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Page No 208:
Question 1:
Calculate median of the following data:
145 | 130 | 200 | 210 | 198 | 234 | 159 | 160 | 178 |
257 | 260 | 300 | 345 | 360 | 390 |
Answer:
Arranging the data in ascending order.
130, 145, 159, 160, 178, 198, 200, 210, 234, 257, 260, 300, 345, 360, 390
Page No 208:
Question 2:
Find out median of the following information:
Marks | : | 10, | 70, | 50, | 20, | 95, | 55, | 42, | 60, | 48, | 80 |
Answer:
Arranging the data in ascending order.
10, 20, 42, 48, 50, 55, 60, 70, 80, 95
Here, the number of observations is even.
Thus, median marks is 52.5.
Page No 208:
Question 3:
We have the following frequency distribution of the size of 51 households . Calculate the arithmetic mean and the median.
Size | 2 | 3 | 4 | 5 | 6 | 7 | Total |
Number of households | 2 | 3 | 9 | 21 | 11 | 5 | 51 |
Answer:
Size (X) |
Households (f) |
fX |
c.f |
|
2 |
2 |
4 |
2 |
|
3 |
3 |
9 |
5 |
|
4 |
9 |
36 |
14 |
|
5 |
21 |
105 |
35 |
|
6 |
11 |
66 |
46 |
|
7 |
5 |
35 |
51 |
|
|
Æ©f = 51 |
Æ©fx = 255 |
|
|
Now, we need to look at the column of cumulative frequency. The item just exceeding the 26thitem is 35, which corresponds to 5.
Hence, the median is 5.
Page No 208:
Question 4:
Find out median:
(a)
Serial No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Values | 2 | 4 | 10 | 8 | 15 | 20 | 12 | 25 | 30 |
X | 5 | 10 | 15 | 20 | 25 |
f | 2 | 4 | 10 | 8 | 15 |
Answer:
(a) For the calculation of median the given values are first arranged in ascending order as follows.
Serial Number | Values | Ascending Order |
1 2 3 4 5 6 7 8 9 |
2 4 10 8 15 20 12 25 30 |
2 4 8 10 12 15 20 25 30 |
Thus, median is 12.
(b)
X |
f |
c.f |
|
5 |
2 |
2 |
|
10 |
4 |
6 |
|
15 |
6 |
12 |
|
20 |
8 |
20 |
|
25 |
10 |
30 |
|
Now, we need to look at the column of cumulative frequency. The item just exceeding the 15.5th item is 20, which corresponds to 20.
Thus, median is 20.
Page No 208:
Question 5:
Find out median , first quartile and third quartile of the following series:
Height (in inches) | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |
No of Persons | 2 | 3 | 6 | 15 | 10 | 5 | 4 | 3 | 1 |
Answer:
Height (X) |
Persons (f) |
c.f |
|
58 |
2 |
2 |
|
59 |
3 |
5 |
|
60 |
6 |
11 |
|
61 |
15 |
26 |
|
62 |
10 |
36 |
|
63 |
5 |
41 |
|
64 |
4 |
45 |
|
65 |
3 |
48 |
|
66 |
1 |
49 |
|
(i)Median
This corresponds to height 61. Thus, the median is 61.
(ii) First Quartile
This corresponds to height 61. Thus, first quartile is 61.
(iii) Third Quartile
This corresponds to height 63. Thus, third quartile is 63.
Page No 209:
Question 6:
The marks obtained by 68 students in an examination are given below. Compute the median.
Marks | Below 20 | 20-40 | 40-60 | 60-80 | Above 80 |
No of students | 0 | 5 | 22 | 25 | 16 |
Answer:
Marks |
Students (f) |
c.f |
|
Below 20 |
0 |
0 |
|
20 – 40 |
5 |
5 |
|
40 – 60 |
22 |
27 |
|
60 –80 |
25 |
52 |
|
Above 80 |
16 |
68 |
|
Median class is given by the size of the item, i.e. the item, which is the 34th item. This corresponds to the class interval 60−80, so this is the median class.
Thus, median is 65.6.
Page No 209:
Question 7:
Calcualte the mean of the following distribution of daily wages of workers in a factory:
Daily Wages (in â¹) | : | 100-120 | 140-160 | 160-180 | 180-200 | Total |
No of Workers | : | 10 | 30 | 15 | 5 | 80 |
Also, calculate the median for the distribution of wages given above.
Answer:
Wages
|
Worker (f) |
X |
fX |
c.f |
100 – 120 |
10 |
110 |
1100 |
10 |
120-140 | 20 | 130 | 2600 | 30 |
140 – 160 |
30 |
150 |
4500 |
60 |
160 – 180 |
15 |
170 |
2550 |
75 |
180 – 200 |
5 |
190 |
950 |
80 |
|
80 |
|
Æ© fX = 11700 |
|
Note: In the above calculation we have assumed a class interval (120-140) and its corresponding frequency as 20.
Page No 209:
Question 8:
The following table gives the marks obtained by 65 students in statistics in a certain examination . Calculate the median.
Marks | No of Students |
More than 70 | 8 |
More than 60 | 18 |
More than 50 | 40 |
More than 40 | 45 |
More than 30 | 50 |
More than 20 | 63 |
More than 10 | 65 |
Answer:
Marks | Students c.f |
More than 70 More than 60 More than 50 More than 40 More than 30 More than 20 More than 10 |
8 18 40 45 50 63 65 |
Now the more than cumulative frequency distribution can be converted into class intervals as follows.
C.I. |
f |
c.f |
|
10 – 20 |
2 |
2 |
|
20 – 30 |
13 |
15 |
|
30 – 40 |
5 |
20 |
|
40 – 50 |
5 |
25 |
|
50 – 60 |
22 |
47 |
|
60 – 70 |
10 |
57 |
|
70 and above |
8 |
65 |
|
|
Æ©f = 65 |
|
|
Median class is given by the size of the item, i.e. the item, which is the 32.5th item. This corresponds to the class interval 50−60, so this is the median class.
Thus, the median marks is 53.4.
Page No 209:
Question 9:
Calculate the Arithmetic mean and median of the following frequency distribution:
Class Interval | 10-20 | 20-40 | 40-70 | 70-120 | 120-200 | Total |
Frequency | 4 | 10 | 26 | 8 | 2 | 50 |
Answer:
Class |
f |
X (Midpoint) |
fX |
c.f |
10 – 20 |
4 |
15 |
60 |
4 |
20 – 40 |
10 |
30 |
300 |
14 |
40 – 70 |
26 |
55 |
1430 |
40 |
70 – 120 |
8 |
95 |
760 |
48 |
120 – 200 |
2 |
160 |
320 |
50 |
|
Æ©f = 50 |
|
Æ©fX = 2870 |
|
Median class is given by the size of the item, i.e. the item, which is the 25th item. This corresponds to the class interval 40−70, so this is the median class.
Thus, mean and median are 57.4 and 52.69 respectively.
Page No 209:
Question 10:
An analysis for more efficiency in a factory , indicating the distribution of ages of workers was as follows:
Age (in years) | 16-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-64 |
Frequency | 15 | 46 | 49 | 32 | 28 | 14 |
(b) Draw a histogram and indicate mode therein.
Answer:
(a)
Age |
Frequency (f) |
Midpoint (m) |
fm |
16 – 19 |
15 |
17.5 |
262.5 |
20 – 29 |
46 |
24.5 |
1127 |
30 – 39 |
49 |
34.5 |
1690.5 |
40 – 49 |
32 |
44.5 |
1424 |
50 – 59 |
28 |
54.5 |
1526 |
60 – 64 |
14 |
62 |
868 |
|
|
|
|
Mean
Median
Note that the given distribution is in the form of inclusive class intervals. For the calculation of median, 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.
Age |
Frequency |
Cumulative |
15.5 – 19.5 19.5 – 29.5 L1→29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5 |
15 46 49→ f1 32 28 14 |
15 61→c.f 110 142 170 184 |
Median class is given by the
Size of item = item = 92th item.
This corresponds to the class interval of (29.5 − 39.5), so this is the median class.
Hence, the median is 35.826
(b)
Here the data is in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal.
Age |
Frequency |
Adjusted |
15.5 – 19.5 19.5 – 29.5 29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5 |
15 46 49 32 28 14 |
From the above graph, it can be seen that mode is 30.5. However, mean can not be determined graphically.
Page No 209:
Question 11:
The marks obtained by 100 students in an examination are given below. Compute the median, 1st and 3rd Quartiles.
Marks | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 |
No of students | 14 | 16 | 18 | 23 | 18 | 8 | 3 |
Answer:
Marks |
f |
c.f |
|
30 – 35 |
14 |
14 |
|
35 – 40 |
16 |
30 |
|
40 – 45 |
18 |
48 |
|
45 – 50 |
23 |
71 |
|
50 – 55 |
18 |
89 |
|
55 – 60 |
8 |
97 |
|
60 – 65 |
3 |
100 |
|
N=100
Median class is given by the size of the item, i.e. the item, which is the 50th item. This corresponds to the class interval 45−50, so this is the median class.
First quartile class is given by the size of the item, i.e. the item, which is the 25th item. This corresponds to the class interval 35−40, so this is the first quartile class.
Third quartile class is given by the size of the item, i.e. the item, which is the 75th item. This corresponds to the class interval 50−55, so this is the third quartile class.
Page No 209:
Question 12:
Compute mode from the following series:
Size of items | : | 2 | 3 | 4 | 5 | 6 | 7 |
Frequency | : | 3 | 8 | 10 | 12 | 16 | 14 |
Size of items | : | 8 | 9 | 10 | 11 | 12 | 13 |
Frequency | : | 10 | 8 | 17 | 5 | 4 | 1 |
Answer:
Size of Items |
Frequency |
|
2 |
3 |
|
3 |
8 |
|
4 |
10 |
|
5 |
12 |
|
6 |
16 |
|
7 |
14 |
|
8 |
10 |
|
9 |
8 |
|
10 |
17 |
→ Modal class |
11 |
5 |
|
12 |
4 |
|
13 |
1 |
|
Here, mode is equal to 10 as it has the highest frequency.
Page No 210:
Question 13:
Calcualte mode for the following data:
No. of persons | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Families | 26 | 113 | 120 | 95 | 60 | 42 | 21 | 14 | 5 | 4 |
Answer:
No. of Persons (X) |
Families (f) |
|
1 |
26 |
|
2 |
113 |
|
3 |
120 |
Modal class |
4 |
95 |
|
5 |
60 |
|
6 |
42 |
|
7 |
21 |
|
8 |
14 |
|
9 |
5 |
|
10 |
4 |
|
Here, mode is equal to 3 as it has the highest frequency.
Page No 210:
Question 14:
Find out the Mode from any of the following two distributions:
(a)
X | : | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
f | : | 6 | 10 | 16 | 14 | 10 | 5 | 2 |
(b)
Marks | : | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
No of candidates | : | 6 | 29 | 87 | 181 | 247 |
Marks | : | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
No of candidates | : | 263 | 113 | 49 | 9 | 2 |
Answer:
(a)
X |
f |
|
30 – 40 |
6 |
|
40 – 50 |
10 |
|
50 – 60 |
16 |
Modal class |
60 – 70 |
14 |
|
70 – 80 |
10 |
|
80 – 90 |
5 |
|
90 – 100 |
2 |
|
(b)
Since this is an inclusive class interval series, to calculate mode, first we need to convert the inclusive class intervals into the exclusive class intervals by 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.
Marks |
f |
|
–0.5 – 9.5 |
6 |
|
9.5 – 19.5 |
29 |
|
19.5 – 29.5 |
87 |
|
29.5 – 39.5 |
181 |
|
39.5 – 49.5 |
247 |
|
49.5 – 59.5 |
263 |
Modal class |
59.5 – 69.5 |
113 |
|
69.5 – 79.5 |
49 |
|
79.5 – 89.5 |
9 |
|
89.5 – 99.5 |
2 |
|
By inspection, we can spot that (49.5 – 59.5) has the highest frequency, i.e.263. So, this is the modal class.
Page No 210:
Question 15:
Life of electric lamps is given in the following table. Calculate the median and the mode.
Life in hours | Number of lamps |
Below 400 | 4 |
400-800 | 12 |
800-1200 | 40 |
1200-1600 | 41 |
1600-2000 | 27 |
2000-2400 | 13 |
2400-2800 | 9 |
Above 2800 | 4 |
Answer:
Class Interval |
f |
c.f. |
|
Below 400 |
4 |
4 |
|
400 – 800 |
12 |
16 |
|
800 – 1200 |
40 |
56 |
|
1200 – 1600 |
41 |
97 |
|
1600 – 2000 |
27 |
124 |
|
2000 – 2400 |
13 |
137 |
|
2400 – 2800 |
9 |
146 |
|
above 2800 |
4 |
150 |
|
Median class is given by the size of the item, i.e. the item, which is the 75th item. This corresponds to the class interval 1200−1600, so this is the median class.
By inspection, we can spot that (1200–1600) has the highest frequency, i.e. 41. So, this is the modal class.
Page No 210:
Question 16:
Calculate Mean, Median and Mode fromt he following data:
Marks (mid points) | No of Students |
59 | 1 |
61 | 2 |
63 | 9 |
65 | 48 |
67 | 131 |
69 | 102 |
71 | 40 |
73 | 17 |
Total = 350 |
Answer:
Converting the series into exclusive series
C.I. |
m |
f. |
fm |
c.f |
|
58 – 60 |
59 |
1 |
59 |
1 |
|
60 – 62 |
61 |
2 |
122 |
3 |
|
62 – 64 |
63 |
9 |
567 |
12 |
|
64 – 66 |
65 |
48 |
3120 |
60 |
|
66 – 68 |
67 |
131 |
8777 |
191 |
|
68 – 70 |
69 |
102 |
7038 |
293 |
|
70 – 72 |
71 |
40 |
2840 |
333 |
|
72 – 74 |
73 |
17 |
1241 |
350 |
|
|
|
Æ©f = 350 |
Æ©fm = 23764 |
|
|
Page No 211:
Question 17:
Following is the distribution of marks of 50 students in a class:
Marks (more than) | : | 0 | 10 | 20 | 30 | 40 | 50 |
No of students | : | 50 | 46 | 40 | 20 | 10 | 3 |
Answer:
Convert the series into exclusive class interval
C.I. |
more than c.f. |
f |
less than c.f |
|
0 – 10 |
50 |
4 |
4 |
|
10 – 20 |
46 |
6 |
10 |
|
20 – 30 |
40 |
20 |
30 |
→ Median class |
30 – 40 |
20 |
10 |
40 |
|
40 – 50 |
10 |
7 |
47 |
|
50 – 60 |
3 |
3 |
50 |
|
N = 50
If 60% students pass, it means 40% students fail. Thus, we have to find D4.
4thdecile class is given by the size of i.e. 20th item which falls in 30 cumulative frequency group.
Page No 211:
Question 18:
The age in completed years of 50 persons is given below:
32 | 61 | 52 | 56 | 22 | 49 | 97 | 35 | 30 | 30 | 95 | 67 | 42 | 20 |
31 | 64 | 20 | 10 | 62 | 60 | 27 | 53 | 31 | 90 | 54 | 25 | 43 | 47 |
35 | 21 | 43 | 75 | 45 | 22 | 36 | 13 | 46 | 23 | 51 | 11 | 15 | 39 |
50 | 42 | 77 | 73 | 81 | 40 | 40 | 55 |
Prepare frequency table taking 10-19, 20-29, ,.......... class intervals and calculate modal age.
Answer:
Age |
Tall marks |
No. of persons (f) |
Exclusive Class Interval |
|
10 – 19 |
4 |
9.5 – 19.5 |
|
|
20 – 29 |
|
8 |
19.5 – 29.5 |
|
30 – 39 |
|
9 |
29.5 – 39.5 |
|
40 – 49 |
10 |
39.5 – 49.5 |
Modal class |
|
50 – 59 |
|
7 |
49.5 – 59.5 |
|
60 – 69 |
5 |
59.5 – 69.5 |
|
|
70 – 79 |
|
3 |
69.5 – 79.5 |
|
80 – 89 |
1 |
79.5 – 89.5 |
|
|
90 – 99 |
|
2 |
89.5 – 99.5 |
|
|
|
Æ©f =49 |
|
|
By inspection, we can spot that (39.5–49.5) has the highest frequency, i.e. 10. So, this is the modal class.
Thus, the modal age is 42 years.
Page No 211:
Question 19:
Determine the value of mode for the following data by using the formula:
Mode = 3 Median − 2 Mean.
Marks | No. of students |
Less than 10 | 5 |
Less than 20 | 15 |
Less than 30 | 98 |
Less than 40 | 242 |
Less than 50 | 367 |
Less than 60 | 405 |
Less than 70 | 425 |
Less than 80 | 438 |
Less than 90 | 439 |
Answer:
Marks |
c.f. |
f |
m |
fm |
|
0 – 10 |
5 |
5 |
5 |
25 |
|
10 – 20 |
15 |
10 |
15 |
150 |
|
20 – 30 |
98 |
83 |
25 |
2075 |
|
30 – 40 |
242 |
144 |
35 |
5040 |
|
40 – 50 |
367 |
125 |
45 |
5625 |
|
50 – 60 |
405 |
38 |
55 |
2090 |
|
60 – 70 |
425 |
20 |
65 |
1300 |
|
70 – 80 |
438 |
13 |
75 |
975 |
|
80 – 90 |
439 |
1 |
85 |
85 |
|
|
|
Æ©f = 439 |
|
Æ©fm =17365 |
|
Median class is given by the size of the item, i.e. the item, which is the 219.5th item. This corresponds to the class interval 30−40, so this is the median class.
Now,
Mode = 3 Median – 2 Mean
= 3 × 38.43 – 2× 39.55
= 115.29 – 79.1
Mode = 36.19
Page No 211:
Question 20:
For the data given below find graphically the following :
(a) First and Third quartile
(b) The central 50% limit of the age .
(c) The number of workers falling in the age group of 28 to 57 years.
Age in years | : | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 |
No.of workers | : | 5 | 10 | 15 | 25 | 65 | 40 |
Age in years | : | 50-54 | 55-59 | 60-64 | 65-69 |
No.of workers | : | 23 | 10 | 5 | 2 |
Answer:
(a)
In order to find quartiles graphically, we first convert the inclusive series into exclusive series and then convert the frequency distribution into cumulative frequency distribution as follows:
Age | Age | Frequency (f) |
Cumulative Frequency (c.f.) |
19.5 – 24.5 24.5 – 29.5 29.5 – 34.5 34.5 – 39.5 39.5 – 44.5 44.5 – 49.5 49.5 – 54.5 54.5 – 59.5 59.5 – 64.5 64.5 – 69.5 |
Less than 24.5 Less than 29.5 Less than 34.5 Less than 39.5 Less than 44.5 Less than 49.5 Less than 54.5 Less than 59.5 Less than 64.5 Less than 69.5 |
5 10 15 25 65 40 23 10 5 2 |
5 15 30 55 120 160 183 193 198 200 |
Thus, the first and third quartiles are 38.5 and 48.25 respectively.
(b) The two limits (i.e. Q1 and Q3) with which central 50% of items lie are 38.5 and 48.25.
(c)
From the graph it is clear that total Number of workers of age 57 years = 190
Number of workers falling in the age group of 28 to 57 years = 190 11=179
Page No 211:
Question 21:
Draw a 'less than' ogive from the following data and hence find out the value of Median.
Class | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 |
Frequency | 6 | 9 | 13 | 23 | 19 | 15 | 9 | 6 |
Answer:
For constructing a less than ogive, first the given frequency distribution is converted into a less than cumulative frequency distribution as follows:
Class | Frequency (f) |
Cumulative Frequency (c.f.) |
Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 Less than 55 Less than 60 |
6 9 13 23 19 15 9 6 |
6 15 28 51 70 85 94 100 |
We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.
Hence, median is 39.78.
Page No 212:
Question 22:
The following table gives the distribution of the wages of 65 employees in a factory.
Wages (in â¹ ) | : | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120 |
(Equal to more than)
Number of employees
|
: | 65 | 57 | 47 | 31 | 17 | 7 | 2 | 0 |
Answer:
Class Interval |
More than Cumulative Frequency | Frequency | Wages | Less than Cumulative Frequency |
5060 6070 7080 8090 90100 100110 110120 120130 |
65 57 47 31 17 7 2 0 |
6557= 8 5747=10 4731=16 3117=14 177= 10 72=5 20=2 0 |
Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 Less than 110 Less than 120 Less than 130 |
8 18 34 48 58 63 65 65 |
Number of employees earning Rs 75 = 24
Number of employees earning atleast Rs 63 = 10
Number of employees earning between Rs 63 and Rs 75 = 24 10 =14
Therefore, 14 employees are earning between Rs 63 and Rs 75.
Page No 212:
Question 23:
Draw the histogram and estimate the value of mode from the following data:
â
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
No of students | 0 | 2 | 3 | 7 | 13 | 11 | 9 | 2 | 1 |
Answer:
Hence, mode is 47.5.
Page No 212:
Question 24:
Represent the following data by means of a histogram and find out mode.
Weekly wages | : | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 |
No of Workers | : | 7 | 19 | 27 | 15 | 12 | 12 | 8 |
Answer:
Hence, mode is 22.
Page No 212:
Question 25:
For a particular set of data in individual incomes the 'less than type' Ogive and the 'more than type' Ogive were found to intersect at â¹ 715. Discuss the significance of 'â¹ 715' for the given set of data.
Answer:
In a graphical presentation of a data, the place where the 'less than type' ogive and 'more than type' ogive intersect is the middle most point of the data. Thus, we may say that the point at which the two ogives intersect represents the median of the given data.
In the given question, the 'less than' and the 'more than' ogives intersect at Rs715.
Therefore, 715 is the median of the given data. It divides the whole data into two equal parts.
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