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Page No 9.74:
Question 1:
Find out the median.
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Marks Obtained | 10 | 12 | 14 | 17 | 18 | 20 | 21 | 30 | 32 |
Answer:
10, 12, 14, 17, 18, 20, 21 30, 32
N = 9
Thus, Median is given by the size of the 5th item. Therefore, Median of the data so given is 18.
Page No 9.74:
Question 2:
Find the value of the median from the following data: 15, 35, 48, 46, 50, 43, 55, 49.
Answer:
First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:
15, 35, 43, 46, 48, 49, 50, 55
N = 8
Since the number of items in the series is even, thus the following formula is used to calculate the median.
Thus, Median is 47.
Page No 9.75:
Question 3:
Calculate the value of median: 25, 20, 15, 45, 18, 7, 10, 64, 38, 12.
Answer:
First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:
7, 10, 12, 15, 18, 20, 25, 38, 45, 64
N = 10
Since the number of items in the series is even, thus the following formula is used to calculate the median.
Thus, Median is 19.
Page No 9.75:
Question 4:
From the following series of marks obtained by 10 candidates in an examination. Find the median: 26, 14, 30, 18, 11, 35, 41, 12, 32.
Answer:
First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:
11, 12, 14, 18, 22, 26, 30, 32, 35, 41
N = 10
Since the number of items in the series is even, thus the following formula is used to calculate the median.
Thus, Median is 24.
Page No 9.75:
Question 5:
Calculate the value of median from the following data:
Income (â¹) | 12,00 | 1,800 | 5,000 | 2,500 | 3,000 | 1,600 | 3,500 |
No. of Persons | 12 | 16 | 2 | 10 | 3 | 15 | 7 |
Answer:
Income | Number of Person (f) |
Cumulative Frequency (c.f.) |
1200 1600 |
12 15 |
12 27 |
1800 | 16 | 43 |
2500 3000 3500 5000 |
10 3 7 2 |
53 56 63 65 |
N=65 |
Median = size of item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 33rd is 43 (in the c.f. column), which is corresponding to 1800. â Hence, median is 1800.
Page No 9.75:
Question 6:
Find out the nedian size from the following:
X | 160 | 150 | 152 | 161 | 156 |
f | 5 | 8 | 6 | 3 | 7 |
Answer:
X | f | Cumulative Frequency (cf) |
150 152 |
8 6 |
8 14 |
156 | 7 | 21 |
160 161 |
5 3 |
26 29 |
N=29 |
Median = size of item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 15th is 21 (in the c.f. column), which is corresponding to 156. â Hence, median is 156.
Page No 9.75:
Question 7:
Find out the median size from the following:
Size | 10−20 | 20−30 | 30−40 | 40−50 |
Frequency | 42 | 25 | 58 | 40 |
Answer:
Size | Frequency (f) |
Cumulative Frequency (c.f.) |
10 − 20 20 − 30 |
42 25 |
42 67 |
30 − 40 | 58 (f) | 125 |
40 − 50 | 40 | 165 |
Median class is given by the size of item, i.e. item, which is 82.5th item.
This corresponds to the class interval of 30 − 40, so this is the median class.
Page No 9.75:
Question 8:
Find the median of the following data:
Age | 20−25 | 25−30 | 30−35 | 35−40 | 40−45 | 45−50 | 50−55 | 55−60 |
No. Persons | 50 | 70 | 100 | 180 | 150 | 120 | 70 | 60 |
Answer:
Age | Frequency (f) | Cumulative Frequency (c.f.) |
20 − 25 25 − 30 30 − 35 |
50 70 100 |
50 120 220 (c.f.) |
35 − 40 | 180 (f) | 400 |
40 − 45 45 − 50 50 − 55 55 − 60 |
150 120 70 60 |
550 670 740 800 |
Median class is given by the size of item, i.e. item, which is 400th item.
This corresponds to the class interval of 35 − 40, so this is the median class.
Page No 9.75:
Question 9:
Calculate the median from the following data:
Marks | Frequency |
Less than 10 | 5 |
Less than 20 | 20 |
Less than 30 | 45 |
Less than 40 | 80 |
Less than 50 | 100 |
Less than 60 | 115 |
Less than 70 | 125 |
Less than 80 | 130 |
Answer:
Marks | Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
Less than 10 Less than 20 Less than 30 |
0 − 10 10 − 20 20 − 30 |
5 15 25 |
5 20 45 c.f |
Less than 40 | 30 − 40 | (f) 35 | 80 |
Less than 50 Less than 60 Less than 70 Less than 80 |
40 − 50 50 − 60 60 − 70 70 − 80 |
20 15 10 5 |
100 115 125 130 |
Median class is given by the size of item, i.e. item, which is 65th item.
This corresponds to the class interval of 30 − 40, so this is the median class.
Page No 9.76:
Question 10:
Find the median form the following:
Marks (More than) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of Students | 100 | 92 | 78 | 44 | 32 | 18 | 15 | 13 |
Answer:
Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
0 − 10 10 − 20 |
8 14 |
8 22 (c.f.) |
20 − 30 | 34 (f) | 56 |
30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
12 14 3 2 13 |
68 82 85 87 100 |
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of 20 − 30, so this is the median class.
Page No 9.76:
Question 11:
Compute median from the following data:
Mid-values | 37.5 | 42.5 | 47.5 | 52.5 | 57.5 |
No. of Students | 30 | 20 | 15 | 13 | 22 |
Answer:
For the calculation of median, the given mid-values must be converted into class intervals using the following 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. In this manner, we obtain the following distribution.
Mid Values | Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
37.5 42.5 |
35 − 40 40 − 45 |
30 20(f) |
30(c.f.) 50 |
47.5 | 45 − 50 | 15 | 65 |
52.5 57.5 |
50 − 55 55 − 60 |
13 22 |
78 100 |
N = ∑f =100 |
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of (40 − 45), so this is the median class.
Page No 9.76:
Question 12:
Calculate median from the following figures:
Class-intervals | 10−29 | 30−39 | 40−49 | 50−59 | 59−60 | 60−69 |
Frequencies | 12 | 19 | 20 | 21 | 15 | 13 |
Answer:
Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
Inclusive Class Interval |
Exclusive Class Interval |
Frequency (f) |
Cumulative Frequency (c.f.) |
10 − 19 20 − 29 |
9.5 − 19.5 19.5 − 29.5 |
12 19 |
12 31 (c.f.) |
30 − 39 | 29.5 − 39.5 | 20 (f) | 51 |
40 − 49 50 − 59 60 − 69 |
39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 |
21 15 13 |
72 87 100 |
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of (29.5 − 39.5), so this is the median class.
Page No 9.76:
Question 13:
Calculate median from following data:
X | Below 10 | 10−20 | 20−30 | 30−40 | 40−50 | 50 and Above |
Frequency | 3 | 7 | 15 | 9 | 6 | 4 |
Answer:
X | Frequency (f) |
Cumulative Frequency (c.f.) |
0 − 10 10 − 20 |
3 7 |
3 10 (c.f.) |
20 − 30 | 15 (f) | 25 |
30 − 40 40 − 50 50 − 60 |
9 6 4 |
34 40 44 |
Median class is given by the size of item, i.e. item, which is 22nd item.
This corresponds to the class interval of (20 − 30), so this is the median class.
Page No 9.76:
Question 14:
From the following data, compute median:
Marks | 0−10 | 10−20 | 20−40 | 40−60 | 60−80 | 80−100 |
No. of Students | 8 | 10 | 22 | 25 | 10 | 5 |
Answer:
Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
0 − 10 10 − 20 |
8 10 |
8 18 (c.f.) |
20 − 40 | 22 (f) | 40 |
40 − 60 60 − 80 80 − 100 |
25 10 5 |
65 75 80 |
N = ∑f =80 |
Median class is given by the size of item, i.e. item, which is 40th item.
This corresponds to the class interval of (20 − 40), so this is the median class.
Page No 9.76:
Question 15:
An Incomplete distribution is given below:
Variables | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 | Total |
Frequency | 12 | 30 | ? | 65 | ? | 25 | 18 | 229 |
Answer:
Given, Median = 46
Let the missing frequencies be f1 & f2.
Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
10 − 20 20 − 30 30 − 40 |
12 30 f1 |
12 42 42 + f1 (c.f.) |
40 − 50 | 65(f) | 107 + f1 |
50 − 60 60 − 70 70 − 80 |
f2 25 18 |
107 + f1 + f2 132 + f1 + f2 229 |
N = ∑f =229 |
Median class is given by the size of item, i.e. item, which is 114.5th item.
This corresponds to the class interval of (40 − 50) as median is 46.
132 + f1 + f2 + 18 = 229
or, f2 = 229 − 132 − 18 − 33.5
f2 = 45.5
Therefore, f1 is 33.5 and f2 is 45.5.
Page No 9.76:
Question 16:
Find the missing frequency form the following distribution, if median is 35 and N = 170.
Variables | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 10 | 20 | ? | 40 | ? | 25 | 15 |
Answer:
Given, Median = 35
N=170
Let the missing frequencies be f1 & f2.
Class Interval | Frequency (f) |
Cumulative Frequency |
0 − 10 10 − 20 20 − 30 |
10 20 f1 |
10 30 30 + f1 (c.f.) |
30 − 40 | 40 (f) | 70 + f1 |
40 − 50 50 − 60 60 − 70 |
f2 25 15 |
70 + f1 + f2 95 + f1 + f2 110 + f1 + f2 |
N = ∑f =170 |
Median class is given by the size of item, i.e. item, which is 85th item.
This corresponds to the class interval of (30 − 40) as median is 35.
110 + f1 + f2 = 170
or, f2 = 170 − 110 − 35
f2 = 25
Therefore, f1 is 35 and f2 is 25.
Page No 9.77:
Question 17:
Determine the value of median from the following data with the help of: (i) 'less than' and 'More than' Ogive Method; (ii) 'Less than' Ogive method; (iii) 'More than' Ogive method.
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
No. of Students | 10 | 15 | 25 | 30 | 10 | 10 |
Answer:
Marks (Class Interval) |
Students (f) |
Less than c.f. |
More than c.f. |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 |
10 15 25 30 10 10 |
10 25 50 80 90 100 |
100 90 75 50 20 10 |
From the point of intersection, a perpendicular is drawn on the x-axis. This line cuts the x-axis at 30. Hence, the median is 30 marks
Median = 30 marks
Page No 9.77:
Question 18:
From the following, determine lower quartile and upper quartile:
X | 12 | 16 | 17 | 21 | 28 | 19 | 30 | 32 |
Answer:
12, 16, 17, 19, 21, 28, 30, 32
N = 8
Hence, lower and upper quartile are 16.25 and 29.5 respectively.
Page No 9.77:
Question 19:
Calculate lower and upper quartiles.
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Marks | 18 | 20 | 25 | 17 | 9 | 11 | 23 | 37 | 38 | 42 |
Answer:
9, 11, 17, 18, 20, 23, 25, 37, 38, 42
N = 10
Hence, lower and upper quartile are 15.5 and 37.25 respectively.
Page No 9.77:
Question 20:
Calculate the median, lower quartile and upper quartile from the following data:
Marks | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |
No. of Students | 2 | 3 | 6 | 15 | 10 | 5 | 4 | 3 | 1 |
Answer:
Marks | Number of students (f) |
Cumulative Frequency (c.f.) |
58 59 60 61 62 63 64 65 66 |
2 3 6 15 10 5 4 3 1 |
2 5 11 26 36 41 45 48 49 |
N=49 |
Median
Median = size of item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th is 26 (in the c.f. column), which is corresponding to 61. â
Hence, median is 61.
Lower Quartile
This corresponds to 26 in the cumulative frequency.
Hence, lower quartile is 61 marks.
Upper Quartile
This corresponds to 41 in the cumulative frequency.
Hence, upper quartile is 63 marks.
Page No 9.77:
Question 21:
From the following series, calculate lower quartile and upper quartile.
Variable | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
Frequency | 16 | 18 | 22 | 21 | 24 | 14 | 11 | 9 |
Answer:
Variable | Frequency (f) |
Cumulative Frequency (c.f.) |
5 | 16 | 16 |
10 | 18 | 34 |
15 | 22 | 56 |
20 | 21 | 77 |
25 | 24 | 101 |
30 | 14 | 115 |
35 | 11 | 126 |
40 | 9 | 135 |
N=135 |
Lower Quartile
This corresponds to 34 in the cumulative frequency.
Hence, lower quartile is 10.
Upper Quartile
This corresponds to 115 in the cumulative frequency.
Hence, upper quartile is 30.
Page No 9.77:
Question 22:
Calculate upper and lower quartile from the following data:
Variable | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 10 | 20 | 35 | 40 | 25 | 25 | 15 |
Answer:
Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) | |
0 − 10 10 − 20 |
10 20 |
10 30 |
|
20 − 30 | 35 | 65 | Q1 |
30 − 40 | 40 | 105 | |
40 − 50 | 25 | 130 | Q3 |
50 − 60 60 − 70 |
25 15 |
155 170 |
|
N=170 |
This lies in the class interval (20-30).
Now,
Page No 9.77:
Question 23:
Age of 11 students of class XI is given below. Find the modal age by: (i) Observation Method; (ii) Frequency distribution Method.
Age (in years) | 15 | 16 | 16 | 17 | 15 | 16 | 18 | 17 | 15 | 17 | 17 |
Answer:
Age (X) |
Tally Marks |
Frequency (f) |
|
15 16 |
3 3 |
||
17 | 4 | Modal Class | |
18 | 1 | ||
N=11 |
Since 17 occurs the highest number of times in the series i.e. 4 times.
Hence, Mode = 17
Page No 9.78:
Question 24:
Find modal item of the followings set of numbers:
2 | 5 | 2 | 3 | 5 | 5 | 6 | 4 | 5 | 3 | 5 | 2 | 5 | 7 | 1 |
Answer:
X | Tally Marks |
Frequency (f) |
|
1 2 3 4 |
1 3 2 1 |
||
5 | 6 | →Modal Class | |
6 7 |
1 1 |
||
N=15 |
Since 5 occurs the highest number of times in the series i.e. 6 times.
Hence, Mode = 5
Page No 9.78:
Question 25:
From the followind data, calculate the value of mode.
Salary (in â¹) | 2,000 | 2,100 | 2,400 | 2,900 | 3,100 | 3,300 |
No. of Workers | 3 | 5 | 10 | 19 | 8 | 4 |
Answer:
Salary (X) |
Frequency (f) |
|
2000 2100 2400 |
3 5 10 |
|
2900 | 19 | Modal Class |
3100 3300 |
8 4 |
The data given in the question is a discrete series. Therefore, using the inspection method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner, 2900 is regarded as the modal value, as it has the highest frequency (of 19 times).
Therefore, mode (Z) is 2900.
Page No 9.78:
Question 26:
Compute the mode from the following by: (i) Observation Method; (ii) Grouping Method.
Hight (in inches) | 60 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 |
No. of Persons | 5 | 13 | 18 | 20 | 21 | 30 | 23 | 12 | 4 | 2 |
Answer:
(i) Observation Method
Height (X) |
No. of persons (f) |
|
60 62 63 64 65 |
5 13 18 20 21 |
|
66 | 30 | Modal Class |
67 68 69 70 |
23 12 4 2 |
Using the observation method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner,66 is regarded as the modal value, as it has the highest frequency (of 30 times).
Therefore, mode (Z) is 66 inches.
(ii) Grouping Method
For the given distribution, the grouping table is as follows.
On the basis of this grouping table, an analysis table is prepared. For each column of the grouping table, we analyse which item/group of items correspond to the highest frequency.
Analysis Table
Column | 60 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 |
I II III IV V VI VII |
â | â â â â |
â â â â â â â |
â â â â |
â â |
|||||
1 | 4 | 7 | 4 | 2 |
From the analysis table, it is clear that 66 repeats the maximum number of times. Thus, mode is 66.
Page No 9.78:
Question 27:
Find out mode from the following data:
Class-Interval | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
Frequency | 5 | 7 | 15 | 18 | 16 | 9 | 6 | 3 |
Answer:
Class Interval | Frequency (f) | |
0 − 5 5 − 10 10 − 15 |
5 7 15 |
f0 |
15 − 20 | 18 | f1 |
20 − 25 25 − 30 30 − 35 35 − 40 |
16 9 6 3 |
f2 |
By inspection, we can say that the modal class is (15 – 20) as it has the highest frequency of 18.
Page No 9.78:
Question 28:
Calculate the mode from the following data:
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
No. of Students | 8 | 11 | 9 | 25 | 12 | 16 |
Answer:
Class Interval | Frequency (f) |
|
10 − 20 20 − 30 30 − 40 |
8 11 9 |
f0 |
40 − 50 | 25 | f1 |
50 − 60 60 − 70 |
12 16 |
f2 |
By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 25.
Page No 9.78:
Question 29:
Calculate the mode from the following data:
Marks(below) | 15 | 30 | 45 | 60 | 75 | 90 |
No. of Students | 10 | 30 | 60 | 84 | 90 | 100 |
Answer:
Class Interval | Number of students (c.f) |
Frequency (f) |
0 − 15 15 − 30 |
10 30 |
10 30 − 10 = 20 (f0) |
30 − 45 | 60 | 60 − 30 = 30 (f1) |
45 − 60 60 − 75 75 − 90 |
84 90 100 |
84 − 60 = 24 (f2) 90 − 84 = 6 100− 90 = 10 |
By inspection, we can say that the modal class is (30 – 45) as it has the highest frequency of 30.
Page No 9.78:
Question 30:
Find out mode of the following frequency distribution:
Marks (More than) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
No. of Students | 40 | 38 | 33 | 25 | 15 | 7 | 5 | 2 | 0 |
Answer:
Class Interval | Number of students (c.f.) |
Frequency (f) |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 |
40 38 33 25 15 7 5 2 0 |
2(=40-38) 5(=38-33) 8(=33-25) (f0) 10(=25-15) (f1) 8(=15-7) (f2) 2(=7-5) 3(=5-2) 2(=2-0) 0 |
By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 10.
Page No 9.78:
Question 31:
Find out the mode value from the following data:
Mid-value | 15 | 25 | 35 | 45 | 55 | 65 | 75 | 85 |
Frequency | 5 | 8 | 12 | 16 | 28 | 15 | 3 | 2 |
Answer:
Mid Value | Class Interval | Frequency (f) |
15 25 35 45 55 65 75 85 |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 |
5 8 12 16 f0 28 f1 15 f2 3 2 |
By inspection, we can say that the modal class is (50 – 60) as it has the highest frequency of 28.
Page No 9.79:
Question 32:
Find out the modal value from the following data:
Marks | 0−9 | 10−19 | 20−29 | 30−39 | 40−49 | 50−59 |
No. of Students | 3 | 7 | 15 | 25 | 10 | 4 |
Answer:
Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
Inclusive Class Interval |
Exclusive Class Interval |
Frequency (f) |
0 − 9 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 |
-0.5 − 9.5 9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 |
3 7 15 f0 25 f1 10 f2 4 |
By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.
Hence, mode is 33.5.
Page No 9.79:
Question 33:
Calculate mode of the following data:
X | Below 5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 | Above 40 |
Y | 20 | 24 | 32 | 28 | 20 | 16 | 34 | 10 | 8 |
Answer:
X | Frequency (f) |
Below 5 5 − 10 |
20 24 f0 |
10 − 15 | 32 f1 |
15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 above 40 |
28 f2 20 16 34 10 8 |
By inspection, we can say that the modal class is (10 – 15) as it has the highest frequency of 32.
Page No 9.79:
Question 34:
Variable | 0−10 | 10−20 | 20−40 | 40−50 | 50−70 |
No. of Students | 5 | 12 | 40 | 32 | 28 |
Answer:
In the given question, class intervals are not equal. For the calculation of mode, first the class intervals are made equal and frequencies are adjusted assuming that frequencies are equally distributed. In this manner, we get the following distribution.
Class Interval | Frequency (f) |
0 − 10 10 − 20 20 − 30 30 − 40 |
5 12 20 20 f0 |
40 − 50 | 32 f1 |
50 − 60 60 − 70 |
14 f2 14 |
By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 32.
Page No 9.79:
Question 35:
Compute graphically, the modal value of the given data:
Age | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 |
No. of Persons | 2 | 5 | 7 | 5 | 2 |
Answer:
Hence, the mode of the series is 25 years.
Page No 9.79:
Question 36:
The monthly profits in rupees (in thousands) of 100 shops are given below:
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
No. of Students | 3 | 5 | 9 | 3 | 2 |
Answer:
Hence, the mode of the series is 34 marks.
Page No 9.79:
Question 37:
Find lower quartile, median and upper quartile from the data given below
Class-interval (More than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Frequency | 100 | 99 | 96 | 85 | 64 | 31 | 9 |
Answer:
Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
10 − 20 20 − 30 30 − 40 40 − 50 |
1 3 11 21 |
1 4 15 36 |
50 − 60 | 33 (f) | 69 |
60 − 70 70 − 80 |
22 9 |
91 100 |
Lower Quartile
This lies in the class interval (40-50).
Now,
Thus, the value of lower quartile is 44.76.
Median
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of 50 − 60, so this is the median class.Uppe
Upper Quartile
This lies in the class interval (60-70).
Thus, the value of upper quartile is 62.72.
Page No 9.79:
Question 38:
From the following data of the ages of different persons, determine the modal age.
Age (in years) | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
No. of Persons | 4 | 26 | 32 | 10 | 9 | 6 | 3 |
Answer:
Age(in years) (X) |
Number of Persons (f) |
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
4 26 f0 32 f1 10 f2 9 6 3 |
By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 32.
Page No 9.80:
Question 39:
Find mean, median, and mode for the following data
Classes | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
Frequencies | 4 | 15 | 10 | 7 | 3 | 1 |
Answer:
Class Interval | Mid Values (m) |
Frequency (f) |
Cumulative Frequency (c.f.) |
fm | |
0 − 10 | 5 | 4(f0) | 4 | 20 | |
10 − 20 | 15 | 15(f1) | 19 | 225 | → Modal Class |
20 − 30 | 25 | 10(f2) | 29 | 250 | |
30 − 40 40 − 50 50 − 60 |
35 45 55 |
7 3 1 |
36 39 40 |
245 135 55 |
|
N = ∑f=40 | ∑fm =930 |
Median class is given by the size of item, i.e. item, which is 20th item.
This corresponds to the class interval of (20 − 30), so this is the median class.
Mode
By inspection, we can say that the modal class is (10 – 20) as it has the highest frequency of 15.
Page No 9.80:
Question 40:
Calculate median from the following:
Marks (less than) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
No. of Students | 3 | 11 | 16 | 32 | 58 | 70 | 76 | 79 | 80 |
Answer:
Class Interval | Cumulative Frequency (c.f.) |
Frequency (f) |
0 − 5 5 − 10 10 − 15 15 − 20 |
3 11 16 32 (c.f.) |
3 8 5 16 |
20 − 25 | 58 | 26 (f) |
25 − 30 30 − 35 35 − 40 40 − 45 |
70 76 79 80 |
12 6 3 1 |
N=80 |
Median class is given by the size of item, i.e. item, which is 40th item.
This corresponds to the class interval of 20 − 25, so this is the median class.
Page No 9.80:
Question 41:
Calculate the mean, the mode and the median from the following frequency distribution.
Height (in inches) | 60−61 | 61−62 | 62−63 | 63−64 | 64−65 | 65−66 | 66−67 | 67−68 | 68−69 |
Frequency | 4 | 16 | 8 | 24 | 35 | 18 | 19 | 16 | 10 |
Answer:
Height | Mid Values (m) |
Frequency (f) |
Cumulative Frequency (c.f.) |
fm | |
60 − 61 61 − 62 62 − 63 63 − 64 |
60.5 61.5 62.5 63.5 |
4 16 8 24 (f0) |
4 20 28 52 (c.f.) |
242 984 500 1524 |
|
64 − 65 | 64.5 | 35 (f1) | 87 | 2257.5 | Modal Class |
65 − 66 66 − 67 67 − 68 68 − 69 |
65.5 66.5 67.5 68.5 |
18 (f2) 19 16 10 |
105 124 140 150 |
1179 1263.5 1080 685 |
|
N = ∑f=150 | ∑fm =9715 |
Median class is given by the size of item, i.e. item, which is 75th item.
This corresponds to the class interval of (64 − 65), so this is the median class.
Mode
By inspection, we can say that the modal class is (64 – 65) as it has the highest frequency of 35.
Page No 9.80:
Question 42:
In the frequency distribution of 100 students gives below. The number of students corresponding to marks groups 10-20 and 30-40 are missing from the table. However, the median is known to be 23. Find the missing frequencies
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 |
No. of Students | 8 | ? | 40 | ? | 10 |
Answer:
Given, Median = 23
N=100
Let the missing frequencies be f1 and f2.
Marks | Frequency (f) | Cumulative Frequency (cf) |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 |
8 f1 40 f2 10 |
8 8+f1 48+f1 48+f1+f2 58+f1+f2 |
N = ∑f =100 |
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of (20 − 30) as median is 23.
58 + f1 + f2 = 100
or, f2 = 100 − 58 − 30
f2 = 12
Therefore, f1 is 30 and f2 is 12.
Page No 9.80:
Question 43:
Calculate the value of median and mode from the following distribution.
Mid-point | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |
Frequency | 8 | 10 | 15 | 25 | 40 | 20 | 15 | 7 |
Answer:
For the calculation of median, the given mid-values must be converted into class intervals using the following 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. In this manner, we obtain the following distribution.
Mid Points | Class Interval | Frequency (f) |
Cumulative Frequency (c.f.) |
20 30 40 50 |
15 − 25 25 − 35 35 − 45 45 − 55 |
8 10 15 25 (f0) |
8 18 33 58 (c.f.) |
60 | 55 − 65 | 40 (f1) | 98 |
70 80 90 |
65 − 75 75 − 85 85 − 95 |
20 (f2) 15 7 |
118 133 140 |
N = ∑f =140 |
Median
Median class is given by the size of item, i.e. item, which is 70th item.
This corresponds to the class interval of (55 − 65), so this is the median class.
Mode
By inspection, we can say that the modal class is (55 − 65) as it has the highest frequency of 40.
Page No 9.80:
Question 44:
From the following data, find the value of mode.
Size | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
Frequency | 2 | 3 | 5 | 7 | 14 | 12 | 8 | 7 | 3 |
Answer:
Size | Frequency | |
5 10 15 20 |
2 3 5 7 |
|
25 | 14 | Modal Class |
30 35 40 45 |
12 8 7 3 |
The data given in the question is a discrete series. Therefore, using the inspection method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner, 25 is regarded as the modal value, as it has the highest frequency (of 14 times).
Therefore, mode (Z) is 25.
Page No 9.80:
Question 45:
Calculate the mean, median and mode of the following data:
Monthly Profit | Frequency |
Less than 10 | 4 |
Less than 20 | 20 |
Less than 30 | 35 |
Less than 40 | 55 |
Less than 50 | 62 |
Less than 60 | 67 |
Answer:
Class Interval | Cumulative Frequency (c.f.) |
Frequency (f) |
Mid Values (m) | fm | |
0 − 10 10 − 20 |
4 20 (c.f) |
4 16 |
5 15 |
20 240 |
|
20 − 30 | 35 | 15 (f0) | 25 | 375 | |
30 − 40 | 55 | 20 (f1) | 35 | 700 | Modal class |
40 − 50 50 − 60 |
62 67 |
7 (f2) 5 |
45 55 |
315 275 |
|
N = ∑f=67 | ∑fm =1925 |
Median class is given by the size of item, i.e. item, which is 33.5th item.
This corresponds to the class interval of (20 − 30), so this is the median class.
Mode
By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 20.
Page No 9.81:
Question 46:
Find the median of the following data:
Valuable | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 7 | 12 | 18 | 25 | 16 | 14 | 8 |
Answer:
Variable | Frequency (f) | Cumulative Frequency (c.f.) |
0 − 10 10 − 20 20 − 30 |
7 12 18 |
7 19 37 (c.f.) |
30 − 40 | 25 (f) | 62 |
40 − 50 50 − 60 60 − 70 |
16 14 8 |
78 92 100 |
N=100 |
Median class is given by the size of item, i.e. item, which is 50th item.
This corresponds to the class interval of (30 − 40), so this is the median class.
Page No 9.81:
Question 47:
Find out mode from the following data:
Size | 0−4 | 4−6 | 6−10 | 10−15 | 15−20 | 20−30 | 30−35 | 35−40 |
Frequency | 2 | 4 | 3 | 5 | 2 | 20 | 6 | 8 |
Answer:
In the given question, class intervals are not equal. For the calculation of mode, first the class intervals are made equal and frequencies are adjusted assuming that frequencies are equally distributed. In this manner, we get the following distribution.
Class Interval | Frequency (f) |
0 − 10 10 − 20 20 − 30 30 − 40 |
9 7 (f0) 20 (f1) 14 (f2) |
By inspection, we can say that the modal class is (20 – 30) as it has the highest frequency of 20.
Page No 9.81:
Question 48:
Calculatate the Median and Mode form the following data:
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
No. of Students | 2 | 18 | 30 | 45 | 35 | 20 | 6 | 3 |
Answer:
Marks |
Number of Students (f) |
Cumulative Frequency (c.f.) |
0 − 10 10 − 20 20 − 30 |
2 18 30 (f0) |
2 20 50 (c.f) |
30 − 40 | 45 (f1) | 95 |
40 − 50 50 − 60 60 − 70 70 − 80 |
35 (f2) 20 6 3 |
130 150 156 159 |
N=159 |
Median
Median class is given by the size of item, i.e. item, which is 79.5th item.
This corresponds to the class interval of (30 − 40), so this is the median class.
Mode
By inspection, we can say that the modal class is (30 − 40) as it has the highest frequency of 45.
Page No 9.81:
Question 49:
Find out the median of the following frequency distribution:
Marks | 50 | 55 | 60 | 65 | 70 | 75 |
No. of Students | 5 | 7 | 6 | 10 | 5 | 8 |
Answer:
Marks |
Number of Students (f) |
Cumulative Frequency (c.f.) |
50 55 60 65 70 75 |
5 7 6 10 5 8 |
5 12 18 28 33 41 |
N= 41 |
Median = size of item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 21st is 28 (in the c.f. column), which is corresponding to 65. âHence, median is 65.
Page No 9.81:
Question 50:
Calculate mode from the following data:
Income | 15−24 | 25−34 | 35−44 | 45−54 | 55−64 | 65−74 |
No. of Workers | 8 | 10 | 15 | 25 | 40 | 20 |
Answer:
Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
Inclusive Class Interval |
Exclusive Class Interval |
Frequency (f) |
0 − 9 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 |
-0.5 − 9.5 9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 |
3 7 15 f0 25 f1 10 f2 4 |
By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.
Hence, mode is 33.5.
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