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Page No 4.49:
Question 1:
Following are the figures of marks obtained by 40 students. You ae required to arrange them in ascending and in descending order.
15 | 18 | 16 | 14 | 10 | 6 | 5 | 3 | 8 | 7 |
22 | 18 | 14 | 19 | 17 | 8 | 6 | 4 | 10 | 3 |
12 | 16 | 15 | 13 | 11 | 10 | 18 | 22 | 14 | 19 |
11 | 18 | 22 | 14 | 25 | 21 | 17 | 8 | 9 | 10 |
Answer:
Marks in Ascending Order:
3, 3, 4, 5, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 10, 11, 11, 12, 13, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 18, 18 ,19, 19, 21, 22, 22, 22, 25
Marks in Descending Order:
25, 22, 22, 22, 21, 19, 19, 18, 18, 18, 18, 17, 17, 16, 16, 15, 15, 14, 14, 14, 14, 13, 12, 11, 11, 10, 10, 10, 10, 9, 8, 8, 8, 7, 6, 6, 5, 4, 3, 3
Page No 4.49:
Question 2:
Heights (in inches) of 35 students of a class in given below. Classify the following data in a discrete frequency series
58 | 60 | 72 | 68 | 58 | 55 | 58 | 72 | 55 | 68 | 66 | 60 |
60 | 55 | 76 | 66 | 58 | 72 | 68 | 60 | 55 | 68 | 55 | 58 |
62 | 66 | 72 | 58 | 60 | 68 | 55 | 58 | 65 | 72 | 68 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Height (in inches) |
Tally Marks | Frequency |
55 58 60 62 65 66 68 72 |
$\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}\mathrm{{\rm I}}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}\mathrm{{\rm I}}}\mathrm{{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}$ $\mathrm{{\rm I}}$ $\mathrm{{\rm I}}$ $\mathrm{{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}\mathrm{{\rm I}}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}\mathrm{{\rm I}}}$ |
6 7 5 1 1 3 6 6 |
$\sum _{}f=35$ |
Page No 4.49:
Question 3:
The following are the marks of the 30 students in statistics. prepare a frequency distribution taking the class−intervals as 10−20, 20−30 and so on.
12 | 33 | 23 | 25 | 18 | 35 | 37 | 49 | 54 | 51 | 37 | 15 | 33 | 42 | 45 |
47 | 55 | 69 | 65 | 63 | 46 | 29 | 18 | 37 | 46 | 59 | 29 | 35 | 45 | 27 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Marks | Tally Marks | Number of students (Frequency) |
10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 |
$\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}$ $\mathrm{{\rm I}{\rm I}{\rm I}}$ |
4 5 7 7 4 3 |
$\sum _{}f=30$ |
Page No 4.50:
Question 4:
Following are the marks (out of 100) obtained by 50 students in statistics:
70 | 55 | 51 | 42 | 57 | 45 | 60 | 47 | 63 | 53 |
33 | 65 | 39 | 82 | 55 | 64 | 58 | 61 | 65 | 42 |
50 | 52 | 53 | 45 | 45 | 25 | 36 | 59 | 63 | 39 |
65 | 54 | 49 | 54 | 64 | 75 | 42 | 41 | 52 | 35 |
30 | 35 | 15 | 48 | 26 | 20 | 40 | 55 | 46 | 18 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Marks | Tally Marks | Number of students (Frequency) |
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 |
- $\mathrm{{\rm I}{\rm I}}$ $\mathrm{{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}$ $\mathrm{{\rm I}{\rm I}}$ $\mathrm{{\rm I}}$ |
0 2 3 7 12 14 9 2 1 |
$\sum _{}f=50$ |
Page No 4.50:
Question 5:
Prepare a frequency table taking class intervals 20−24, 25−29, 30−34 and so on, from the following data:
21 | 20 | 55 | 39 | 48 | 46 | 36 | 54 | 42 | 30 |
29 | 42 | 32 | 40 | 34 | 31 | 35 | 37 | 52 | 44 |
39 | 45 | 37 | 33 | 51 | 53 | 52 | 46 | 43 | 47 |
41 | 26 | 52 | 48 | 25 | 34 | 37 | 33 | 36 | 27 |
54 | 36 | 41 | 33 | 23 | 39 | 28 | 44 | 45 | 38 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Class Intervals | Tally Marks | Frequency |
20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 |
$\mathrm{{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\overline{)\mathrm{{\rm I}{\rm I}{\rm I}{\rm I}}}\mathrm{{\rm I}{\rm I}}$ $\mathrm{{\rm I}}$ |
3 5 8 11 8 7 7 1 |
$\sum _{}f=50$ |
Page No 4.50:
Question 6:
From the following data, calculate the lower limit of the first class and upper limit of the last class.
Daily Wages | Less than 120 | 120−140 | 140−160 | 160−180 | Above 180 |
No. of Workers | 35 | 12 | 10 | 40 | 13 |
Answer:
In the given question, the size of each class is 20. Thus, maintaining the uniformity, we take the first class as 100-120 and the last class as 180-200. Therefore, the lower limit of the first class-interval is 120 − 20 = 100 and the upper limit of the last class is 180 + 20 = 200.
Page No 4.50:
Question 7:
Calculate the missing class-intervals from the following distribution:
Class-Interval | Below 20 | 20−50 | 50−90 | 90−140 | More than 140 |
Frequency | 7 | 12 | 16 | 2 | 4 |
Answer:
In the given question, the class-intervals are of different width. Thus, the missing class-intervals cannot be determined.
Page No 4.50:
Question 8:
Convert the following 'more than' cumulative frequency distribution into a 'less than' cumulative frequency distribution
Class-Interval (More than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
Frequency | 124 | 119 | 107 | 84 | 55 | 31 | 12 | 2 |
Answer:
The 'more than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.
Class-Interval | Frequency |
10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 |
124 − 119 = 5 119 − 107 = 12 107 − 84 = 23 84 − 55 = 29 55 − 31 = 24 31 − 12 = 19 12 − 2 = 10 2 |
From the above distribution, we can make the 'less than' cumulative frequency distribution as follows.
Class-Interval | Frequency |
Less then 20 Less then 30 Less then 40 Less then 50 Less then 60 Less then 70 Less then 80 Less then 90 |
5 5 + 12 = 17 17 + 23 = 40 40 +29 = 69 69 + 24 = 93 93 + 19 = 112 112 + 10 = 122 122 + 2 = 124 |
Page No 4.50:
Question 9:
Convert the following cumulative frequency series into simple frequency series
Marks | No. of Students |
Less than 20 | 10 |
Less than 40 | 18 |
Less than 60 | 25 |
Less than 80 | 45 |
Less than 100 | 55 |
Answer:
The 'less than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.
Class-Interval | Frequency |
0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 |
10 18 − 10 = 8 25 − 18 = 7 45 − 25 = 20 55 − 45 = 10 |
Page No 4.51:
Question 10:
Prepare a frequency distribution from the following data:
Mid-Points | 25 | 35 | 45 | 55 | 65 | 75 |
Frequency | 6 | 10 | 9 | 12 | 8 | 5 |
Answer:
The class interval can be calculated from the mid-points using the following adjustment formula.
The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.
For the given data, the class interval is calculated by the following value of adjustment.
$\mathrm{Value}\mathrm{of}\mathrm{adjustment}=\frac{35-25}{2}=5$
Thus, we add and subtract 5 to each mid-point to obtain the class interval.
For instance:
The lower limit of first class = 25 – 5 = 20
Upper limit of first class = 25 + 5 = 30
Thus, the first class interval is (20 − 30). Similarly, we can calculate the remaining class intervals.Frequency Distribution
Mid Value | Class Interval | Frequency |
25 35 45 55 65 75 |
20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
6 10 9 12 8 5 |
${\sum}_{}^{}f=50$ |
Page No 4.51:
Question 11:
The ages of 20 husbands and wives are given below. Form a two-way frequency table showing the relationship between the ages of husbands and wives with the class-intervals 20−25, 25−30, etc.
S.No | Age of Husband | Age of Wife | S. No. | Age of Husband | Age of Wife |
1 | 28 | 23 | 11 | 27 | 24 |
2 | 37 | 30 | 12 | 39 | 34 |
3 | 42 | 40 | 13 | 23 | 20 |
4 | 25 | 26 | 14 | 33 | 31 |
5 | 29 | 25 | 15 | 36 | 29 |
6 | 47 | 31 | 16 | 32 | 35 |
7 | 37 | 35 | 17 | 22 | 23 |
8 | 35 | 25 | 18 | 29 | 27 |
9 | 23 | 21 | 19 | 38 | 34 |
10 | 41 | 38 | 20 | 48 | 47 |
Answer:
The given data can be presented in the form of a bivariate frequency distribution as follows.
Age of wife (Y) |
Age of husband (X) | Total | |||||
20 − 25 | 25 − 30 | 30 − 35 | 35 − 40 | 40 − 45 | 45 − 50 | ||
20 − 25 | 5 | ||||||
25 − 30 | 5 | ||||||
30 − 35 | 5 | ||||||
35 − 40 | 3 | ||||||
40 − 45 | 1 | ||||||
45 − 50 | 1 | ||||||
Total | 3 | 5 | 2 | 6 | 3 | 1 | 20 |
Page No 4.51:
Question 12:
The following data give the points scored in a tennis match by two player X and Y at the end of twenty games:
(10,12) | (2,11) | (7,9) | (15,19) | (17,21) | (12,8) | (16,10) | (14,14) | (22,18) | (16,4) |
(15,16) | (22,20) | (19,15) | (7,18) | (11,11) | (12,18) | (10,10) | (5,13) | (11,7) | (10,10) |
Answer:
The given data can be presented in the form of a bivariate frequency distribution as follows.
Player Y |
Player X | Total | ||||
0 − 4 | 5 − 9 | 10 − 14 | 15 − 19 | 20 − 24 | ||
0 − 4 | 1 | |||||
5 − 9 | 3 | |||||
10 − 14 | 8 | |||||
15 − 19 | 6 | |||||
20 − 24 | 2 | |||||
Total | 1 | 3 | 8 | 6 | 2 | 20 |
Page No 4.51:
Question 13:
In a survey, it was found that 64 families bought milk in the following quantities in a particular month. Prepare a frequency distribution with classes as 5−9, 10−14 etc.
19 | 16 | 22 | 9 | 22 | 12 | 39 | 19 | 6 | 24 | 16 | 18 | 7 | 17 | 20 | 25 |
10 | 24 | 20 | 21 | 10 | 7 | 18 | 28 | 14 | 23 | 25 | 34 | 22 | 14 | 33 | 23 |
13 | 36 | 11 | 26 | 11 | 37 | 30 | 13 | 22 | 21 | 32 | 21 | 31 | 17 | 16 | 23 |
15 | 27 | 17 | 21 | 23 | 14 | 24 | 8 | 9 | 15 | 29 | 20 | 18 | 28 | 26 | 12 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Quantity | Tally Marks | Frequency |
5 − 9 | 6 | |
10 − 14 | 11 | |
15 − 19 | 13 | |
20 − 24 | 18 | |
25 − 29 | 8 | |
30 − 34 | 5 | |
35 − 39 | 3 | |
40 − 44 | ||
${\sum}_{}^{}f=64\phantom{\rule{0ex}{0ex}}$ |
Page No 4.51:
Question 14:
You are given below a mid value series, convert it into a continuous series.
Mid-Value | 15 | 25 | 35 | 45 | 55 |
Frequency | 8 | 12 | 15 | 10 | 5 |
Answer:
The class interval can be calculated from the mid-points using the following adjustment formula.
The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.
For the given data, the class interval is calculated by the following value of adjustment.
$\mathrm{Value}\mathrm{of}\mathrm{adjustment}=\frac{25-15}{2}=5$
Thus, we add and subtract 5 to each mid-point to obtain the class interval.
For instance:
The lower limit of first class = 15 – 5 = 10
Upper limit of first class = 15 + 5 = 20
Thus, the first class interval is (10 − 20). Similarly, we can calculate the remaining class intervals.Mid value | Class Interval | Frequency |
15 25 35 45 55 |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 |
8 12 15 10 5 |
${\sum}_{}^{}f=50$ |
Page No 4.52:
Question 15:
For the data given below, prepare a frequency distribution table with classes 100−110, 110-120, etc.
125 | 108 | 112 | 126 | 110 | 132 | 136 | 130 | 140 | 155 |
120 | 130 | 136 | 138 | 125 | 111 | 112 | 125 | 140 | 148 |
147 | 137 | 145 | 150 | 142 | 135 | 137 | 132 | 165 | 154 |
Answer:
In the form of frequency distribution, the given data can be arranged as follows.
Class Interval | Telly Marks | Frequency |
100 − 110 | 1 | |
110 − 120 | 4 | |
120 − 130 | 5 | |
130 − 140 | 10 | |
140 − 150 | 6 | |
150 − 160 | 3 | |
160 − 170 | 1 | |
$\sum _{}^{}f=30$ |
Page No 4.52:
Question 16:
Prepare a bivariate frequency distribution for the following data for 20 students:
Marks in Maths | 10 | 11 | 10 | 11 | 11 | 14 | 12 | 12 | 13 | 10 |
Marks in Statistics | 20 | 21 | 22 | 21 | 23 | 23 | 22 | 21 | 24 | 23 |
Marks in Maths | 13 | 12 | 11 | 12 | 10 | 14 | 14 | 12 | 13 | 10 |
Marks in Statistics | 24 | 23 | 22 | 23 | 22 | 22 | 24 | 20 | 24 | 23 |
Answer:
The data can be presented in the form of a bivariate distribution as follows.
Marks in Statistics | Marks in Maths | |||||
10 | 11 | 12 | 13 | 14 | Total | |
20 | 2 | |||||
21 | 3 | |||||
22 | 5 | |||||
23 | 6 | |||||
24 | 4 | |||||
Total | 5 | 4 | 5 | 3 | 3 | 20 |
Page No 4.52:
Question 17:
In a school, no student has scored less than 25 marks or more than 60 marks in an examination. Their cumulative frequencies are as follows:
Less than | 60 | 55 | 50 | 45 | 40 | 35 | 30 |
Total frequency | 120 | 115 | 90 | 75 | 65 | 45 | 32 |
Answer:
Marks | Cumulative Frequency |
Less than 60 Less than 55 Less than 50 Less than 45 Less than 40 Less than 35 Less than 30 |
120 115 90 75 65 45 32 |
This can be presented in the form of a simple frequency distribution as follows.
Marks | Number of Students |
25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60 |
32 45 − 32 = 13 65 − 45 = 20 75 − 65 = 10 90 − 75 = 15 115 − 90 = 25 120 − 115 = 5 |
Page No 4.52:
Question 18:
Marks scored by 50 students are given below:
40 | 45 | 38 | 24 | 46 | 42 | 45 | 18 | 53 | 64 |
45 | 32 | 52 | 54 | 78 | 65 | 52 | 64 | 66 | 43 |
48 | 55 | 50 | 43 | 48 | 20 | 27 | 65 | 37 | 55 |
51 | 55 | 62 | 66 | 38 | 16 | 60 | 58 | 46 | 35 |
72 | 62 | 54 | 58 | 30 | 36 | 43 | 82 | 46 | 53 |
(b) Represent the marks in the form of discrete frequency distribution.
(c) Construct an inclusive frequency distribution with first class as 10−19. Also construct class boundaries.
(d) Construct a frequency distribution with exclusive class-intervals, taking the lowest class as 10−20.
(e) Convert the exclusive series constructed in (d) into 'less than' and 'more than' cumulative frequency distribution.
Answer:
(a)
Marks arranged in ascending order | |
16 18 20 24 27 30 32 35 36 37 38 38 40 42 43 43 43 45 45 45 46 46 46 48 48 |
50 51 52 52 53 53 54 54 55 55 55 58 58 60 62 62 64 64 65 65 66 66 72 78 82 |
(b)
Discrete Frequency Distribution
Marks | Frequency (f) |
16 18 20 24 27 30 32 35 36 37 38 40 42 43 45 46 48 50 51 52 53 54 55 58 60 62 64 65 66 72 78 82 |
1 1 1 1 1 1 1 1 1 1 2 1 1 3 3 3 2 1 1 2 2 2 3 2 1 2 2 2 2 1 1 1 |
$\sum _{}^{}f=50$ |
(c)
Inclusive Frequency Distribution
Class Interval (Marks) |
No. of Students |
10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 |
2 3 7 13 13 9 2 1 |
$\sum _{}^{}f=50$ |
(d)
Exclusive Frequency Distribution
Class Interval | No. of Students |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 −90 |
2 3 7 13 13 9 2 1 |
$\sum _{}^{}f=50$ |
(e)
Less than frequency distribution
Marks | Cumulative Frequency |
Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 |
2 2 + 3 = 5 5 + 7 = 12 12 + 13 = 25 25 + 13 = 38 38 + 9 = 47 47 + 2 = 49 49 + 1 = 50 |
More than frequency distribution
Marks | Cumulative Frequency |
More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 More than 80 |
50 50 − 2 = 48 48 − 3 = 45 45 − 7 = 38 38 − 13 = 25 25 − 13 = 12 12 − 9 = 3 3 − 2 =1 |
Page No 4.52:
Question 19:
From the following data of the ages of different persons, prepare less than and more than cumulative frequency distribution.
Age (in years) | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
No. of Persons | 5 | 12 | 10 | 6 | 4 | 11 | 2 |
Answer:
Age (in years) |
No. of Persons |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
5 12 10 6 4 11 2 |
$\sum _{}f=50$ |
Less than frequency distribution
Age | Cumulative frequency |
Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 |
5 5 + 12 = 17 17 + 10 = 27 27 + 6 = 33 33 + 4 = 37 37 + 11 = 48 48 + 2 = 50 |
More than frequency distribution
Age | Cumulative frequency |
More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 |
50 50 − 5 = 45 45 − 12 = 33 33 − 10 = 23 23 − 6 = 17 17 − 4 = 13 13 − 11 = 2 |
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