Chapter 24 : Measures of Central Tendency (Mean, Median, Quartiles and Mode)  Selina Solutions for Class 10 Maths ICSE
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Chapter 24  Measures of Central Tendency (Mean, Median, Quartiles and Mode) Excercise Ex. 24(A)
Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
(i)
(ii)
Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
(a) Here n = 9
(b)
If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63
Find the mean of the natural numbers from 3 to 12.
Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Here n = 10
(a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
(a) The mean of 7, 11, 6, 5 and 6
(b)
If we subtract 2 from each number, then the mean will be 72 = 5
If the mean of 6, 4, 7, 'a' and 10 is 8. Find the value of 'a'
No. of terms = 5
Mean = 8
Sum of numbers = 8 x 5 = 40 .(i)
But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)
From (i) and (ii)
27+a = 40
a = 13
The mean of the number 6, 'y', 7, 'x' and 14 is 8. Express 'y' in terms of 'x'.
No. of terms = 5 and mean = 8
Sum of numbers = 5 x 8 = 40 ..(i)
but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)
from (i) and (ii)
27 + y + x = 40
x + y = 13
y = 13  x
The ages of 40 students are given in the following table:
Age( in yrs) 
12 
13 
14 
15 
16 
17 
18 
Frequency 
2 
4 
6 
9 
8 
7 
4 
Find the arithmetic mean.
Age in yrs x_{i} 
Frequency (f_{i}) 
f_{i}x_{i} 
12 
2 
24 
13 
4 
52 
14 
6 
84 
15 
9 
135 
16 
8 
128 
17 
7 
119 
18 
4 
72 
Total 
40 
614 
If 69.5 is the mean of 72, 70, 'x', 62, 50, 71, 90, 64, 58 and 82, find the value of 'x'.
No. of terms = 10
Mean = 69.5
Sum of the numbers = 69.5 x 10 = 695 ..........(i)
But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82
= 619 + x ......(ii)
from (i) and (ii)
619 + x = 695
x = 76
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of 'f'.
Height (cm) 
50 
55 
58 
60 
65 
70 
71 
No. of plants 
2 
4 
10 
f 
5 
4 
3 
Height (cm) x_{i} 
No. of Plants f_{i} 
f_{i}x_{i} 
50 
2 
100 
55 
4 
220 
58 
10 
580 
60 
f 
60f 
65 
5 
325 
70 
4 
280 
71 
3 
213 
Total 
28+f 
1718 + 60f 
Mean = 60.95
From the data given below, calculate the mean wage, correct to the nearest rupee.
Category 
A 
B 
C 
D 
E 
F 
Wages (Rs/day) 
50 
60 
70 
80 
90 
100 
No. of workers 
2 
4 
8 
12 
10 
6 
(i) If the number of workers in each category is doubled, what would be the new mean wage?
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?
(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%, what would be the new mean wage?
Wages (Rs/day) (x) 
No. of Workers (f) 
fx 
50 
2 
100 
60 
4 
240 
70 
8 
560 
80 
12 
960 
90 
10 
900 
100 
6 
600 
Total 
42 
3360 
(i) Mean remains the same if the number of workers in each category is doubled.
Mean = 80
(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%
New mean =
(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then
New mean =
The contents of 100 match boxes were checked to determine the number of matches they contained.
No. of matches 
35 
36 
37 
38 
39 
40 
41 
No. of boxes 
6 
10 
18 
25 
21 
12 
8 
(i) calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches.
No. of matches (x) 
No. of boxes (f) 
fx 
35 
6 
210 
36 
10 
360 
37 
18 
666 
38 
25 
950 
39 
21 
819 
40 
12 
480 
41 
8 
328 
Total 
100 
3813 
(i)
(ii) In the second case,
New mean = 39 matches
Total contents = 39 x 100 = 3900
But total number of matches already given = 3813
Number of new matches to be added = 3900  3813 = 87
If the mean of the following distribution is 3, find the value of p.
x 
1 
2 
3 
5 
p + 4 
f 
9 
6 
9 
3 
6 
In the following table, ∑f = 200 and mean = 73. Find the missing frequencies f_{1}, and f_{2}.
x 
0 
50 
100 
150 
200 
250 
f 
46 
f_{1} 
f_{2} 
25 
10 
5 
Find the arithmetic mean (correct to the nearest wholenumber) by using stepdeviation method.
x 
5 
10 
15 
20 
25 
30 
35 
40 
45 
50 
f 
20 
43 
75 
67 
72 
45 
39 
9 
8 
6 
Find the mean (correct to one place of decimal) by using shortcut method.
x 
40 
41 
43 
45 
46 
49 
50 
f 
14 
28 
38 
50 
40 
20 
10 
Chapter 24  Measures of Central Tendency (Mean, Median, Quartiles and Mode) Excercise Ex. 24(B)
The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.
Age  Years 
16  18 
18  20 
20  22 
22 24 
2426 
No. of Students 
2 
7 
21 
17 
3 
Age in years C.I. 
x_{i} 
Number of students (f_{i}) 
x_{i}f_{i} 
16  18 
17 
2 
34 
18  20 
19 
7 
133 
20  22 
21 
21 
441 
22  24 
23 
17 
391 
24  26 
25 
3 
75 
Total 

50 
1074 
The following table gives the weekly wages of workers in a factory.
Weekly Wages (Rs) 
No. of Workers 
5055 
5 
5560 
20 
6065 
10 
6570 
10 
7075 
9 
7580 
6 
8085 
12 
8590 
8 
Calculate the mean by using:
(i) Direct Method
(ii) Short  Cut Method
(i) Direct Method
Weekly Wages (Rs) 
MidValue x_{i} 
No. of Workers (f_{i}) 
f_{i}x_{i} 
5055 
52.5 
5 
262.5 
5560 
57.5 
20 
1150.0 
6065 
62.5 
10 
625.0 
6570 
67.5 
10 
675.0 
7075 
72.5 
9 
652.5 
7580 
77.5 
6 
465.0 
8085 
82.5 
12 
990.0 
8590 
87.5 
8 
700.0 
Total 

80 
5520.00 
(ii) Short  cut method
Weekly wages (Rs) 
No. of workers (f_{i}) 
Midvalue x_{i} 
A = 72.5 d_{i}=xA 
f_{i}d_{i} 
5055 
5 
52.5 
20 
100 
5560 
20 
57.5 
15 
300 
6065 
10 
62.5 
10 
100 
6570 
10 
67.5 
5 
50 
7075 
9 
A=72.5 
0 
0 
7580 
6 
77.5 
5 
30 
8085 
12 
82.5 
10 
120 
8590 
8 
87.5 
15 
120 
Total 
80 


280 
The following are the marks obtained by 70 boys in a class test:
Marks 
No. of boys 
30  40 
10 
40  50 
12 
50  60 
14 
60  70 
12 
70  80 
9 
80  90 
7 
90  100 
6 
Calculate the mean by:
(i) Short  cut method
(ii) Step  deviation method
(i) Short  cut method
Marks 
No. of boys (f_{i}) 
Midvalue x_{i} 
A = 65 d_{i}=xA 
f_{i}d_{i} 
30  40 
10 
35 
30 
300 
40  50 
12 
45 
20 
240 
50  60 
14 
55 
10 
140 
60  70 
12 
A = 65 
0 
0 
70  80 
9 
75 
10 
90 
80  90 
7 
85 
20 
140 
90  100 
6 
95 
30 
180 
Total 
70 


270 
(ii) Step  deviation method
Marks 
No. of boys (f_{i}) 
Midvalue x_{i} 
A = 65

f_{i}u_{i} 
30  40 
10 
35 
3 
30 
40  50 
12 
45 
2 
24 
50  60 
14 
55 
1 
14 
60  70 
12 
A = 65 
0 
0 
70  80 
9 
75 
1 
9 
80  90 
7 
85 
2 
14 
90  100 
6 
95 
3 
18 
Total 
70 


27 
Here A = 65 and h = 10
Find mean by step  deviation method:
C. I. 
6370 
7077 
7784 
8491 
9198 
98105 
105112 
Freq 
9 
13 
27 
38 
32 
16 
15 
C. I. 
Frequency (f_{i}) 
Midvalue x_{i} 
A = 87.50

f_{i}u_{i} 
63  70 
9 
66.50 
3 
27 
70  77 
13 
73.50 
2 
26 
77  84 
27 
80.50 
1 
27 
84  91 
38 
A = 87.50 
0 
0 
91  98 
32 
94.50 
1 
32 
98  105 
16 
101.50 
2 
32 
105  112 
15 
108.50 
3 
45 
Total 
150 


29 
Here A = 87.50 and h = 7
The mean of the following frequency distribution is . Find the value of 'f'.
C. I. 
0  10 
10  20 
20  30 
30  40 
40  50 
freq 
8 
22 
31 
f 
2 
C. I. 
frequency 
Midvalue (x_{i}) 
f_{i}x_{i} 
010 
8 
5 
40 
1020 
22 
15 
330 
2030 
31 
25 
775 
3040 
f 
35 
35f 
4050 
2 
45 
90 
Total 
63+f 

1235+35f 
Using stepdeviation method, calculate the mean marks of the following distribution.
C.I 
5055 
5560 
6065 
6570 
7075 
7580 
8085 
8590 
Frequency 
5 
20 
10 
10 
9 
6 
12 
8 
Let the assumed mean A= 72.5
C.I 
f_{i} 
Mid value (x_{i}) 
d_{i}=x_{i} ; A 
f_{i}d_{i} 
5055 
5 
52.5 
20 
100 
5560 
20 
57.5 
15 
300 
6065 
10 
62.5 
10 
100 
6570 
10 
67.5 
5 
50 
7075 
9 
72.5 
0 
0 
7580 
6 
77.5 
5 
30 
8085 
12 
82.5 
10 
120 
8590 
8 
87.5 
15 
120 
Total 
80 

280 

Using the information given in the adjoining histogram, calculate the mean.
C.I. 
Frequency 
Mid value x 
fx 
1525 
10 
20 
200 
2535 
20 
30 
600 
3545 
25 
40 
1000 
4555 
15 
50 
750 
5565 
5 
60 
300 
Total 
75 

2850 
If the mean of the following observations is 54, find the value of 'p'.
Class 
0  20 
20  40 
40  60 
60  80 
80  100 
Frequency 
7 
p 
10 
9 
13 
Class 
Frequency (f) 
Mid Value (x) 
fx 
0  20 
7 
10 
70 
20  40 
p 
30 
30p 
40  60 
10 
50 
500 
60  80 
9 
70 
630 
80  100 
13 
90 
1170 
Total 
39 + p 
2370 + 30p 
Here mean = 54 ..(ii)
from (i) and (ii)
The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f_{1} and f_{2}.
Class 
020 
2040 
4060 
6080 
80100 
100120 
Freq 
5 
f_{1} 
10 
f_{2} 
7 
8 
Class 
Freq (f) 
Mid value 
fx 
020 
5 
10 
50 
2040 
f_{1} 
30 
30f_{1} 
4060 
10 
50 
500 
6080 
f_{2} 
70 
70f_{2} 
80100 
7 
90 
630 
100120 
8 
110 
880 
Total 
30+f_{1}+f_{2} 

2060+30f_{1}+70f_{2} 
Now, and
from (i)
using (i) and (ii)
Calculate the mean of the distribution, given below, using the short cut method :
Mark 
1120 
2130 
3140 
4150 
5160 
6170 
7180 
No. of students 
2 
6 
10 
12 
9 
7 
4 
Chapter 24  Measures of Central Tendency (Mean, Median, Quartiles and Mode) Excercise Ex. 24(C)
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
The middle term is 4 which is the 5^{th} term.
Median = 4
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
Arranging the given data in descending order:
28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5
The middle terms are 24 and 24, 5^{th} and 6^{th} terms
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
Arranging in ascending order:
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37
(i) Middle term is 10^{th} term i.e. 29
Median = 29
(ii) Lower quartile =
(iii) Upper quartile =
(iv) Interquartile range = q_{3}  q_{1} =35  26 = 9
From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Interquartile range
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Arrange in ascending order:
0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95
(i) Median is the mean of 8^{th} and 9^{th} term
(ii) Upper quartile =
(iii) Interquartile range =
The ages of 37 students in a class are given in the following table:
Age (in years) 
11 
12 
13 
14 
15 
16 
Frequency 
2 
4 
6 
10 
8 
7 
Find the median.
Age (in years) 
Frequency 
Cumulative Frequency 
11 
2 
2 
12 
4 
6 
13 
6 
12 
14 
10 
22 
15 
8 
30 
16 
7 
37 
Number of terms = 37
Median =
Median = 14
The weight of 60 boys are given in the following distribution table:
Weight (kg) 
37 
38 
39 
40 
41 
No. of boys 
10 
14 
18 
12 
6 
Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
Weight (kg) x 
no. of boys f 
cumulative frequency 
37 
10 
10 
38 
14 
24 
39 
18 
42 
40 
12 
54 
41 
6 
60 
Number of terms = 60
(i) median = the mean of the 30^{th} and the 31^{st} terms
(ii) lower quartile (Q_{1}) =
(iii) upper quartile (Q_{3}) =
(iv) Interquartile range = Q_{3}  Q_{1 }= 40  38 = 2
Estimate the median for the given data by drawing an ogive:
Class 
010 
1020 
2030 
3040 
4050 
frequency 
4 
9 
15 
14 
8 
Class 
Frequency 
Cumulative Frequency 
010 
4 
4 
1020 
9 
13 
2030 
15 
28 
3040 
14 
42 
4050 
8 
50 
Number of terms = 50
Through mark of 25.5 on the yaxis, draw a line parallel to xaxis which meets the curve at A. From A, draw a perpendicular to xaxis, which meets xaxis at B.
The value of B is the median which is 28.
By drawing an ogive, estimate the median for the following frequency distribution:
Weight (kg) 
1015 
1520 
2025 
2530 
3035 
No. of boys 
11 
25 
12 
5 
2 
Weight (kg) 
No. of boys 
Cumulative Frequency 
1015 
11 
11 
1520 
25 
36 
2025 
12 
48 
2530 
5 
53 
3035 
2 
55 
Number of terms = 55
Through mark of 28 on the yaxis, draw a line parallel to xaxis which meets the curve at A. From A, draw a perpendicular to xaxis, which meets xaxis at B.
The value of B is the median which is 18.4 kg
From the following cumulative frequency table, find:
(i) median
(ii) lower quartile
(iii) upper quartile
Marks(less than) 
10 
20 
30 
40 
50 
60 
70 
80 
90 
100 
Cumulative frequency 
5 
24 
37 
40 
42 
48 
70 
77 
79 
80 
Marks (less than) 
Cumulative frequency 
10 
5 
20 
24 
30 
37 
40 
40 
50 
42 
60 
48 
70 
70 
80 
77 
90 
79 
100 
80 
Number of terms = 80
\Median = 40^{th} term.
(i) Median = Through 40^{th} term mark draw a line parallel to the xaxis which meets the curve at A. From A, draw a perpendicular to xaxis which meets it at B.
Value of B is the median = 40
(ii) Lower quartile (Q_{1}) = 20^{th} term = 18
(iii) Upper Quartile (Q_{3}) = 60^{th} term = 66
In a school, 100 pupils have heights as tabulated below:
Height (in cm) 
No. of pupils 
121  130 
12 
131  140 
16 
141  150 
30 
151  160 
20 
161  170 
14 
171  180 
8 
Find the median height by drawing an ogive.
Height (in cm) 
No. of pupils 
Cumulative Frequency 
121  130 
12 
12 
131  140 
16 
28 
141  150 
30 
58 
151  160 
20 
78 
161  170 
14 
92 
171  180 
8 
100 
Number of terms = 100
Through 50^{th} term mark draw a line parallel to the xaxis which meets the curve at A. From A, draw a perpendicular to xaxis which meets it at B.
Value of B is the median = 148
Median height = 148cm
Chapter 24  Measures of Central Tendency (Mean, Median, Quartiles and Mode) Excercise Ex. 24(D)
Find the mode of the following data:
(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6
(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8
(i) Mode = 7
Since 7 occurs 4 times
(ii) Mode = 11
Since it occurs 4 times
The following table shows the frequency distribution of heights of 50 boys:
Height (cm) 
120 
121 
122 
123 
124 
Frequency 
5 
8 
18 
10 
9 
Find the mode of heights.
Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.
Find the mode of following data, using a histogram:
Class 
010 
1020 
2030 
3040 
4050 
Frequency 
5 
12 
20 
9 
4 
Mode is in 2030, because in this class there are 20 frequencies.
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:
Expenditure (Rs) 
No. of students 
2025 
4 
2530 
7 
3035 
23 
3540 
18 
4045 
6 
4550 
2 
Mode is in 3035 because it has the maximum frequency.
Find the median and mode for the set of numbers:
2, 2, 3, 5, 5, 5, 6, 8 and 9
which is 5.
Mode = 5 because it occurs maximum number of times.
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
(i) What are his modal marks?
(ii) What are his median marks?
(iii) What are his total marks?
(iv) What are his mean marks?
Arranging the given data in ascending order:
7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19
(i) Mode = 16 as it occurs maximum number of times.
(ii)
(iii)Total marks = 7+10+12+12+14+15+16+16+16+17+19 =
154
(iv)
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
(i)
(ii) Median = mean of 8^{th} and 9^{th} term
(iii) Mode = 5 as it occurs maximum number of times.
At a shooting competition the score of a competitor were as given below:
Score 
0 
1 
2 
3 
4 
5 
No. of shots 
0 
3 
6 
4 
7 
5 
(i) What was his modal score?
(ii) What was his median score?
(iii) What was his total score?
(iv) What was his mean score?
Score x 
No. of shots f 
fx 
0 
0 
0 
1 
3 
3 
2 
6 
12 
3 
4 
12 
4 
7 
28 
5 
5 
25 
Total 
25 
80 
(i) Modal score = 4 as it has maximum frequency 7.
(ii)
(iii) Total score = 80
(iv)
Chapter 24  Measures of Central Tendency (Mean, Median, Quartiles and Mode) Excercise Ex. 24(E)
The following distribution represents the height of 160 students of a school.
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
 The median height.
 The interquartile range.
 The number of students whose height is above 172 cm.
Taking Height of student along xaxis and cumulative frequency along yaxis we will draw an ogive.
(i)
Through mark for 80, draw a parallel line to xaxis which meets the curve; then from the curve draw a vertical line which meets the xaxis at the mark of 157.5.
(ii)Since, number of terms = 160
(iii)Through mark for 172 on xaxis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the yaxis at the mark of 145.
The number of students whose height is above 172 cm
= 160  144 = 16
Draw ogive for the data given below and from the graph determine: (i) the median marks.
(ii) the number of students who obtained more than 75% marks.
Marks 
10  19 
20 29 
30  39 
40  49 
50  59 
60  69 
70  79 
80  89 
90  99 
No. of students 
14 
16 
22 
26 
18 
11 
6 
4 
3 
Marks 
No. of students 
Cumulative frequency 
9.5  19.5 
14 
14 
19.5  29.5 
16 
30 
29.5  39.5 
22 
52 
39.5  49.5 
26 
78 
49.5  59.5 
18 
96 
59.5  69.5 
11 
107 
69.5  79.5 
6 
113 
79.5  89.5 
4 
117 
89.5  99.5 
3 
120 
Scale:
1cm = 10 marks on X axis
1cm = 20 students on Y axis
(i)
Through mark 60, draw a parallel line to xaxis which meets the curve at A. From A, draw a perpendicular to xaxis meeting it at B.
The value of point B is the median = 43
(ii) Total marks = 100
75% of total marks = marks
The number of students getting more than 75% marks = 120  111 = 9 students.
Mean of 1, 7, 5, 3, 4 and 4 =
m=4
Now, mean of 3, 2, 4, 2, 3, 3 and p = m1 = 41 = 3
Therefore, 17+p = 3 x n …. Where n = 7
17+p = 21
p = 4
Arranging in ascending order:
2, 2, 3, 3, 3, 3, 4, 4
Mean = 4^{th} term = 3
Therefore, q = 3
In a malaria epidemic, the number of cases diagnosed were as follows:
Date July 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
Num 
5 
12 
20 
27 
46 
30 
31 
18 
11 
5 
0 
1 
On what days do the mode and upper and lower quartiles occur?
Date 
Number 
C.f. 
1 
5 
5 
2 
12 
17 
3 
20 
37 
4 
27 
64 
5 
46 
110 
6 
30 
140 
7 
31 
171 
8 
18 
189 
9 
11 
200 
10 
5 
205 
11 
0 
205 
12 
1 
206 
(i) Mode = 5^{th} July as it has maximum frequencies.
(ii) Total number of terms = 206
Upper quartile =
Lower quartile =
The income of the parents of 100 students in a class in a certain university are tabulated below.
Income (in thousand Rs) 
08 
816 
1624 
2432 
3240 
No. of students 
8 
35 
35 
14 
8 
(i) Draw a cumulative frequency curve to estimate the median income.
(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.
(iii) Calculate the Arithmetic mean.
We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at B.
Through B, a vertical line is drawn which meets OX at M.
OM = 17.6 units
Hence, median income = 17.6 thousands
The marks of 20 students in a test were as follows:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.
Calculate:
(i) the mean (ii) the median (iii) the mode
Arranging the terms in ascending order:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20
Number of terms = 20
(i)
(ii)
(iii) Mode = 15 as it has maximum frequencies i.e. 3
The marks obtained by 120 students in a mathematics test is given below:
Marks 
No. of students 
010 
5 
1020 
9 
2030 
16 
3040 
22 
4050 
26 
5060 
18 
6070 
11 
7080 
6 
8090 
4 
90100 
3 
Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:
(i)the median
(ii)the number of students who obtained more than 75% in test.
(iii)the number of students who did not pass in the test if the pass percentage was 40.
(iv)the lower quartile
Marks 
No. of students 
c.f. 
010 
5 
5 
1020 
9 
14 
2030 
16 
30 
3040 
22 
52 
4050 
26 
78 
5060 
18 
96 
6070 
11 
107 
7080 
6 
113 
8090 
4 
117 
90100 
3 
120 
(i)
Through mark 60.5, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis meeting it at B.
The value of point B is the median = 43
(ii) Number of students who obtained up to 75% marks in the test = 110
Number of students who obtained more than 75% marks in the test = 120  110 = 10
(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x=40, y=52)
(iv) Lower quartile = Q_{1} =
Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.
Weight 
Frequency 
4045 
5 
4550 
17 
5055 
22 
5560 
45 
6065 
51 
6570 
31 
7075 
20 
7580 
9 
Use your ogive to estimate the following:
(i) The percentage of students weighing 55 kg or more
(ii) The weight above which the heaviest 30% of the student fall
(iii) The number of students who are
(a) underweight
(b) overweight,
if 55.70 kg is considered as standard weight.
Weight 
Frequency 
C. f. 
4045 
5 
5 
4550 
17 
22 
5055 
22 
44 
5560 
45 
89 
6065 
51 
140 
6570 
31 
171 
7075 
20 
191 
7580 
9 
200 
(i) Number of students weighing more than 55 kg = 20044 = 156
Therefore, percentage of students weighing 55 kg or more
(ii) 30% of students =
Heaviest 60students in weight = 9 + 21 + 30 = 60
weight = 65 kg ( from table)
(iii) (a) underweight students when 55.70 kg is standard = 46 (approx) from graph
(b) overweight students when 55.70 kg is standard = 200 55.70 = 154 (approx) from graph
The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
Marks obtained 
5 
6 
7 
8 
9 
10 
No. of students 
3 
9 
6 
4 
2 
1 
Marks obtained(x) 
No. of students (f) 
c.f. 
fx 
5 
3 
3 
15 
6 
9 
12 
54 
7 
6 
18 
42 
8 
4 
22 
32 
9 
2 
24 
18 
10 
1 
25 
10 
Total 
25 
171 
Number of terms = 25
(i) Mean =
(ii)
(iii) Mode = 6 as it has maximum frequencies i.e. 6
The mean of the following distribution is 52 and the frequency of class interval 3040 is 'f'. Find f.
C.I 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
Freq 
5 
3 
f 
7 
2 
6 
13 
C.I. 
Frequency(f) 
Mid value (x) 
fx 
1020 
5 
15 
75 
2030 
3 
25 
75 
3040 
f 
35 
35f 
4050 
7 
45 
315 
5060 
2 
55 
110 
6070 
6 
65 
390 
7080 
13 
75 
975 
Total 
36+f 
1940+35f 
The monthly income of a group of 320 employees in a company is given below:
Monthly Income (thousands) 
No. of employees

67 
20 
78 
45 
89 
65 
910 
95 
1011 
60 
1112 
30 
1213 
5 
Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine :
(i) the median wage.
(ii) number of employees whose income is below Rs 8500.
(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Monthly Income (thousands) 
No. of employees (f) 
Cumulative frequency 
67 
20 
20 
78 
45 
65 
89 
65 
130 
910 
95 
225 
1011 
60 
285 
1112 
30 
315 
1213 
5 
320 
Total 
320 
Number of employees = 320
(i)
Through mark 160, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis meeting it at B.
The value of point B is the median = Rs 9.3 thousands
(ii) The number of employees with income below Rs 8500 = 95 (approx from the graph)
(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)
Therefore number of employees (senior employees) = 320305 =15
(iv) Upper quartile =
A mathematics aptitude test of 50 students was recorded as follows:
Marks 
No. of students

5060 
4 
6070 
8 
7080 
14 
8090 
19 
90100 
5 
Draw a histogram for the above data using a graph paper and locate the mode.
(i)Draw the histogram
(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.
(iii) From P, draw a perpendicular to xaxis meeting at Q.
(iv) Value of Q is the mode = 82 (approx)
Marks obtained by 200 students in an examination are given below:
Marks 
No. of students

010 
5 
1020 
11 
2030 
10 
3040 
20 
4050 
28 
5060 
37 
6070 
40 
7080 
29 
8090 
14 
90100 
6 
Draw an ogive of the given distribution on a graph paper taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph:
(i) the median marks.
(ii) number of students who failed if minimum marks required to pass is 40
(iii) if scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Marks 
No. of students

Cumulative frequency 
010 
5 
5 
1020 
11 
16 
2030 
10 
26 
3040 
20 
46 
4050 
28 
74 
5060 
37 
111 
6070 
40 
151 
7080 
29 
180 
8090 
14 
194 
90100 
6 
200 
Number of students = 200
(i)
Through mark 100, draw a parallel line to xaxis which meets the curve at A, From A draw a perpendicular to xaxis meeting it at B.
The value of point B is the median = 57 marks (approx)
(ii) The number of students who failed (if minimum marks required to pass is 40)= 46 (approx from the graph)
(iii) The number of students who secured grade one in the examination = 200  188 = 12 (approx from the graph)
The marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.
If mean of the distribution is 7.2, find a and b.
Find the mode and the median of the following frequency distribution.
Since the frequency for x = 14 is maximum.
So Mode = 14.
According to the table it can be observed that the value of x from the 13^{th} term to the 17^{th} term is 13.
So the median = 13.
The median of the observations 11, 12, 14, (x  2) (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
The number 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
(Use a graph paper for this question). The daily pocket expenses of 200 students in a school are given below :
Pocket expenses (in Rs) 
05 
510 
1015 
1520 
2025 
2530 
3035 
3540 
No. of students (frequency) 
10 
14 
28 
42 
50 
30 
14 
12 
Draw a histogram representing the above distribution and estimate the mode from the graph.
Histogram is as follows:
In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.
From E, draw a perpendicular to xaxis meeting at L.
Value of L is the mode. Hence, mode = 21.5
The marks obtained by 100 students in a mathematics test are given below :
Marks 
010 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
8090 
90100 
No. of students 
3 
7 
12 
17 
23 
14 
9 
6 
5 
4 
Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both the axes.
Use the ogive to estimate :
(i) Median
(ii) Lower quartile
(iii) Number of students who obtained more than 85% marks in the test.
(iv) Number of students failed, if the pass percentage was 35.
Marks 
Number of students (Frequency) 
Cumulative Frequency 
010 
3 
3 
1020 
7 
10 
2030 
12 
22 
3040 
17 
39 
4050 
23 
62 
5060 
14 
76 
6070 
9 
85 
7080 
6 
91 
8090 
5 
96 
90100 
4 
100 
The ogive is as follows:
The mean of following numbers is 68. Find the value of 'x'.
45, 52, 60, x, 69, 70, 26, 81 and 94.
Hence, estimate the median.
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Here the number of observations i. e is 10, which is even.'
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