2
Therefore, the arithmetic mean of the given data is 2.
Question 11: If the mean of the following distribution is 7.5, then find the value of P.
x | 3 | 5 | 7 | 9 | 11 | 13 |
f | 6 | 8 | 15 | P | 8 | 4 |
Solution:
Table for calculating sum is:
x_{i} | f_{i} | f_{i}x_{i} |
3 | 6 | 18 |
5 | 8 | 40 |
7 | 15 | 105 |
9 | P | 9P |
11 | 8 | 88 |
13 | 4 | 52 |
∑f_{i} = 41 + P | ∑f_{i}x_{i} = 303 + 9P |
Mean = ∑f_{i}x_{i}/∑f_{i}
7.5 = (303 + 9P)/(41 + P)
(41 + P)(7.5) = 303 + 9P
(41 × 7.5) + 7.5P = 303 + 9P
307.5 – 303 = 9P – 7.5P
⇒ 1.5P = 4.5
⇒ P = 4.5/1.5
⇒ P = 3
Hence, the value of P is 3.
Question 12: If mean of the following frequency distribution is 1.46, then find the unknown frequencies.
x | 0 | 1 | 2 | 3 | 4 | 5 | Sum |
f | 46 | – | – | 25 | 10 | 5 | 200 |
Solution:
Let x and y be the missing frequencies.
Table for calculating sum is:
x_{i} | f_{i} | f_{i}x_{i} |
0 | 46 | 0 |
1 | x | x |
2 | y | 2y |
3 | 25 | 75 |
4 | 10 | 40 |
5 | 5 | 25 |
∑f_{i} = 86 + x + y | ∑f_{i}x_{i} = 140 + x + 2y |
Given, sum of frequencies = 200
Thus, 86 + x + y = 200
x + y = 200 – 86
x + y = 114….(i)
Mean = ∑f_{i}x_{i}/∑f_{i}
1.45 = (140 + x + y)/200
140 + x + y = 292
x + 2y = 292 – 140
x + 2y = 152 ….(ii)
Subtracting (i) from (ii),
x + 2y – (x + y) = 152 – 114
y = 38
Substituting y = 38 in (i),
x + 38 = 114
x = 114 – 38
x = 76
Hence, the missing frequencies are 76 and 38.
RBSE Maths Chapter 17: Exercise 17.3 Textbook Important Questions and Solutions
Question 13: Find the arithmetic mean of the following frequency distribution.
Class | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
Frequency | 6 | 8 | 10 | 9 | 7 |
Solution:
Class | Frequency (f_{i}) | Mid-value (x_{i}) | f_{i}x_{i} |
0 – 6 | 6 | 3 | 18 |
6 – 12 | 8 | 9 | 72 |
12 – 18 | 10 | 15 | 150 |
18 – 24 | 9 | 21 | 189 |
24 – 30 | 7 | 27 | 189 |
∑f_{i} = 40 | ∑f_{i}x_{i} = 618 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 618/40
= 15.45
Therefore, the arithmetic mean of the given data is 15.45.
Question 14: Find the arithmetic mean of the following frequency distribution.
Marks obtained (x) | 100 – 120 | 120 -140 | 140 – 160 | 160 – 180 | 180 – 200 |
Number of students (f) | 10 | 20 | 20 | 15 | 5 |
Solution:
Marks obtained | Number of students (f_{i}) | Mid-value (x_{i}) | f_{i}x_{i} |
100 – 120 | 10 | 110 | 1100 |
120 – 140 | 20 | 130 | 2600 |
140 – 160 | 20 | 150 | 3000 |
160 – 180 | 15 | 170 | 2550 |
180 – 200 | 5 | 190 | 950 |
∑f_{i} = 70 | ∑f_{i}x_{i} = 10200 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 10200/70
= 145.71
Therefore, the arithmetic mean of the given data is 145.71.
Question 15: The salaries of the workers of a factory are as follows.
Monthly salary (in Rs.) | 1000 – 1200 | 1200 – 1400 | 1400 – 1600 | 1600 – 1800 | 1800 – 2000 |
Number of workers | 10 | 20 | 20 | 15 | 5 |
Find the arithmetic mean of the salaries.
Solution:
Monthly salary (in Rs.) | Number of workers (f_{i}) | Mid-value (x_{i}) | f_{i}x_{i} |
100 – 120 | 10 | 1100 | 11000 |
120 – 140 | 20 | 1300 | 26000 |
140 – 160 | 20 | 1500 | 30000 |
160 – 180 | 15 | 1700 | 25500 |
180 – 200 | 5 | 1900 | 9500 |
∑f_{i} = 70 | ∑f_{i}x_{i} = 102000 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 102000/70
= 1457.14
Therefore, the arithmetic mean of the salary of workers is Rs. 1457.14.
RBSE Maths Chapter 17: Exercise 17.4 Textbook Important Questions and Solutions
Question 16: Find the mean of the following frequency distribution by taking assumed mean.
x | 800 | 820 | 860 | 900 | 920 | 980 | 1000 |
f | 7 | 14 | 19 | 25 | 20 | 10 | 5 |
Solution:
Let assumed mean = A = 900 and h = 20
x_{i} | f_{i} | d_{i} = x_{i} – A | u_{i} = d_{i}/h | f_{i}u_{i} |
800 | 7 | -100 | -5 | -35 |
820 | 14 | -80 | -4 | -56 |
860 | 19 | -40 | -2 | -38 |
900 = A | 25 | 0 | 0 | 0 |
920 | 20 | 20 | 1 | 20 |
980 | 10 | 80 | 4 | 40 |
1000 | 5 | 100 | 5 | 25 |
∑f_{i} = 100 | ∑f_{i}u_{i} = -44 |
By step-deviation method,
Arithmetic mean = A + [(∑f_{i}u_{i})/∑f_{i}] × h
= 900 + (-44/100) × 20
= 900 – 8.8
= 891.2
Therefore, the mean of the given distribution is 891.2.
Question 17: Find the mean of the following frequency distribution by taking the assumed mean.
Expenditure on water (in Rs.) | Number of houses |
15 – 20 | 7 |
20 – 25 | 5 |
25 – 30 | 7 |
30 – 35 | 8 |
35 – 40 | 9 |
40 – 45 | 11 |
45 – 50 | 7 |
50 – 55 | 5 |
55 – 60 | 4 |
60 – 65 | 4 |
65 – 70 | 3 |
Solution:
Expenditure on water (in Rs.) | Number of houses (fi) | Mid-value (x_{i}) | d_{i} = x_{i} – A | u_{i} = d_{i}/h | f_{i}u_{i} |
15 – 20 | 7 | 17.5 | -25 | -5 | -35 |
20 – 25 | 5 | 22.5 | -20 | -4 | -20 |
25 – 30 | 7 | 27.5 | -15 | -3 | -21 |
30 – 35 | 8 | 32.5 | -10 | -2 | -16 |
35 – 40 | 9 | 37.5 | -5 | -1 | -9 |
40 – 45 | 11 | 42.5 = A | 0 | 0 | 0 |
45 – 50 | 7 | 47.5 | 5 | 1 | 7 |
50 – 55 | 5 | 52.5 | 10 | 2 | 10 |
55 – 60 | 4 | 57.5 | 15 | 3 | 12 |
60 – 65 | 4 | 62.5 | 20 | 4 | 16 |
65 – 70 | 3 | 67.5 | 25 | 5 | 15 |
∑f_{i} = 70 | ∑f_{i}u_{i} = -41 |
A = 42.5 and h = 5
By step-deviation method,
Arithmetic mean = A + [(∑f_{i}u_{i})/∑f_{i}] × h
= 42.5 + (-41/70) × 5
= 42.5 – 2.93
= 39.57
Therefore, the mean of the given distribution is 39.57.
Question 18: Find the mean of the following distribution by taking assumed mean as 25.
Class interval | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
f | 6 | 10 | 13 | 7 | 4 |
Solution:
Class | f_{i} | Mid-value (x_{i}) | d_{i} = x_{i} – A | f_{i}d_{i} |
0 – 10 | 6 | 5 | -20 | -120 |
10 – 20 | 10 | 15 | -10 | -100 |
20 – 30 | 13 | 25 = A | 0 | 0 |
30 – 40 | 7 | 35 | 10 | 70 |
40 – 50 | 4 | 45 | 20 | 80 |
∑f_{i} = 40 | ∑f_{i}d_{i} = -70 |
By assumed mean method,
Mean = A + (∑f_{i}d_{i})/∑f_{i}
= 25 + (-70/40)
= 25 – 1.75
= 23.25
Therefore, the arithmetic mean of the given distribution is 23.25.
RBSE Maths Chapter 17: Exercise 17.5 Textbook Important Questions and Solutions
Question 19: Find the median of the following variates.
25, 34, 33, 13, 20, 26, 36, 28, 19, 34
Solution:
Given,
25, 34, 33, 13, 20, 26, 36, 28, 19, 34
Arranging in the ascending order:
13, 19, 20, 25, 26, 28, 33, 34, 34, 36
Number of variates = n = 10
Median = (1/2) [(n/2)th + (n/2 + 1)th observation]
= (1/2) [(10/2)th + (10/2 + 1)th observation]
= (5th + 6th observation)/2
= (26 + 28)/2
= 54/2
= 27
Hence, the median is 27.
Question 20: Find the median of the following data.
19, 25, 59, 48, 35, 31, 30, 32, 51
If 25 is replaced by 52, then find a new median.
Solution:
Given,
19, 25, 59, 48, 35, 31, 30, 32, 51
Arranging in the ascending order:
19, 25, 30, 31, 32, 35, 48, 51, 59
Number of observations = n = 9
Median = [(n + 1)/2]th observation
= [(9 + 1)/2]th observation
= 5th observation
= 32
Now, 25 is replaced with 52 in the given data.
Ascending order of the data is:
19, 30, 31, 32, 35, 48, 51, 52, 59
Median = [(n + 1)/2]th observation
= [(9 + 1)/2]th observation
= 5th observation
= 35
Therefore, the first median is 31 and the new median is 35.
Question 21: The marks obtained by students of a class are given below. Find their median.
Marks | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
Number of students | 2 | 8 | 16 | 26 | 20 | 16 | 7 | 4 |
Solution:
Marks | Number of students | Cumulative frequency |
15 | 2 | 2 |
20 | 8 | 10 |
25 | 16 | 26 |
30 | 26 | 52 |
35 | 20 | 72 |
40 | 16 | 88 |
45 | 7 | 95 |
50 | 4 | 99 |
N = 99 |
N/2 = 99/2 = 49.5
Cumulative frequency above 49.5 is 52.
The corresponding variable value of 52 is 30.
Therefore, the median is 30.
RBSE Maths Chapter 17: Exercise 17.6 Textbook Important Questions and Solutions
Question 22: Marks obtained by 100 students are given in the following table. Find the median from these.
Marks obtained | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
Number of students | 6 | 20 | 44 | 26 | 3 | 1 |
Solution:
Cumulative frequency table is:
Marks obtained | Number of students | Cumulative frequency |
20 – 30 | 6 | 6 |
30 – 40 | 20 | 26 = cf |
40 – 50 | 44 = f | 70 |
50 – 60 | 26 | 96 |
60 – 70 | 3 | 99 |
70 – 80 | 1 | 100 |
N = 100 |
N/2 = 100/2 = 50
Cumulative frequency just above 50 is 70.
The class interval corresponding to 70 is 40 – 50.
Thus, median class = 40 – 50
Lower limit of median class = l = 40
Frequency of median class = f = 44
Cumulative frequency above the median class = cf = 26
Class height = h = 10
Median = l + [(N/2 – cf)/f] × h
= 40 + [(50 – 26)/44] × 10
= 40 + (240/44)
= 40 + 5.45
= 45.45
Hence, the median is 45.45.
Question 23: Find the median from the following frequency distribution.
Class interval | 0 – 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 | 40 – 48 |
f_{i} | 42 | 30 | 50 | 22 | 8 | 5 |
Solution:
Marks obtained | Number of students | Cumulative frequency |
0 – 8 | 42 | 42 |
8 – 16 | 30 | 72 = cf |
16 – 24 | 50 = f | 122 |
24 – 32 | 22 | 144 |
32 – 40 | 8 | 152 |
40 – 48 | 5 | 157 |
N = 157 |
N/2 = 157/2 = 78.5
Cumulative frequency just above 78.5 is 122.
The class interval corresponding to 122 is 16 – 24.
Thus, median class = 16 – 24
Lower limit of median class = l = 16
Frequency of median class = f = 50
Cumulative frequency above the median class = cf = 72
Class height = h = 8
Median = l + [(N/2 – cf)/f] × h
= 16 + [(78.5 – 72)/50] × 8
= 16 + (6.5 × 8 /50)
= 16 + (52/50)
= 16 + 1.04
= 17.04
Hence, the median is 17.04.
RBSE Maths Chapter 17: Exercise 17.7 Textbook Important Questions and Solutions
Question 24: Find the mode of the following distribution.
(i) 2 5 7 5 3 1 5 8 7 5
(ii) 2 4 6 2 6 6 7 8
(iii) 2.5 2.5 2.1 2.5 2.7 2.8 2.5
Solution:
(i) Given,
2 5 7 5 3 1 5 8 7 5
5 has maximum frequency i.e. it is repeated 4 times.
Hence, mode of data is 5.
(ii) Given,
2 4 6 2 6 6 7 8
6 has maximum frequency i.e. it is repeated 3 times.
Hence, mode of data is 6.
(iii) Given,
2.5 2.5 2.1 2.5 2.7 2.8 2.5
2.5 has maximum frequency i.e. it is repeated 4 times.
Hence, mode of data is 2.5.
Question 25: Find the mode of the following frequency distribution.
x | 1.1 | 1.2 | 1.3 | 1.4 | 1.5 | 1.6 |
f | 20 | 50 | 80 | 60 | 15 | 8 |
Solution:
From the given,
1.3 has the maximum frequency , i.e. 80.
Hence, mode of the given distribution is 80.
Question 26: The ages (in years) of 20 students of a class are as follows.
15 16 13 14 14 13 15 14 13 13
14 12 15 14 16 13 14 14 13 15
Find mode by representing these in frequency distribution.
Solution:
Age (in years) | Frequency |
12 | 1 |
13 | 6 |
14 | 7 |
15 | 4 |
16 | 2 |
Thus, the number of students with 14 years age are maximum, that means the maximum frequency is 7.
Hence, the mode is 14.
Question 27: Find the mode from the following frequency distribution.
Class | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 |
Frequency | 3 | 7 | 16 | 12 | 9 | 5 | 3 |
Solution:
From the given,
Maximum frequency is 16 which corresponds to the class 20 – 25.
Modal class = 20 – 25
Lower limit of modal class = l = 20
Frequency of the class preceding the modal class = f_{0} = 7
Frequency of the modal class = f_{1} = 16
Frequency of the class succeeding the modal class = f_{2} = 12
Class height = h = 5
Mode = l + [(f_{1} – f_{0})/ (2f_{1} – f_{0} – f_{2})] × h
= 20 + [(16 – 7)/ (2 × 16 – 7 – 12)] × 5
= 20 + [9/(32 – 19)] × 5
= 20 + (45/13)
= 20 + 3.46
= 23.46 (approx)
Hence, the mode is 23.46.
Question 28: Find the mode of the following distribution.
Height (in cm) | 52 – 55 | 55 – 58 | 58 – 61 | 61 – 64 |
Number of students | 10 | 20 | 25 | 10 |
Solution:
Maximum frequency is 25 which corresponds to the class 58 – 61.
Modal class = 58 – 61
Lower limit of modal class = l = 58
Frequency of the class preceding the modal class = f_{0} = 20
Frequency of the modal class = f_{1} = 25
Frequency of the class succeeding the modal class = f_{2} = 10
Class height = h = 3
Mode = l + [(f_{1} – f_{0})/ (2f_{1} – f_{0} – f_{2})] × h
= 58 + [(25 – 20)/ (2 × 25 – 20 – 10)] × 3
= 58 + [5/(50 – 30)] × 3
= 58 + (15/20)
= 58 + 0.75
= 58.75 (approx)
Hence, the mode is 58.75.
RBSE Maths Chapter 17: Additional Important Questions and Solutions
Question 1: Positional mean is:
(A) Arithmetic mean
(B) Geometric mean
(C) Harmonic mean
(D) Median
Solution:
Correct answer: (D)
We know that, median of any data can be found after arranging either in ascending or descending order.
Thus, the median depends on the number of observations.
Therefore, median is considered as the positional mean.
Question 2: Mode value of any series is:
(A) Middle value
(B) Value whose frequency is maximum
(C) Minimum frequency value
(D) Limit value
Solution:
Correct answer: (B)
We know that, if an observation occurs most number of times, then it is called mode.
Thus, the value with maximum frequency is the mode.
Question 3: The median of following series is:
520, 20, 340, 190, 35, 800, 1210, 50, 80
(A) 1210
(B) 520
(C) 190
(D) 35
Solution:
Correct answer: (C)
Given,
520, 20, 340, 190, 35, 800, 1210, 50, 80
Arranging the given data in ascending order:
20, 35, 50, 80, 190, 340, 520, 800, 1210
Number of observations = n = 9
Median = [(n + 1)/2]th observation
= [(9 + 1)/2]th observation
= 5th observation
= 190
Therefore, the median is 190.
Question 4: The marks obtained by four students in statistics are 53, 75, 42, 70. Mean of their marks is:
(A) 42
(B) 64
(C) 60
(D) 56
Solution:
Correct answer: (C)
Given,
Marks of the students in Statistics are:
53, 75, 42, 70
Arithmetic mean = Sum of observations/Number of observations
= Sum of marks/Number of students
= (53 + 75 + 42 + 70)/4
= 240/4
= 60
Question 5: If the arithmetic mean of 5, 7, 9, x is 9, then the value of x is:
(A) 11
(B) 15
(C) 18
(D) 16
Solution:
Correct answer: (B)
Given,
Arithmetic mean of 5, 7, 9, x = 9
We know that,
Arithmetic mean = Sum of observations/Number of observations
9 = (5 + 7 + 9 + x)/4
⇒ 36 = 21 + x
⇒ x = 36 – 21
⇒ x = 15
Question 6: The mode of the distribution 3, 5, 7, 4, 2, 1, 4, 3, 4 is:
(A) 7
(B) 4
(C) 3
(D) 1
Solution:
Correct answer: (B)
Given,
3, 5, 7, 4, 2, 1, 4, 3, 4
Here, 4 occurred the maximum number of times than other values.
Hence, 4 is the mode of the given data.
Question 7: Define arithmetic mean and give two demerits.
Solution:
Arithmetic mean is the value obtained by dividing the sum of observations by the total number of observations of the given data.
Arithmetic mean = Sum of observations/Number of observations
Demerits of arithmetic mean:
(i) The value obtained by calculating the mean may not be realistic. For example, the average number of students is 80.5 or 15.1 and so on.
(ii) It is influenced by the extreme observations.
Question 8: Give main uses of median.
Solution:
(i) The purpose of median is to segregate the lower half from the higher half of the data.
(ii) Median is the only average which can be used while dealing with qualitative data.
(iii) This can be represented using an ogive curve hence it can be found by observing the graph without actually calculating using formula.
Question 9: Find the mean and mode of the following frequency distribution.
Score | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
Number of students | 4 | 28 | 42 | 20 | 6 |
Solution:
Arithmetic Mean:
Marks obtained | Number of students (f_{i}) | Mid-value (x_{i}) | f_{i}x_{i} |
20 – 30 | 4 | 25 | 100 |
30 – 40 | 28 | 35 | 980 |
40 – 50 | 42 | 45 | 1890 |
50 – 60 | 20 | 55 | 1100 |
60 – 70 | 6 | 65 | 390 |
∑f_{i} = 100 | ∑f_{i}x_{i} = 4460 |
Mean = ∑f_{i}x_{i}/∑f_{i}
= 4460/100
= 44.6
Therefore, the arithmetic mean of the given data is 44.6.
Mode:
From the given data,
Maximum frequency is 42 which corresponds to the class 40 – 50.
Thus, the modal class = 40 – 50
Lower limit of modal class = l = 40
Frequency of the class preceding the modal class = f_{0} = 28
Frequency of the modal class = f_{1} = 42
Frequency of the class succeeding the modal class = f_{2} = 20
Class height = h = 10
Mode = l + [(f_{1} – f_{0})/ (2f_{1} – f_{0} – f_{2})] × h
= 40 + [(42 – 28)/ (2 × 40 – 28 – 20)] × 10
= 40 + [14/(80 – 48)] × 10
= 40 + (140/32)
= 40 + 4.375
= 44.375
Therefore, the mode of the given distribution is 44.375.
Question 10: The runs scored by players of a cricket team are as follows.
57, 17, 26, 91, 115, 26, 83, 41, 57, 0, 26
Find their AM, median and mode.
Solution:
Given,
57, 17, 26, 91, 115, 26, 83, 41, 57, 0, 26
Arithmetic mean = Sum of observations/Number of observations
= Sum of runs/Number of players
= (57 + 17 + 26 + 91 + 115 + 26 + 83 + 41 + 57 + 0 + 26)/11
= 539/11
= 49
Arranging the runs in an ascending order:
0, 17, 26, 26, 26, 41, 57, 57, 83, 91, 115
Number of players = n = 11
Median = [(n + 1)/2]th observation
= [(11 + 1)/2]th observation
= 6th observation
= 41
26 runs scored by the maximum number of players.
Thus, mode = 26
Hence, AM is 49, median is 41 and mode is 26.