Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 18  Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18A
Solution 1
Solution 2
Solution 3
We form the table as below:
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
1030 
15 
20 
300 
3050 
18 
40 
720 
5070 
25 
60 
1500 
7090 
10 
80 
800 
90110 
2 
100 
200 

𝛴f_{i} = 70 

𝛴(f_{i} × x_{i}) = 3520 
Therefore, mean
Solution 4
We form the table as below:
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
020 
6 
10 
60 
2040 
8 
30 
240 
4060 
10 
50 
500 
6080 
12 
70 
840 
80100 
6 
90 
540 
100120 
5 
110 
550 
120140 
3 
130 
390 

𝛴f_{i} = 50 

𝛴(f_{i} × x_{i}) = 3120 
Therefore, mean
Solution 5
We form the table as below:
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
06 
10 
3 
30 
612 
11 
9 
99 
1218 
7 
15 
105 
1824 
4 
21 
84 
2430 
4 
27 
108 
3036 
3 
33 
99 
3642 
1 
39 
39 

𝛴f_{i} = 40 

𝛴(f_{i} × x_{i}) = 564 
Therefore, mean number of days
Solution 6
We form the table as below:
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
500700 
6 
600 
3600 
700900 
8 
800 
6400 
9001100 
10 
1000 
10000 
11001300 
9 
1200 
10800 
13001500 
7 
1400 
9800 

𝛴f_{i} = 40 

𝛴(f_{i} × x_{i}) = 40600 
Therefore, mean daily expenses
Solution 7
We form the table as below:
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
6468 
6 
66 
396 
6872 
8 
70 
560 
7276 
10 
74 
740 
7680 
12 
78 
936 
8084 
3 
82 
246 
8488 
1 
86 
86 

𝛴f_{i} = 40 

𝛴(f_{i} × x_{i}) = 2964 
Therefore, mean heartbeats per minute
Solution 8
Here, class size, h = 10.
Let the assumed mean be A = 35.
For calculating the mean age, we prepare the following table:
Class 
Frequency f_{i} 
Class mark x_{i} 

(f_{i} × u_{i}) 
010 
105 
5 
3 
315 
1020 
222 
15 
2 
444 
2030 
220 
25 
1 
220 
3040 
138 
35 
0 
0 
4050 
102 
45 
1 
102 
5060 
113 
55 
2 
226 
6070 
100 
65 
3 
300 

𝛴f_{i} = 1000 


𝛴(f_{i} × u_{i}) = 351 
Therefore,
Thus, the mean age of persons visiting the marketing centre on that day is 31.5 years.
Solution 9
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
020 
12 
10 
120 
2040 
15 
30 
450 
4060 
32 
50 
1600 
6080 
x 
70 
70x 
80100 
13 
90 
1170 

𝛴f_{i} = 72 + x 

𝛴(f_{i} × x_{i}) = 3340 + 70x 
Arithmetic mean
Solution 10
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
1113 
3 
12 
36 
1315 
6 
14 
84 
1517 
9 
16 
144 
1719 
13 
18 
234 
1921 
f 
20 
20f 
2123 
5 
22 
110 
2325 
4 
24 
96 

𝛴f_{i} = 40 + f 

𝛴(f_{i} × x_{i}) = 704 + 20f 
Arithmetic mean
Solution 11
Solution 12
Solution 13
Solution 14
We have
_{}
Class 
Frequency _{} 
Mid Value

_{} 
0  20 
7 
10 
70 
20  40 
_{} 
30 
30_{} 
40  60 
12 
50 
600 
60  80 
_{} =18  
70 
1260  70_{} 
80  100 
8 
90 
720 
100  120 
5 
110 
550 
_{} = 50 
_{} 
_{}
Solution 15
Class 
Frequency f_{i} 
Class mark x_{i} 
(f_{i} × x_{i}) 
080 
20 
40 
800 
80160 
25 
120 
3000 
160240 
x 
200 
200x 
240320 
y 
280 
280y 
320400 
10 
360 
3600 

𝛴f_{i} = 55 + x + y 

𝛴(f_{i} × x_{i}) = 7400 + 200x + 280y 
Now,
Also, arithmetic mean = 188
Solution 16
Solution 17
We have, Let A = 25 be the assumed mean.
Marks 
Frequency _{} 
Mid value _{} 
Deviation _{} 
_{} 
0  10 10  20 20  30 30  40 40  50 50  60 
12 18 27 20 17 6 
5 15 25 = A 35 45 55 
20 10 0 10 20 30 
240 180 0 200 340 180 
_{} = 100 
_{} = 300 
_{}
Hence, mean marks per student = 28
Solution 18
Let the assumed mean be 150, h = 20
Marks 
Frequency _{} 
Mid value _{} 
Deviation d_{i} = _{}  150 
_{} di 
100  120 120  140 140  160 160  180 180  200 
10 20 30 15 5 
110 130 150=A 170 190 
40 20 0 20 40 
400 400 0 300 200 
_{} = 80 
_{} di=300 
_{}
Hence, Mean = 146.25
Solution 19
Let A = 50 be the assumed mean, we have
Marks 
Frequency _{} 
Mid value _{} 
Deviation _{} 
_{} 
0  20 20  40 40  60 60  80 80  100 100  120 
20 35 52 44 38 31 
10 30 50 = A 70 90 110 
40 20 0 20 40 60 
800 700 0 880 1520 1860 
_{}= 220 
_{} 
_{}
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Let h = 20 and assume mean = 550, we prepare the table given below:
Age 
Frequency _{} 
Mid value _{} 
_{} 
_{} 
500  520 520  540 540  560 560  580 580  600 600  620 
14 9 5 4 3 5 
510 530 550 = A 570 590 610 
2 1 0 1 2 3 
27 9 0 4 6 15 
_{} = 40 
_{} 
Thus, A = 550, h = 20, and _{} = 40, _{}
_{}
Hence, the mean of the frequency distribution is 544.
Solution 26
The given series is an inclusive series, making it an exclusive series, we have
Class 
Frequency _{} 
Mid value _{} 
_{} 
_{} 
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 
4 14 22 16 6 5 3 
27 32 37 42 = A 47 52 57 
3 2 1 0 1 2 3 
12 28 22 0 6 10 9 
_{} = 70 
_{} 
Thus, A = 42, h = 5, _{} = 70 and _{}
_{}
Hence, Mean = 39.36 years
Solution 27
The given series is an inclusive series making it an exclusive series,we get
class 
Frequency _{} 
Mid value _{} 
_{} 
_{} 
4.5  14.5 14.5  24.5 24.5  34.5 34.5  44.5 44.5  54.5 54.5  64.5 
6 11 21 23 14 5 
9.5 19.5 29.5=A 39.5 49.5 59.5 
2 1 0 1 2 3 
12 11 0 23 28 15 
_{} = 80 
_{} 
Thus, A = 29.5, h = 10, _{} = 80 and _{}
_{}
Hence, Mean = 34.87 years
Solution 28
Solution 29
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18B
Solution 1
Solution 2
We prepare the frequency table, given below
Marks 
No. of students _{} 
C.F. 
0  7 7  14 14  21 21  28 28  35 35  42 42  49 
3 4 7 11 0 16 9 
3 7 14 25 25 41 50 
N = _{} = 50 
Now, _{}
The cumulative frequency is 25 and corresponding class is 21  28.
Thus, the median class is 21  28
l = 21, h = 7, f = 11, c = C.F. preceding class 21  28 is 14 and = 25
_{}
Hence the median is 28.
Solution 3
We prepare the frequency table given below:
Daily wages 
Frequency _{} 
C.F. 
0  100 100  200 200  300 300  400 400  500 
40 32 48 22 8 
40 72 120 142 150 
N = _{} = 150 
Now, N = 150, therefore _{}
The cumulative frequency just greater than 75 is 120 and corresponding class is 200  300.
Thus, the median class is 200  300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and _{}
_{}
Hence the median of daily wages is Rs. 206.25.
Solution 4
We prepare the frequency table, given below:
Class 
Frequency _{} 
C.F 
5  10 10  15 15  20 20  25 25  30 30  35 3540 40  45 
5 6 15 10 5 4 2 2 
5 11 26 36 41 45 47 49 
_{} = 49 
Now, N = 49_{ }
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15  20.
Thus, the median class is 15  20
_{}l = 15, h = 5, f = 15
c = CF preceding median class = 11 and_{}
_{}
Median of frequency distribution is 19.5
Solution 5
We prepare the cumulative frequency table as given below:
Consumption 
Frequency _{} 
C.F 
65  85 85  105 105  125 125  145 145  165 165  185 185  205 
4 5 13 20 14 7 4 
4 9 22 42 56 63 67 
N = _{} = 67 
Now, N = 67_{}
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125  145.
Thus, the median class is 125  145
_{}l = 125, h = 20, _{}and c = CF preceding the median class = 22, = 33.5
_{}
Hence median of electricity consumed is 136.5
Solution 6
Frequency table is given below:
Height 
Frequency _{} 
C.F 
135  140 140  145 145  150 150  155 155  160 160  165 165  170 170  175 
6 10 18 22 20 15 6 3 
6 16 34 56 76 91 97 100 
N = _{} =100 
N = 100, _{}
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150  155
Thus, the median class is 150  155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
_{}
Hence, Median = 153.64
Solution 7
The frequency table is given below. Let the missing frequency be x.
Class 
Frequency _{} 
C.F 
0  10 10  20 20  30 30  40 40  50 
5 25 x 18 7 
5 30 30 + x 48 + x 55 + x 
Median = 24 _{}Median class is 20  30
_{}
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
_{}
Hence, the missing frequency is 25.
Solution 8
Solution 9
Solution 10
Let _{}be the frequencies of class intervals 0  10 and 40  50
_{}
Median is 32.5 which lies in 30  40, so the median class is 30  40
l = 30, h = 10, f = 12, N = 40 and _{}
_{}
Solution 11
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Class 
Frequency _{} 
C.F 
18.5  25.5 25.5  32.5 32.5  39.5 39.5  46.5 46.5  53.5 53.5  60.5 
35 96 68 102 35 4 
35 131 199 301 336 340 
f_{i} = N = 340 
_{}
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5  39.5.
_{}Median class is 32.5  39.5
_{}l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
_{}
Hence median is 36.5 years
Solution 12
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
Wages per day (in Rs) 
Frequency _{} 
C.F 
60.5  70.5 70.5  80.5 80.5  90.5 90.5  100.5 100.5  110.5 110.5  120.5 
5 15 20 30 20 8 
5 20 40 70 90 98 
f_{i} = N =98 
_{}
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5  100.5.
_{}median class is 90.5  100.5
_{}l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
_{}
Hence, Median = Rs 93.50
Solution 13
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks 
Frequency _{} 
C.F 
0.5  5.5 5.5  10.5 10.5  15.5 15.5  20.5 20.5  25.5 25.5  30.5 30.5  35.5 35.5  40.5 40.5  45.5 
7 10 16 32 24 16 11 5 2 
7 17 33 65 89 105 116 121 123 
f_{i} = N =123 
_{}
The cumulative frequency just greater than 61.5 is 65.
_{}The corresponding median class is 15.5  20.5.
Then the median class is 15.5  20.5
_{}l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
_{}
Hence, Median = 19.95
Solution 14
Marks 
Frequency _{} 
C.F 
0  10 10  20 20  30 30  40 40  50 50  60 60  70 70  80 
12 20 25 23 12 24 48 36 
12 32 57 80 92 116 164 200 
N = _{} 
_{}
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50  60.
Thus the median class is 50  60
_{}l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, = 100
_{}
Hence, Median = 53.33
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18C
Solution 1
Solution 2
Solution 3
Solution 4
As the class 26  30 has maximum frequency so it is modal class
_{}
Hence, mode = 28.5
Solution 5
As the class 1500  2000 has maximum frequency, so it os modal class
_{}
Hence the average expenditure done by maximum number of workers = Rs. 1820
Solution 6
As the class 5000  10000 has maximum frequency, so it is modal class
_{}
Hence, mode = Rs. 7727.27
Solution 7
As the class 15  20 has maximum frequency so it is modal class.
_{}
_{}
Hence mode = 17.3 years
Solution 8
As the class 85  95 has the maximum frequency it is modal class
_{}
Hence, mode = 85.71
Solution 9
The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get
Class 
Frequency 
0.5  5.5 5.5 10.5 10.5  15. 5 15.5  20.5 20.5  25. 5 25.5  30.5 30.5  35.5 35.5  40.5 40.5  45.5 45.5  50.5 
3 8 13 18 28 20 13 8 6 3 
As the class 20.5  25.5 has maximum frequency, so it is modal class
_{}
_{}
_{}
Hence, mode = 23.28
Solution 10
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18D
Solution 1
Mean:
Class 
Frequency f_{i} 
Cumulative frequency 
Class mark x_{i} 
(f_{i} × x_{i}) 
010 
4 
4 
5 
20 
1020 
4 
8 
15 
60 
2030 
7 
15 
25 
175 
3040 
10 
25 
35 
350 
4050 
12 
37 
45 
540 
5060 
8 
45 
55 
440 
6070 
5 
50 
65 
325 

𝛴f_{i} = 50 


𝛴(f_{i} × x_{i}) = 1910 
Therefore, mean
Median:
Cumulative frequency just greater than 25 is 37 and the corresponding class is 4050.
Thus, the median class = 3040
Mode:
Mode = 3(Median)  2(Mean)
= 3(40)  2(38.2)
= 120  76.4
= 43.6
Solution 2
Solution 3
Solution 4
Solution 5
Let the assumed mean A be 145.Class interval h = 10.
Class 
Frequency _{} 
MidValue _{} 
_{} 
_{} 
C.F. 
120130 130140 140150 150160 160170 
2 8 12 20 8 
125 135 145=A 155 165 
2 1 0 1 2 
4 8 0 20 16 
2 10 22 42 50 
N = 50 
_{} 
(i)Mean _{}
_{}
(ii)N = 50, _{}
Cumulative frequency just after 25 is 42
Corresponding median class is 150  160
Cumulative frequency before median class, c = 22
Median class frequency f = 20
_{}
(iii)Mode = 3 median  2 mean
= 3 151.5  2 149.8 = 454.5  299.6
= 154.9
Thus, Mean = 149.8, Median = 151.5, Mode = 154.9
Solution 6
Class 
Frequency_{} 
Midvalue _{} 
_{} 
_{} 
C.F. 
100120 120140 140160 160180 180200 
12 14 8 6 10 
110 130 150= A 170 190 
2 1 0 1 2 
24 14 0 6 20 
12 26 34 40 50 
N = 50 
_{} 
Let assumed mean A = 150 and h = 20
(i)Mean _{}
_{}
(ii)_{}
Cumulative frequency just after 25 is 26
_{}Corresponding frequency median class is 120  140
So, l = 120, f = 14, _{}h = 20, c = 12
_{ }
(iii)Mode = 3Median  2Mode
= 3(138.6)  2(145.2)
= 415.8  190.4
= 125.4
Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4
Solution 7
Class 
Frequency_{} 
Midvalue _{} 
_{} 
_{} 
C.F. 
100150 150200 200250 250300 300350 
6 7 12 3 2 
125 175 225 275 325 
2 1 0 1 2 
12 7 0 3 4 
6 13 25 28 30 
N = 30 
_{} 
Let assumed mean = 225 and h = 50
(i)Mean = _{}
(ii)_{}
Cumulative frequency just after 15 is 25
_{}corresponding class interval is 200  250
_{}Median class is 200  250
Cumulative frequency c just before this class = 13
_{}
_{}
Hence, Mean = 205 and Median = 208.33
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18E
Solution 1
We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the 'less than type' ogive as follows:
At y = 26.5, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 68 units
Hence, median marks = 68
Solution 2
Number of wickets 
Less than 15 
Less than 30 
Less than 45 
Less than 60 
Less than 75 
Less than 90 
Less than 105 
Less than 120 
Number of bowlers 
2 
5 
9 
17 
39 
54 
70 
80 
We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the 'less than type' ogive as follows:
At y = 40, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 78 units
Hence, median number of wickets = 78
Solution 3
'More than type' distribution is as follows:
We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the 'more than type' ogive as follows:
Solution 4
'More than type' distribution is as follows:
We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the 'more than type' ogive as follows:
Solution 5
'More than type' distribution is as follows:
We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the 'more than type' ogive as follows:
Solution 6
We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the 'more than type' ogive as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 70.5 units
Hence, median production yield = 70.5 kg/ha
Solution 7
'More than type' distribution is as follows:
Production yield 
Number of farms_{} 
More than 65 
24 
More than 60 
54 
More than 55 
74 
More than 50 
90 
More than 45 
96 
More than 40 
100 
On a graph paper, we plot the points A(40, 100), B(45, 96), C(50, 90), D(55, 74), E(60, 54) and F(65, 24).
Join these points to get a 'More than Ogive'.
Solution 8
More than series
We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)
Hence,_{}
Take a point A(0, 115) on the yaxis and draw APxaxis meeting the curve at P, Draw PM _{}xaxis intersecting xaxis at M
Then,OM = 590
Hence median = 590
Solution 9
(i) Less than series:
Marks 
No. of students 
Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 
2 7 13 21 31 56 76 94 98 100 
Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
Join these points free hand to get the curve representing "less than" cumulative curve.
(ii)From the given table we may prepare the 'more than' series as shown below
Marks 
No. of students 
More than 45 More than 40 More than 35 More than 30 More than 25 More than 20 More than 15 More than 10 More than 5 More than 0 
2 6 24 44 69 79 87 93 98 100 
Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)
Join these points free hand to get the required curve.
Here _{}
Two curves intersect at point P(28, 50)
Hence, the median = 28
Solution 10
We may prepare less than series and more than series
(i)Less than series
Height in (cm) 
Frequency 
Less than 140 Less than 144 Less than 148 Less than 152 Less than 156 Less than 160 Less than 164 Less than 168 Less than 172 Less than 176 Less than 180 
0 3 12 36 67 109 173 248 330 416 450 
Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)
(ii)More than series
Height in cm 
C.F. 
More than 140 More than 144 More than 148 More than 152 More than 156 More than 160 More than 164 More than 168 More than 172 More than 176 More than 180 
450 447 438 414 383 341 277 202 120 34 0 
Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)
The curves intersect at (167, 225).
Hence, 167 is the median.
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 18F
Solution 1
Solution 2
Class having maximum frequency is the modal class.
Here, maximum frequency = 27
Hence, the modal class is 40  50.
Thus, the lower limit of the modal class is 40.
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise MCQ
Solution 1
Correct option: (d)
Range is not a measure of central tendency.
Solution 2
Correct option: (a)
Mean cannot be determined graphically.
Solution 3
Correct option: (a)
Since mean is the average of all observations, it is influenced by extreme values.
Solution 4
Correct option: (c)
Mode can be obtained graphically from a histogram.
Solution 5
Correct option: (d)
Ogives are used to determine the median of a frequency distribution.
Solution 6
Correct option: (b)
The cumulative frequency table is useful in determining the median.
Solution 7
Correct option: (b)
Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.
Solution 8
Solution 9
Solution 10
Correct option: (c)
d_{i}'s are the deviations from A of midpoints of the classes.
Solution 11
Correct option: (b)
While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.
Solution 12
Correct option: (b)
Mode = (3 x median)  (2 x mean)
Solution 13
Correct option: (c)
Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5
Solution 14
Solution 15
Correct option: (c)
Class having maximum frequency is the modal class.
Here, maximum frequency = 30
Hence, the modal class is 30  40.
Solution 16
Solution 17
Solution 18
Correct option: (c)
Mean = 8.9
Median = 9
Mode = 3Median  2Mean
= 3 x 9  2 x 8.9
= 27  17.8
= 9.2
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Correct option: (c)
For a symmetrical distribution, we have
Mean = mode = median
Solution 24
Correct option: (c)
Number of families having income more than Rs. 20000 = 50
Number of families having income more than Rs. 25000 = 37
Hence, number of families having income range 20000 to 25000 = 50  37 = 13
Solution 25
Solution 26
Correct option: (d)
Mean of 20 numbers = 0
Hence, sum of 20 numbers = 0 x 20 = 0
Now, the mean can be zero if
sum of 10 numbers is (S) and the sum of remaining 10 numbers is (S),
sum of 11 numbers is (S) and the sum of remaining 9 numbers is (S),
…….
sum of 19 numbers is (S) and the 20^{th} number is (S), then their sum is zero.
So, at the most, 19 numbers can be greater than zero.
Solution 27
Solution 28
Solution 29
(a)  (s)
The most frequent value in a data is known as mode.
(b)  (r)
Mean cannot be determined graphically.
(c)  (q)
An ogive is used to determine median.
(d)  (p)
Standard deviation is not a measure of central tendency.
Solution 30
Solution 31
Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Test Yourself
Solution 1
Correct option: (b)
The cumulative frequency table is useful in determining the median.
Solution 2
Correct option: (c)
Mean = 27
Median = 33
Mode = 3Median  2Mean
= 3 x 33  2 x 27
= 99  54
= 45
Solution 3
Solution 4
Solution 5
Solution 6
Number of athletes who completed the race in less than 14.6 seconds
= 2 + 4 + 15 + 54
= 75
Solution 7
Solution 8
The frequency table is as follows:
Classes Profit (in lakhs Rs.) 
Frequency Number of shops 
5  10 
2 
10  15 
12 
15  20 
2 
20  25 
4 
25  30 
3 
30  35 
4 
35  40 
3 
The frequency corresponding to the class 20  25 is 4.
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Marks 
Less than 10 
Less than 20 
Less than 30 
Less than 40 
Less than 50 
Number of students 
3 
11 
28 
48 
70 
We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:
Solution 14
Solution 15
Marks obtained 
Less than 20 
Less than 30 
Less than 40 
Less than 50 
Less than 60 
Less than 70 
Less than 80 
Less than 90 
Less than 100 
Number of students 
2 
7 
17 
40 
60 
82 
85 
90 
100 
We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 56.
Hence, median = 56
Solution 16
Solution 17
Solution 18
Solution 19
Mode :
Solution 20
Less Than Series:
Class interval 
Frequency 
Less than 10 
2 
Less than 15 
14 
Less than 20 
16 
Less than 25 
20 
Less than 30 
23 
Less than 35 
27 
Less than 40 
30 
We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.
More Than Series:
Class interval 
Frequency 
More than 5 
30 
More than 10 
28 
More than 15 
16 
More than 20 
14 
More than 25 
10 
More than 30 
7 
More than 35 
3 
We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.
The two curves intersect at L. Draw LM ⊥ OX.
Thus, median = OM = 16
Solution 21
Less Than Series:
Class interval 
Frequency 
Less than 45 
1 
Less than 50 
10 
Less than 55 
25 
Less than 60 
43 
Less than 65 
83 
Less than 70 
109 
Less than 75 
125 
Less than 80 
139 
Less than 85 
149 
We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.
More Than Series:
Class interval 
Frequency 
More than 40 
149 
More than 45 
148 
More than 50 
139 
More than 55 
124 
More than 60 
106 
More than 65 
66 
More than 70 
40 
More than 75 
24 
More than 80 
10 
We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.
Solution 22
Mode :