Frank Solutions for Chapter 22 Statistics Class 9 Mathematics ICSE
Exercise 22
1. Define primary data and secondary data.
Answer
Data collected by investigator himself through personal observations with a definite plan or design in mind is known as primary data.
The data which has been collected previously by someone, other than the investigator but is used by the
2. The marks obtained by the students in a class test are given below:
31, 12, 28, 45, 32, 16, 49, 12, 18, 26, 34, 39, 29, 28, 25, 46, 32, 13, 14, 26, 25, 34, 23, 23, 25, 45, 33, 22, 18, 37, 26, 19, 20, 30, 28, 38, 42, 21, 36, 19, 20, 40, 48, 15, 46, 26, 23, 33, 47, 40.
Arrange the above marks in classes each with a class size of 5 and answer the following:
(i) What is the highest score?
(ii) What is the lowest score?
(iii) What is the range?
(iv) If the pass mark is 20, how many students failed?
(v) How many students got 40 or more marks?
Answer
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(ii) The lowest score is 12
(iii) Range = 49 – 12 = 37
(iv) Given, pass marks is 20
So, all the students in the class 11 – 15 and 16 – 20 must have failed except for the students with score 20
Therefore,
Number of such students = 5 + 7 – 2 = 10
(v) Number of students scoring above 40 is the sum total of students in the classes 41 – 45 and 46 – 50
i.e 3 + 5 = 8
Number of students scoring exactly 40 = 2
Hence,
Number of students scoring 40 or more marks = 8 + 2 = 10
3. Find the class boundaries and class marks of the following classes:
55 – 59, 60 – 64, 65 – 69, 70 – 74, 75 – 79, 80 – 84, 85 – 89, 90 – 94 and 95 – 99
Answer
For the class 55 – 59,
The actual lower limit = 55 – 0.5 = 54.5
The actual upper limit = 59 + 0.5 = 59.5
Therefore,
The class boundaries are 54.5 and 59.5
The class mark = (1/2) (54.5 + 59.5)
We get,
The class mark = 57
Similarly calculating for the other classes we get the following table:
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Answer
Given classes are 2.1 – 4.0, 4.1 – 6.0 and 6.1 – 8.0
Since classes are inclusive,
We have,
Adjustment
Adjustment factor = 0.05
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1, 2, 1, 3, 2, 5, 1, 6, 4, 4, 2, 3, 5, 6, 4, 2, 2, 3, 4, 1, 0, 5, 0, 5, 3, 2, 3, 4, 4, 1, 1, 2, 4, 3, 1, 4
Arrange these data in a discrete frequency distribution table and answer the following:
(i) What is the range of the number of goals scored by AFC?
(ii) How many times did AFC score 3 or more than 3 goals?
(iii) Which variate has the highest frequency?
Answer
The discrete frequency distribution table is shown below
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Minimum goals scored = 0
Hence,
Range of the goals scored = 6 – 0 = 6
(ii) Number of times AFC scored 3 or more goals = 6 + 8 + 4 + 2 = 20
(iii) The variate which has highest frequency is 4
6. Prepare a cumulative frequency distribution table of the marks scored by 60 students in a test are given below:
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The cumulative frequency distribution table of the marks scored by 60 students is shown below:
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The cumulative frequency distribution table is shown below:
748, 567, 890, 231, 150, 458, 356, 762, 386, 824, 525, 663, 724, 841, 315, 641, 156, 712, 156, 317, 814, 547, 879, 456, 463, 664, 175, 584, 515, 487, 871, 511, 522, 454, 247, 819, 412, 326, 445, 311, 321, 545, 344, 266, 351.
Answer
The table for the given data is as follows:
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The frequency distribution table is shown below
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(a) Find the number of students whose weight lie in the interval 40 – 45
(b) Find the interval which has the most number of students
Answer
Frequency distribution table is shown below
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(b) The interval which has the most number of students is 45 – 50
11. For the set of numbers given below, find mean:
(i) 5, 7, 8, 4, 6
(ii) 3. 0, 5, 2, 6, 2
Answer
Mean = (Σx)/N
(i) Mean = (5 + 7 + 8 + 4 + 6)/5
Mean = (30/5)
We get,
Mean = 6
(ii) Mean = (3 + 0 + 5 + 2 + 6 + 2)/6
Mean = 18/6
We get,
Mean = 3
12. Calculate mean of the following:
4, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 11, 3
Answer
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Mean = Σfx/Σf
Mean = 98/14
We get,
Mean = 7
13. The weights of the seven members of a family, in kilograms are given below:
20, 52, 56, 72, 64, 13, 80.
Find mean weight.
Answer
Mean = (Σx/N)
Mean = (20 + 52 + 56 + 72 + 64 + 13 + 80)/7
Mean = 357/7
We get,
Mean = 51 kg
14. A boy scored the following marks in various class tests during a terminal exam, each test being marked out of 20.
17, 15, 16, 7, 10, 14, 12, 19, 16, 12
Find his average mean marks.
Answer
Mean = (Σx/N)
Average mean marks = (17 + 15 + 16 + 7 + 10 + 14 + 12 + 19 + 16 + 12)/10
Average mean marks = 138/10
We get,
Average mean marks = 13.8
15. Individual scores of a school cricket eleven in a match are given below:
10, 9, 31, 45, 0, 4, 8, 15, 12, 0, 6
Find the average score.
Answer
Mean = Σx/N
Average score = (10 + 9 + 31 + 45 + 0 + 4 + 8 + 15 + 12 + 0 + 6)/11
Average score = 140/11
We get,
Average score = 12.7
16. The daily maximum relative humidity (in percent) in Mumbai from May 1 to May 7, 1992 is given below:
64,70, 65, 80, 75,
Find the mean.
Answer
17. The marks obtained by 10 students are listed below:
2, 5, 3, 8, 0, 9, x, 6, 1, 8
If the mean marks is 5, find x.
Answer
18. The heights of 8 students of standard X in centimetres are given below:
148, 162, 160, 154, 170, 162, x, 152
If the mean height is 158, find x.
Answer
19. If the mean of 7, 16, 9, 15, 16, a, 12, 8, b, 11 is 12, write a in terms of b.
Answer
20. A test out of 25 marks was given to 16 students and marks scored are recorded below:
25, 8, 14, 20, 16, 22, 10, 15, 8, 7, 24, 18, 18, 19, 6, 11, 17
(i) Find the mean marks
(ii) In the report, marks were entered out of 50. What is the mean of the recorded marks in the report?
Answer
(i)
21. In history project, marks out of 20 were awarded to 8 students. The marks were as shown below:
14, 16, 18, 14, 16, 14, 12, 16
(i) Find the mean marks.
(ii) Each of the above students was 2 extra marks for submitting the project a week before the due date. What is the revised mean of this group?
Answer
(i) Total marks awarded to 8 students = 14 + 16 + 18 + 14 + 16 + 14 + 12 + 1 = 120
Mean marks = 120/8
= 15
(ii) When extra 2 marks are awarded, new addition total marks scored = 2×8 = 16
∴ New total marks scored = 120 + 16 = 136
∴ revised mean marks = 136/8
= 17
22. The mean of 16 natural numbers is 48. Find the resulting mean, if each of the number is
(a) increased by 5
(b) decreased by 8
(c) multiplied by 4
(d) divided by 0.25
(e) increased by 50
(f) decreased by 10%
Answer
Mean = 48
Total numbers = n = 16
Therefore,
(a) Resulting mean (when each number is increased by 5) = 48 + 5 = 53
(b) Resulting mean (when each number is decreased by 8) = 48 – 8 = 40
(c) Resulting mean (when each number is multiplied by 4) = 48×4 = 192
(d) Resulting mean (when each number is divided by 0.25) = 48 + 0.25 = 192
(e) Resulting mean (when each number is increased by 50%)
= 48 + 50% of 48
= 48 + 24
= 72
(f) Resulting mean (when each number is decreased by 10%)
= 48 – 10% of 48
= 48 – 4.8
= 43. 2
23. The mean of 4 observations is 20. If one observation is excluded, the mean of the remaining observations becomes 15. Find the excluded observation.
Answer
24. The mean monthly income of 8 men in Rs 8079.75. A man whose monthly income is Rs 8280 has also been taken into consideration. Calculate the mean monthly income of all the men.
Answer
25. The mean of 200 observations is 20. It is found that the value of 180 is wrongly copied as 280. Find the actual mean.
Answer
26. Find the median of the following sets of numbers.
(i) 15, 8, 14, 20, 13, 12, 16
(ii) 25, 11, 15, 10, 17, 6, 5, 12.
Answer
(i)
(ii)
27. Calculate the median of the following sets of number:
1, 9, 10, 8, 2, 4, 4, 3, 9, 1, 5, 6, 2 and 4.
Answer
28. 3, 8, 10, x, 14, 16, 18, 20 are in ascending order and their median is 13. Calculate the numerical value of x.
Answer
29. The following data has been arranged in ascending order.
0, 1, 2, 3, x + 1, x + 5, 20, 21, 26, 29.
Answer
30. A boy scored the following marks in various class tests during a term, each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16
(i) What is his mean marks?
(ii) What is his median marks?
Answer
(i)
