Question1
Number of Plants 
0  2 
2  4 
4  6 
6  8 
8  10 
10  12 
12  14 
Number of houses 
1 
2 
1 
5 
6 
2 
3 
Which method did you use for finding the mean, and why?
Solution:
Let us solve the problem using direct method
Number of plants 
Number of houses (f_{i}) 
Class mark (x_{i}) 
f_{i} x_{i} 
0  2 2  4 4  6 6  8 8  10 10  12 12  14 
1 2 1 5 6 2 3 
1 3 5 7 9 11 13 
1 6 5 35 54 22 39 
Î£ f_{i} = 20 
Î£ f_{i}x_{i} = 162 
Mean ( = = = 8.1
Thus the mean number of plants per house = 8.1
The method used here is direct method because the numerical values of x_{i} and f_{i} are small.
Question2
Daily wages (in ` ) 
100 â€“ 120 
120 â€“ 140 
140 â€“ 160 
160 â€“ 180 
180 â€“ 200 
Number of workers 
12 
14 
8 
6 
10 
Solution:
(Direct method)
The following problem is solved using direct method
Daily wages (in ` ) 
Number of workers (f_{i}) 
Class mark (x_{i}) 
f_{i} x_{i} 
100  120 120  140 140  160 160  180 180  200 
12 14 8 6 10 
110 130 150 170 190 
1320 1820 1200 1020 1900 
Total 
Î£ f_{i} = 50 
Î£ f_{i}x_{i} = 7260 
The mean of the given data ( = = = 145.2
The mean daily wages of the workers = `145.20.
Question3
Daily pocket allowance (in ` ) 
11 â€“ 13 
13 â€“ 15 
15 â€“ 17 
17 â€“ 19 
19 â€“ 21 
21 â€“ 23 
23 â€“ 25 
Number of workers 
7 
6 
9 
13 
f 
5 
4 
Solution:
Let us solve the problem using assumed mean method
Daily pocket allowance 
Number of children 
x_{i} 
d_{i} = x_{i } A 
f_{i}d_{i} 
11  13 
7 
12 
6 
42 
13  15 
6 
14 
4 
24 

9 
16 
2 
18 

13 
18 
0 
0 

f 
20 
2 
2f 
21  23

5 
22 
4 
20 
23 â€“ 25 
4 
24 
6 
24 
Ã¥ f_{i} = 44 + f 
Ã¥ f_{i}d_{i} =  40 + 2f 
Here âˆ‘ f_{i}d_{i} = 40 + 2f âˆ‘ f_{i} = 44 + f
Mean = A + = 18
= 18 + = 18
Ãž = 18  18
= 0
40 + 2f = 0 â‡’ 2f = 40
f = = 20.
Question4
Numbers of heart beats per minute 
65  68 
68  71 
71  74 
74  77 
77  80 
80  83 
83  36 
Number of women  2  4  3  8  7  4  2 
Solution:



d_{i} = x_{i}  A 

65  68 
2 

9 
18 
68  71 
4 
69.5 
6 
24 
71  74 
3 
72.5 
3 
9 
74  77 
8 
75.5 
0 
0 
77  80 
7 
78.5 
3 
21 
80  83 
4 
81.5 
6 
24 
83  86 
2 
84.5 
9 
18 


Mean ( x) = A + where A = 75.5
= 75.5 +
= 75.5 + 0.4
(x) = 75.9
âˆ´ The mean heart beats per minute = 75.9.
Question5
Number of mangoes 
50 â€“ 52 
53 â€“ 55 
56 â€“ 58 
59 â€“ 61 
62  64 
Number of boxes 
15 
110 
135 
115 
25 
Solution:
Question6
Daily Expenditure 
100 â€“ 150 
150  200 
200 â€“ 250 
250 â€“ 300 
300 â€“ 350 
Number of households 
4 
5 
12 
2 
2 
Solution:
Daily expenditure (in ` ) 


u_{i} = 


























âˆ‘ f_{i} = 25 
âˆ‘ f_{i}u_{i} = 7 
Mean = A +
= 225 +
= 225  14
= 211
âˆ´ The mean of daily expenditure on food = `211.
Question7
Find the mean concentration of SO_{2} in the air.
Concentration of SO_{2} (in ppm) 
Frequency 
0.00  0.04 
4 
0.04  0.08 
9 
0.08  0.12 
9 
0.12  0.16 
2 
0.16  0.20 
4 
0.20  0.24 
2 
Solution:
Let us solve the problem by Stepdeviation method
Concentration of SO_{2}(in ppm) 
Frequency (f_{i}) 
x_{i} 
d_{i} = x_{i}  0.1 
u_{i} = 
f_{i}u_{i} 
0.0  0.04 0.04  0.08 0.08  0.12 0.12  0.16 0.16  0.20 0.20  0.24 
4 9 9 2 4 2 
0.02 0.06 0.1 0.14 0.18 0.22 
0.08 0.04 0 0.04 0.08 0.12 
2 1 0 1 2 3 
8 9 0 2 8 6 
Total 
Î£ f = 30 
Î£ f_{i}u_{i} = 1 
Mean (x) = A + = 0.1 +
= 0.1  (0.00133)
= 0.09867
âˆ´ The mean concentration of SO_{2} in the air is 0.09867 ppm.
Question8
Number of Days 
0 â€“ 6 
6 â€“ 10 
10 â€“ 14 
14 â€“ 20 
20 â€“ 28 
28 â€“ 38 
38 â€“ 40 
Number of students 
11 
10 
7 
4 
4 
3 
1 
Solution:
Number of day 
Number of students (f_{i}) 
x_{i} 
d_{i} = x_{i} â€“ A 
U_{i} = d_{i}/2 
f_{i}u_{i} 
0 â€“ 6 
11 
3 
14 
7 
77 
6 â€“ 10 
10 
8 
9 
4.5 
45 
10 â€“ 14 
7 
12 
5 
2.5 
17.5 
14 â€“ 20 
4 
17 
0 
0 
0 
20 â€“ 28 
4 
24 
7 
3.5 
14 
28 â€“ 38 
3 
33 
16 
8 
24 
38 â€“ 40 
1 
39 
22 
11 
11 
Î£ f_{i} = 40 
Î£ f_{i}u_{i} = 90.5 
Mean (x) = A + Ã— h = 17 +
= 17 â€“ 4.525
= 12.475
The mean number of days a student was absent = 12.48 days.
Question9
Literacy rate (In %) 
45 â€“ 55 
55 â€“ 65 
65 â€“ 75 
75 â€“ 85 
85 â€“ 95 
Number of cities 
3 
10 
11 
8 
3 
Solution:
Mean (x) = A + Ã— h
= 70 +
= 70 â€“ 0.571 = 69.428
= 69.43%
The mean literacy rate (in percentage) = 69.43%.
Question10
Age 
5 â€“ 15 
15 â€“ 25 
25 â€“ 35 
35 â€“ 45 
45 â€“ 55 
55 â€“ 65 
Number of patients 
6 
11 
21 
23 
14 
5 
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mean = A + Ã— h
= 40 +
= 40 â€“ 4.625
= 35.375
Mode = l + Ã— h
Here the maximum class frequency is 23, and the class corresponding frequency is 35 â€“ 45. So the modal class is 35 â€“ 75.
Now, lower limit(l) of modal class = 35
Class size(h) = 10
Frequency (f_{1}) of the modal class = 23
Frequency (f_{0}) of class preceding the model class = 21
Frequency (f_{2}) of class succeeding the model class = 14
Mode = l + Ã— h
= 35 +
= 35 +
= 35 +
= 35 + 1.82
= 36.82.
Question11
Lifetime 
0 â€“ 20 
20 â€“ 40 
40 â€“ 60 
60 â€“ 80 
80  100 
100 â€“ 120 
Frequency 
10 
35 
52 
61 
38 
29 
Solution:
Here maximum frequency is 61, the class corresponding to this 60 â€“ 80. So, the model class is 60 â€“ 80.
Now lower limit of the model class (l) = 60
frequency(f_{1}) of the modal class = 61
frequency(f_{0}) of the preceding the model class = 52
frequency(f_{2}) of class succeeding the model class = 38
class size (h) = 20
Mode = l + Ã— h
= 60 + Ã— 20
= 60 + Ã— 20
= 60 +
= 60 + 5.625
= 65.625
âˆ´ Modal lifetime of the components = 65.63 hours.
Question12
Expenditure (in `) 
Number of families 
1000 â€“ 1500 
24 
1500 â€“ 2000 
40 
2000 â€“ 2500 
33 
2500 â€“ 3000 
28 
3000 â€“ 3500 
30 
3500 â€“ 4000 
22 
4000 â€“ 4500 
16 
4500 â€“ 5000 
7 
Mode l + Ã— h
The highest frequency is 40. The modal class is 1500  2000.
Lower limit (l) of the modal class = 1500
Class size (h) = 500
Frequency of the modal class (f_{1}) = 40
Frequency of the preceding modal class (f_{0}) = 24
Frequency of the succeeding modal class (f_{2}) = 33
Mode = l + Ã— h
Mode = 1500 + Ã— 500
= 1500 +
= 1500 +
= 1500 + 347.83
= 1847.83 âˆ´ The model monthly expenditure of the family = `1847.83
Mean = A +
= 3250 
= 3250  587.50
= 266.65
âˆ´ Mean monthly expenditure = `2662.5.
Question13
Number of student per teacher 
Number of states/ U.T. 
15 â€“ 20 
3 
20 â€“ 25 
8 
25 â€“ 30 
9 
30 â€“ 35 
10 
35 â€“ 40 
3 
40 â€“ 45 
0 
45 â€“ 50 
0 
50 â€“ 55 
2 
Solution:
In this problem, the maximum frequency is 10, and the class corresponding to this frequency is 30  35. So, the modal class is 30  35. Now
Lower limit (l) of modal class = 30
Class size (h)= 5
frequency (f_{1}) of model class = 10
frequency (f_{0}) proceeding the modal class = 9
frequency (f_{2}) succeeding the modal class = 3
Mode = l +
Substituting the values above in the given formula,
= 30 +
= 30 +
= 30 + = = 30.625
âˆ´ The mode of the given data is 30.6
Now calculate mean by applying direct method.
Number of students / teacher 
Number of states/U.T (f_{i}) 
Classmark (x_{i}) 
di = x_{i}  A = x_{i}  325 
f_{i}d_{i} 
1520 
3 
17.5 
15 
45 
2025 
8 
22.5 
10 
80 
2530 
9 
27.5 
5 
45 
3035 
10 
32.5 
0 
0 
3540 
3 
37.5 
5 
15 
4045 
0 
42.5 
10 
0 
4550 
0 
47.5 
15 
0 
50  55 
2 
52.5 
20 
40 
Ã¥ f_{i} = 35  Ã¥ f_{i}d_{i} = 11.5 
= A +
= 32.5 +
= 32.5  3.29 = 29.21
âˆ´ The mean of the given data = 29.21
Thus we interpret that the most states/ U.T. student teacher ratio of 30.6 and on an average this ratio is 29.2.
Question14
Runs scored 
Number of batsman 
3000 â€“ 4000 
4 
4000 â€“ 5000 
18 
5000 â€“ 6000 
9 
6000 â€“ 7000 
7 
7000 â€“ 8000 
6 
8000 â€“ 9000 
3 
9000 â€“ 10000 
1 
10000 â€“ 11000 
1 
Find the mode of the data.
Solution:
The maximum frequency is 18, and the class corresponding to the maximum frequency is 4000 â€“ 5000.
Now the model class is 4000 â€“ 5000
Class size(h) = 1000
lower limit (f) of the model class = 4000
frequency(fo) of the class proceeding the modal class = 4
frequency(f_{1}) of the modal class = 18
frequency(f_{2}) of the class succeeding the modal class = 9
mode = l +
= 4000 +
= 4000 +
= 4000 +
= 4000 + 608.7
= 4608.7
Mode = 4608.7 runs.
Question15
Number of cars 
0  10 
10  20 
20  30 
30  40 
40  50 
50  60 
60  70 
70 â€“ 80 
Frequency 
7 
14 
13 
12 
20 
11 
15 
8 
Solution:
Now,
The model class is 40 â€“ 50
Class size (h) = 10
lower limit (l) of the model class = 40
frequency(f_{1}) of the modal class = 20
frequency(f_{0 })of the class proceeding the modal class = 12
frequency(f_{2 })of the class succeeding the modal class = 11
Mode = l +
= 40 +
= 40 +
= 40 +
= 40 + 4.71
= 44.71 cars
âˆ´ Mode = 44.71 cars.
Question16
Monthly consumption (in units) 
Number of consumers 
65  85 
4 
85  105 
5 
105  125 
13 
125  145 
20 
145  165 
14 
165  185 
8 
185  205 
4 
Solution:
n = 68, = 34. Here 125  145 is the class whose cumulative frequency 42 is greater than 34.
Here
L  lower limit of median class = 125
= 34 where n = number of observations
c.f  cumulative frequency of class proceeding the median class = 22
f  frequency of median class = 20
h  class size = 20
Median = l +
= 125 +
= 125 + 12
= 137
âˆ´ The median of the given data is 137.
The mean can be calculated using assumed mean method.
Monthly consumption (in units) 
Number of consumer (f_{i}) 
Class mark (x_{i}) 
d_{i} = x_{i} f_{i} = x_{i}  135 
f_{i}d_{i} 
6585 
4 
75 
60 
240 
85105 
5 
95 
40 
200 
105125 
13 
115 
20 
260 
125  145 
20 
135 
0 
0 
145  165 
14 
155 
20 
280 
165  185 
8 
175 
40 
320 
185  205 
4 
195 
60 
240 
âˆ‘ f_{i} = 68  âˆ‘ f_{i}d_{i} = 140 
Mean( = A +
= 135 +
= 135 + 2.06 = 137.06
âˆ´ The mean of the given data = 137.06
Let us find the mode as follows.
The highest frequency is 20 which corresponds to the class interval 125  145. So the modal class is 125  145.
Here
Lower limit (l) of the modal class = 125
Class size (h) = 20
frequency (f_{1}) of the modal class = 20
frequency (f_{0}) of the class succeeding the modal class = 13
frequency (f_{2}) of the class succeeding the modal class = 14
mode = l +
= 125 +
= 125 +
= 125 + 10.77
= 135.77
âˆ´ The mode of the data = 135.77
The three measures are approximately the same in this case.
Question17
Class Interval 
Frequency 
0 â€“ 10 
5 
10 â€“ 20 
x 
20 â€“ 30 
20 
30  40 
15 
40 â€“ 50 
y 
50 â€“ 60 
5 
Total 
60 
Class Interval 
Frequency (f_{i}) 
Cumulative frequency 
0  10 
5 
5 
10  20 
x 
5 + x 
20  30 
20 
25 + x 
30  40 
15 
40 + x 
40  50 
y 
40 + x + y 
50  60 
5 
45 + x + y 
Total 
Σ f_{i }= 60 

Here in this problem = n = 45 + x + y
It is given that total frequency is 60, thus n = 60 âˆ´ 45 + x + y = 60 â‡’ x + y = 60  45
= 15(1)
The median is 28.5 which lies in the class 2030
So,
l (lower limit) of the median class = 20
f (frequency) of the median class = 20
cf(cumulative) frequency of the class = 5 + x
proceeding 20  30
h (class size) = 10
median = l +
28.5 = 20 +
28.5 = 20 +
28.5  20 +
8.5 = â‡’ 17.0 = 25  x â‡’ x = 25  17
= 8
substitute x = 8 in (1)
x + y = 15
8 + y = 15
y = 15  8
= 7
âˆ´ The value of x and y are 8 and 7.
Question18
Age 
Number of policy holders 
Below 20 
2 
Below 25 
6 
Below 30 
24 
Below 35 
45 
Below 40 
78 
Below 45 
89 
Below 50 
92 
Below 55 
98 
Below 60 
100 
Solution:
Age 
Number of policy holders 
Cumulative frequency 
Below 20 
2 
2 
Below 25 
6 
6 
Below 30 
24 
24 
Below 35 
45 
45 
Below 40 
78 
78 
Below 45 
89 
89 
Below 50 
92 
92 
Below 55 
98 
98 
Below 60 
100 
100 
Here n = 100, = = 50
The observation lies in the class 35 â€“ 40
Lower limit (l) = 35
cf â€“ cumulative frequency of the class proceeding the interval 35 â€“ 40 = 45.
frequency (f) of the median class 35 â€“ 40 = 33
class size (h) = 5
median = l +
= 35 +
= 35 +
= 35 + 0.758
= 35.76
âˆ´ The median age is 35.76 years.
Question19
Length 
Number of leaves 
118 â€“ 126 
3 
127 â€“ 135 
5 
136 â€“ 144 
9 
145 â€“ 153 
12 
154 â€“ 162 
5 
163 â€“ 171 
4 
172 â€“ 180 
2 
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5  126.5, 126.5  135.5, . . ., 171.5  180.5.)
Solution:
Now n = 40, = = 20
This observation lies in the class 144.5  153.5
l (lower limit of the classes) = 144.5
cf (cumulative frequency of the class proceeding 144.5  153.5 = 17
f (frequency of the median class) = 12
h (class size) = 9
median = l +
= 144.5 +
= 144.5 +
= 144.5 +
= 144.5 + 2.25
= 146.75 mm
âˆ´ The median length of the leaves = 146.75 mm.
Question20
Life time 
Number of lamps 
1500 â€“ 2000 
14 
2000 â€“ 2500 
56 
2500 â€“ 3000 
60 
3000 â€“ 3500 
86 
3500 â€“ 4000 
74 
4000 â€“ 4500 
62 
4500 â€“ 5000 
48 
Find the median life time of a lamp.
Solution:
Life time 
Number of lamps 
Cumulative frequency 
1500  2000 
14 
14 
2000  2500 
56 
14 + 56 = 70 
2500  3000 
60 
70 + 60 = 130 
3000  3500 
86 
130 + 86 = 216 
3500  4000 
74 
216 + 74 = 290 
4000  4500 
62 
290 + 62 = 352 
4500  5000 
48 
352 + 48 = 400 
Σ f_{i} = 400 
n = âˆ‘ f_{i} = 400
= = 200
Thus the observation lies in the class 3000  3500
median = l +
where l  lower limit of the median class = 3000
f = frequency of the median class = 86
cf = cumulative frequency proceeding the median class = 216
h  class size = 500
median = 3000 +
= 3000 +
= 3000 + 406.98
= 3406.98
Thus the median life time of a lamp is 3406.98 hours.
Question21
Number of letters 
1 â€“ 4 
4 â€“ 7 
7 â€“ 10 
10 â€“ 13 
13 â€“ 16 
16 â€“ 19 
Number of surnames 
6 
30 
40 
16 
4 
4 
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution:
Number of letters 
Number of surnames 
Cumulative frequency 
1  4 
6 
6 
4  7 
30 
6 + 30 = 36 
7  10 
40 
36 + 40 = 76 
10  13 
16 
76 + 16 = 92 
13  16 
4 
9 + 4 = 96 
16  19 
4 
96 + 4 = 100 
100 
Here n = 100,
Here the observation lies in the class 7  10
l  lower limit of the class = 7
f  frequency of the median class (7  10) = 40
cf  (cumulative frequency of the class proceeding (7  10) = 36
h  class size = 3
median = l +
= 7 +
= 7 +
= 7 + 1.05
= 8.05
âˆ´ median = 8.05
Median number of letters in the surnames = 8.05
Let us calculate the mean using step deviation method
This mean number of letters in the surname = 8.32Mean = A +
= 11.5 +
= 11.5  3.18
= 8.32
Now we have to find the modal size of the surnames














Here the maximum frequency is 40. The class corresponding to this frequency is 7  10.
l = lower limit of the model class = 7
frequency(f_{1}) of the modal class = 40
frequency(f_{0}) of the class proceeding the modal class = 30
frequency(f_{2}) of the class succeeding the modal class = 16
size of the class (h) = 3
mode = l +
= 7 +
= 7 +
= 7 +
= 7 + = 7 + 0.88
= 7.88
âˆ´ The model size of the surname = 7.88.
Question22
Weight (in kg) 
40 â€“ 45 
45 â€“ 50 
50 â€“ 55 
55 â€“ 60 
60 â€“ 65 
65 â€“ 70 
70 â€“ 75 
Number of students 
2 
3 
8 
6 
6 
3 
2 
Solution:
Weight (in kg) 
Number of students 
Cumulative frequency 
40 â€“ 45 
2 
2 
45 â€“ 50 
3 
2+3 = 5 
50 â€“ 55 
8 
5 + 8 = 13 
55 â€“ 60 
6 
13 + 6 = 19 
60 â€“ 65 
6 
19 + 6 = 25 
65 â€“ 70 
3 
25 + 3 = 28 
70 â€“ 75 
2 
28 + 2 = 30 
n = 30,
Here the observation lies in the class 55 â€“ 60
The lower limit (l) = 55
f (frequency of the median class) = 6
cf (cumulative frequency of the class proceeding the median class = 13
Median =
=
= 55 + (5/3)
= 55 + 1.67
= 56.67 kg.
Question23
Daily Income (in `) 
100 120 
120 140 
140 160 
160 180 
180200 
Number of students 
12 
14 
8 
6 
10 
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
The less than type cumulative frequency distribution is given as follows:
Daily income (in `) 
Number of students 
Cumulative frequency 
less than 120 
12 
12 
less than 140 
14 
12 + 14 = 26 
less than 160 
8 
26 + 8 = 34 
less than 180 
6 
34 + 6 = 40 
less than 200 
10 
40 + 10 = 50 
Draw the ogive curve by plotting the points(120, 12), (140, 26), (160, 34), (180, 40), (200, 50)
First draw the coordinate axes with upper limits of daily income along the horizontal axis and the cumulative frequency along the vertical axis.
Question24
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Weight (in kg) 
Number of students 
Less than 38 
0 
Less than 40 
3 
Less than 42 
5 
Less than 44 
9 
Less than 46 
14 
Less than 48 
28 
Less than 50 
32 
Less than 52 
35 
Solution:
Draw a less than type ogive curve, by plotting the points(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
The xcoordinate of the point 17.5 will be the median.
Question25
Production yield 
50  55 
55  60 
60  65 
65  70 
70  75 
75  80 
Number of fans 
2 
8 
12 
24 
38 
16 
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
Production yield kg / ha 
Number of forms 
Cumulative frequency 


















Now draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).