Solution

STATISTICS MEDIAN 2M AND 5M

Class 10 - Mathematics
Section A
1. Arrange the given data in ascending order
73, 85, 89, 94, 94, 100, 104, 105, 120, 133
n = 10(even)
th th
n n
value+ ( +1) value

Median =
2 2
∴
2
th th
10 10
value+ ( +1) value

=
2 2

5
th
value + 6 th
value
= 2
90+104
= 2
= 97

r
2. The median class of a frequency distribution is 125-145, so l = 25, h = 20 n = 78

atu
The frequency f = 20 and cumulative frequency of the class preceding the median class cf = 22
median = 137
n
−cf

Median = l + ( 2

f
) × h

yL
n
−22

or, 120 + (
2
) × 20 = 137
20

n
− 22 = 137 − 120
2
n
= 17 + 22
2
em
n
= 39
2

n = 78
Hence, the required sum of frequencies = 78
Marks(xi) Frequency(fi) fixi
ad

3.
15 5 75
Ac

20 8 106

22 11 242

24 20 480

25 23 575
V²

30 18 540

33 13 429
M

38 3 114

45 1 45

Total 102 2660

9 9

Here, n = 9, ∑ f = 102, ∑ fi xi = 2660 i

i=l i=l

∑fi xi

i=l 2660
∴ x
¯¯¯
¯
=
9
=
102
= 26.078 = 26.08
∑ fi
i=l

xi fi Cumulative Frequency
4.
1 8 8

2 10 18

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 3 11 29

4 16 45

5 20 65

6 25 90

7 15 105

8 9 114

9 5 119

n = 119 (odd)
n+1
Median is the ( 2
) th Observation
119+1
=( 2
)

= 60th Observation.
From Cumulative Frequencies, we get 60th Observation as 5.

r
Therefore, Median = 5

atu
5. Table:
Marks Frequency c.f.

30 10 10

yL
40 4 14

50 12 26

60 6 32
em
70 1 33

80 8 41
ad

90 4 45
N = 45

45+1
Median = ( th
Ac

)
2

i.e, 23rd Observation

Median = 50
6. It is given that the mean of 100 observations is 50 . The value of the largest observation is 100.
Σf x
Mean =
V²

Σf

Σf x
⇒ 50 = 100

⇒ Σf x = 5000
M

It was later found that it is 110 not 100.

Correct, Σf x = 5000 + 110 - 100 = 5010
′

′
Σf x
Therefore, Correct Mean = Σf
=
5010

100
= 50.1
It is given that the median of 100 observations is 52.
Median will remain same i.e. = 52

7. Marks Obtained Frequency c.f.

0 - 10 8 8

10 - 20 10 18

20 - 30 12 30

30 - 40 22 52

40 - 50 30 82

50 - 60 18 100

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 n = 100 ⇒ n

2
= 50 and cumulative frequency just greater than 50 is 52
Therefore, Median Class is 30 - 40.

8. Class Frequency, f Cumulative Frequency

0 - 10 12 12

10 - 20 8 20

20 - 30 8 28

30 - 40 15 43

40 - 50 3 46

∑f = n = 46
From table we get n = 46. So,
n

2
= = 23
46

Now 23 is close to 28

r
∴ median class is 20 - 30.

atu
9. Classes Frequency c.f.

5-7 55 55

7-9 55 110

yL
9 -11 70 180

11 - 13 150 330

13 - 15 86 416
em
15 - 17 84 500
N

2
=
500

2
= 250
Median class = 11 - 13
ad

∴ Lower Limit of the median class = 11

10. Class Interval Frequency C.F

Ac

0 - 10 5 5

10 - 20 8 13

20 - 30 7 20
V²

30 - 40 12 32

40 - 50 28 60
M

50 - 60 20 80

60 - 70 10 90

70 - 80 10 100

Total N = Σ fi = 100

Here N = 100, then N

2
= 50
which lies in the class 40 - 50 ...(∵ 32 < 50 < 60)
∴ Required Median class interval is 40 - 50

11. Lives in hours of 15 Pieces are

= 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719
Arrange the above data in ascending order
694, 696, 699, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745
N = 15(odd)
N +1
Median = ( 2
) th term

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 15+1
=( 2
) th term
= 8th term
= 716
12. Arrange the value of x is ascending order and then forming c.f table
X F C.F

2 13 13

3 7 20

5 8 28

6 9 37

7 12 49

9 14 63

10 11 74

tur
here, total frequency = 74 which is even, N

2
=
74

2
= 37 and N

2
+ 1 = 38
The values of 37th and 38th item are 6 and 7

La
∴ median = (6 + 7)
1

= 6.5
13. Total number of observations = N = 19 (odd)
th

observation = Value of 10th observation

N +1
⇒ Median = Value of ( )

given, median = 30
2

y
em
⇒ 10th observation is 30.
Now, two values 8 and 32 are added.
Since 8 < 30 and 32 > 30, each one of these two will go on either side of median.
Hence, median is not affected.
ad

⇒ Median = 30

14. Class Interval Frequency Cumulative frequency

Ac

0 - 10 4 4

10 - 20 4 8

20 - 30 8 16
V²

30 - 40 10 26

40 - 50 12 38

50 - 60 8 46
M

60 - 70 4 50
Here, N = 50 ⇒ N

2
= 25
The cumulative frequency just greater than 25 is 26.
Hence, median class is 30 - 40

15. Family size 1-3 3-5 5-7 7-9 9 - 11

Number of families 7 8 2 2 1

C.f. 7 15 17 19 20
Median class 3 - 5
N
−C

Median = l + 2

f
× h
10−7
=3+ 8
× 2
= 3.75

16. Daily wages(in Rs) Number of workers Cumulative Frequency

4 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 125 6 6

130 20 26

135 24 50

140 28 78

145 15 93

150 4 97

160 2 99

180 1 100

Total = 100
n = 100
100 100+2
Median is the mean of ( 2
) th and ( 2
th observations, i.e, 50th and 51th Observations.
)

r
135+140
Median = = 137.5

atu
2

Median wage of a worker in the factory is Rs 137.50

Classes fi c.f.
17.
5 4 4

yL
9 5 9

10 6 15

12 12 27
em
13 11 38

16 6 44
ad

18 4 48

20 2 50
Ac

n
n = 50 ⇒ = 25
2

25 is nearer to 27 cumulative frequency.

Therefore, Median is 12.
18. Median = Mode + (Mean - Mode)2

= 15 + 2
(30 - 15)
V²

= 15 + 2

3
× 15
= 15 + 10
= 25
M

19. The empirical relation between the three modes of central tendency is :
Mode = 3Median - 2 Mean
Since (Mode = 175, Mean = 169)
Therefore,
⇒ 175 = 3Median - 2(169)

⇒ 175 = 3 Median - 338

⇒ 175 + 338 = 3 Median

⇒ 513 = 3 Median

⇒ Median =
513

⇒ Median = 171
Hence, Median of the distribution is 171.

20. Class Interval Frequency

0 - 10 9

10 - 20 11

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 20 - 30 10

30 - 40 8

40 - 50 8

50 - 60 3
Here, modal class = 10 - 20
Upper limit of modal class = 20
21. c = 5 + 7 = 12
a = 18 - 12 = 6
d = 18 + 5 = 23
b = 30 - 23 = 7
22. The given data is an inclusive series. So, we convert it into an exclusive form, as given below.
Class 159.5 - 162.5 162.5 - 165.5 165.5 - 168.5 168.5 - 171.5 171.5 - 174.5

Frequency 15 118 142 127 18

tur
Clearly, the modal class is 165.5 - 168.5 as it has the maximum frequency.
∴ xk = 165.5, h = 3, fk = 142, fk-1 = 118, fk+1 = 127

La
(f −f )

Mode, Mo = x
k k−1

k + {h × }
(2f −f −f )
k k−1 k+1

(142−118)
= 165.5 + {3 × }
(2×142−118−127)

3×24
= 165.5 + { }

= 165.5 +
24

13
39

y
em
= 165.5 + 1.85
= 167.35 cm
This means that height of maximum number of players in the school is 167.35 cm(approx.).
23. Arranging the data in ascending order, we have
ad

33, 35, 39, 40, 41, 42, 65, 68, 69, 72

Number of items (N) = 10 which is even.
th th
N N
Size of ( ) item+Size of ( +1) item
Ac

2 2
∴ M edian =
2
th th
Size of 5 item+Size of 6 item
∴=
2
41+42
=
2
= 41.5 .
24. Given, mode = 36 , median = 43
V²

We know that, relation between mean , median & mode is:-

3 median = mode + 2 mean.......(1)
Or, -2 mean = mode - 3 median
M

Or, -2 mean = 36 - 3 × 43​

Or, - 2 mean = 36 - 129
Or, - 2 mean = - 93
Thus, Mean = 46.5

25. Marks Frequency c.f.

5 4 4

7 3 7

9 2 9

10 8 17

12 4 21

14 4 25

16 3 28

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 17 2 30

19 3 33

20 1 34
N = 34
Median = average of 17th and 18th Observation
10+12
=
2

= 11
Section B
Marks Number of Students
Cumulative frequency
26. (Class) (Frequency)

30-35 14 14

35-40 16 30

r
40-45 18 48

atu
45-50 23 71 (Median class)

50-55 18 89

55-60 8 97

yL
60-65 3 100
Here, N = 100
N
Therefore, 2
= 50, This observation lies in the class 45-50.
em
l (the lower limit of the median class) = 45
cf (the cumulative frequency of the class preceding the median class) = 48
f (the frequency of the median class) = 23
h (the class size) = 5
ad

n
−cf

Median = l + (
2
)h
f

50−48
= 45 + ( ) × 5
Ac

23

10
= 45 + = 45.4
23

So, the median percentage of marks is 45.4.

27. First we prepare the following cumulative table to compute the median
Class Frequency Cumulative frequency
V²

5-10 5 5

10-15 6 11
M

15-20 15 26

20-25 10 36

25-30 5 41

30-35 4 45

35-40 2 47

40-45 2 49

N = 49
We have, N=49
∴ =
N

2
= 24.5
49

The cumulative frequency just greater than N/2 is 26 and the corresponding class is 15 - 20. Thus, 15 - 20 is the median class such
that

7 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 l = 15, f = 15, F = 11 and h = 5
N
−F
24.5−11 13.5
Median = 19.5
2
∴ = l + × h = 15 + × 5 = 15 +
f 15 3

28. Given:
Median value = 33
To find: M
Since the media is 33, it falls in the median class of 30-45
The formula for median= l + ( − cf ) ÷ f × h n

here, l = lower limit

n = total frequency.
cf = cumulative frequency of media class
h = class width
we know from the data,l= 30,
h = 15,
f = 40,

r
cf = 52, and

atu
n (112+m)
=
2 2

Putting the values in the equation, we get:

30 + [(112 + ) - 52] ÷ 40 × 15 = 33
m

2
(3×80)
(8 + m) =

yL
15

8 + m= 16
m=8

29. Class interval (Indusive) Class interval (Exclusive) Frequency Cumulative frequency
em
160-162 159.5-162.5 15 15

163-165 162.5-165.5 118 133 = F

166-168 165.5-168.5 142 = f 275

ad

169-171 168.5-171.5 127 402

172-174 171.5-174.5 18 420

Ac

N = 420
N = 420
N 420
∴
2
=
2
= 210
The cumulative frequency just greater than N
is 275.
V²

so, 165.5-168.5 is the median class.

l = 165.5, f = 142, F = 133 ,h = 168.5 - 165.5 = 3
N
−F
M

∴ Median = l + 2

f
× h
210−133
= 165.5 + 142
× 3
= 165.5 + 77×3

142

= 165.5 + 1.63
= 167.13
30. Converting given distribution continuous by subtracting 1

2
from lower limit and adding 1

2
to upper limit.
Table:
C.I. Frequency Cumulative Frequency

129.5 - 139.5 4 4

139.5 - 149.5 9 13

149.5 - 159.5 18 31

159.5 - 169.5 28 59

169.5 - 179.5 24 83

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Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 179.5 - 189.5 10 93

189.5 - 199.5 7 100

n
n = 100 ⇒ = 50
2

Median class = 159.5 − 169.5,

l = 159.5, c. f . = 31, f = 28, h = 10
n
−c.f .

We know that, Median = l + 2

f
h

50−31
= 159.5 + ( ) × 10
28

19
= 159.5 + × 10
28

= 166.285

= 166.3

31. The frequency distribution table for calculations of median is as follow:

Class Interval Frequency (f) Cumulative frequency

40-45 9 9

r
atu
45-50 5 14

50-55 8 22

55-60 9 31

yL
60-65 6 37

65-70 3 40
We know,
em
N
−cf

Median = l + {h × 2

f
}

Here,
l denotes lower limit of median class
ad

h denotes width of median class

f denotes frequency of median class
cf denotes cumulative frequency of the class preceding the median class
Ac

N denotes sum of frequency

N

2
= 20
cumulative frequency just greater than 20 is 22
Therefore, median class is 50-55
V²

Median class is 50 - 55
So,
l = 50,
M

h = 5,
f = 8,
cf = cf of preceding class = 14
N

2
= 20
O substituting all the above values in the formula, we get
N
−cf

Median = l + {h × 2

f
}

20−14
Median = 50 + {5 × 8
}

6
Median = 50 + {5 × 8
}

Median = 50 + {5 × 3

4
}

Median = 50 + { 15

4
}

Median = 50 + 3.75
⇒ Median = 53.75

9 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 Speed (in km/h) (C.I.) No. of players (fi) c.f.
32.
85-100 11 11

100-115 9 20

115-130 8 28

130-145 5 33

N = 33
N = No. of observations = 33.
N 33
= = 16.5
2 2

16.5 lies in class 100 - 115.

so, class 100-115 is the median class.
l = 100, f = 9, c.f. = 11, h = 115 - 100 = 15
N
( −c⋅f )h

Median= l +
2

r
f

33

atu
( −11)15
2
= 100 +
9
(16.5−11)15
= 100 +
9
5.5×15
= 100 +
9

yL
82.5
= 100 +
9

= 100 + 9.166
= 109.166.
Therefore, the median bowling speed is 109.166 km/h
em
33. Classes Number of students Cumulative Frequency

0 - 10 2 2

10 - 20 12 14
ad

20 - 30 22 36

30 - 40 8 44
Ac

40 - 50 6 50
n = 50,
n

2
=
50

2
= 25
Median Class = 20 - 30
V²

l = 20, f = 22, c.f. = 14,h = 10

n
( −c.f .)

Median = l +
2
× h
f
M

(25−14)
= 20 + × 10
22
11
= 20 + × 10
22

= 20 + 5
= 25

34. Class Interval Frequency Cumulative Frequency

0 - 10 6 6

10 - 20 16 22

20 - 30 30 52

30 - 40 9 61

40 - 50 4 65
Here, N = 65 ⇒ N

2
= 32.5

The cumulative frequency just greater than 32.5 is 52.

10 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 Hence, median class is 20 - 30.
∴ l = 20, h = 10, f = 30, cf = cf of preceding class = 22
N
( −cf )

Now, Median = l + {h ×
2
}
f

(32.5−22)
= 20 + {10 × }
30

10.5
= 20 + {10 × }
30

= 20 + 3.5
= 23.5
Thus, the median of the data is 23.5.

35. Class interval Frequency Cumulative frequency

15-25 8 8

25-35 10 18

35-45 15 33

r
atu
45-55 25 58(F)

55-65 40(f) 98

65-75 20 118

yL
75-85 15 133

85-95 7 140

N = 140
em
N = 140
N 140
∴ = = 70
2 2

The cumulative frequency just greater than N

2
is 98.
ad

∴ median class is 55-65 .

l = 55, f = 40, F = 58, h = 65 − 55 = 10
N
−F

∴ Median = l +
2
× h
Ac

70−58
= 55 + × 10
40

= 55 + 3

= 58

36. Let f1 and f2 be the frequencies of class intervals 0 - 10 and 40 - 50.

V²

f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40
⇒ f1 + f2 = 9

Median is 32.5 which lies in 30 - 40, so the median class is 30 - 40.

M

l = 30, h = 10, f = 12, N = 40 and c = f1 + 5 + 9 = (f1 + 14)

N
( −c)

Now, median = l + [h ×
2
]
f

20− f −14
1
⇒ 32.5 = [30 + (10 × )]
12

6−f
1
= [30 + (10 × )]
12

30−5f1
= [30 + ( )]
6

30−5f
1
= 2.5
6

30 - 5f1 = 15
5f1 = 15 ⇒ f 1 = 3

f1 = 3 and f2 = (9 - 3) = 6
37. a. According to the question,
Class interval Frequency Cumulative Frequency

11 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 10.5 - 20.5 141 141

20.5 - 30.5 221 362

30.5 - 40.5 439 801

40.5 - 50.5 529 1330

50.5 - 60.5 495 1825

60.5 - 70.5 322 2147

70.5 - 80.5 153 2300

b. N = 2300 ⇒ N

2
= 1150

cumulative frequency just greater than 1150 is 1330.

therefore, median class is 40.5 − 50.5
40.5+50.5 91
Class Mark of median class = 2
=
2
= 45.5

c. According to table,

r
maximum frequency = 529

atu
therefore, modal class is 40.5 − 50.5
the cumulative frequency of modal class is 1330.

Weight Cumulative
Number of students
38. (in kg) frequency

yL
40-45 2 2
45-50 3 5
50-55 8 13
em
55-60 6 19
60-65 6 25
65-70 3 28
70-75 2 30
ad

Now, n = 30
So, =n

2
30
= 15
2

This observation lies in the class 55-60,

Ac

So, 55-60 is the median class.

Therefore,
l = 55
h=5
V²

f=6
cf = 13
n
−cf
15−13
Median = l + (
2
∴ ) × h = 55 + ( ) × 5
M

f 6

10 5
= 55 + = 55 +
6 3

= 55 + 1.67 = 56.67
Hence, the median weight of the students is 56.67 kg.
39. Calculation of Median:
Class Interval Frequency c.f.

9.5 - 19.5 2 2

19.5 - 29.5 4 6

29.5 - 39.5 8 14

39.5 - 49.5 9 23

49.5 - 59.5 4 27

59.5 - 69.5 2 29

12 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 69.5 - 79.5 1 30
n = 30, n

2
= 15, Median class = 39.5 - 49.5
l = 39.5, c. f . = 14, f = 9, h = 10

15−14
Median = 39.5 + ( 9
) × 10

= 39.5 + 1

9
× 10

= 39.5 + 1.11 = 40.61

40. Class frequency cummulative frequency

0-10 2 2

10-20 5 7

20-30 x 7+x

30-40 12 19+x

40-50 17 36+x

r
atu
50-60 20 56+x

60-70 y 56+x+y

70-80 9 65+x+y

yL
80-90 7 72+x+y

90-100 4 76+x+y
Given, total frequency = 100
em
So, 76 + x + y = 100
Or, x + y = 100 - 76
Or, x + y = 24.......(1)
Median class = 50 - 60
ad

l = 50; = 50; cf = 36 + x; h = 10; f = 20; Median = 52.5

n

2
n/2−cf
Median = l + [ f
]× h
Ac

50−(36+x)
Or, 52.5 = 50 + [ 20
]× 10
Or, 52.5 = 50 + 25 - 18 - x

Or, 52.5 = 57 - x

∴
x
= 57 - 52.5
V²

2
x
⇒
2
= 4.5
Or, x = 9
Substituting x=9 in equation (1), we get :-
M

9 + y = 24
Or, y = 24 - 9
Or, y = 15
Hence, x = 9 & y = 15

41. C.I. f c.f.

80 - 90 9 9

90 - 100 17 26

100 - 110 19 45

110 - 120 45 90

120 - 130 33 123

130 - 140 15 138

140 - 150 12 150

13 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 n = 150 ⇒ n

2
= 50
Median Class = 110 - 120
l = 110, f = 45, c.f. = 45, h = 10
n
−cf

we know that, Median = l + h

2
×
f

75−45
= 110 + 45
× 10
= 116.67

42. Class Interval Frequency Cumulative Frequency

0 - 10 8 8

10 - 20 16 24

20 - 30 36 60

30 - 40 34 94

40 - 50 6 100

r
N
Here, N = 100 ⇒

atu
= 50
2

The cumulative frequency just greater than 50 is 60.

Hence, median class is 20 - 30.
∴ l = 20, h = 10, f = 36, cf = cf of preceding class is 24

yL
N
( −cf )

Now, Median = l + {h ×
2
}
f

(50−24)
= 20 + {10 × }
36

26
= 20 + {10 × }
em
36

= 20 + 7.22
= 27.2
Thus, the median of the data is 27.2.
ad

43. Let us first construct the table of cumulative frequency as shown below:

Height (in cm) Number of students (fi) cumulative Frequency (c.f.)

Ac

130-135 4 4

135-140 11 15

140-145 12 27
V²

145-150 7 34

150-155 10 44

155-160 6 50
M

∑ fi = 50
N 50
Here, 2
=
2
= 25, which lies in cumulative frequency of 27.
∴ The median class is 140-145.
So, l = 140, c.f. = 15, f = 12 and h = 5
N
− c.f.
Now, as median (Me) = l + 2

f
× h

25−15
∴ Required median = 140 + 12
× 5

= 140 + 10

12
× 5

= 140 + 4.17
= 144.17
44. Observations in ascending order are:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Number of observations = n = 10 (even)
th th
n n
Value of ( ) observation + Value of ( +1) observation
Median
2 2
⇒ =
2

14 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 10
th 10
th
Value of ( ) observation + Value of ( +1) observation
2 2
⇒ 63 =
?

⇒ 126 = Value of 5th observation + Value of 6th observation

⇒ 126 = x + (x + 2)

⇒ 126 = 2x + 2

⇒ 2x = 124

⇒ x = 62

45. We shall first convert the given data to continuous classes. Then, the data become
Length (in mm) Number of leaves Cumulative frequency

117.5-126.5 3 3

126.5-135.5 5 8

135.5-144.5 9 17

144.5-153.5 12 29

r
153.5-162.5 5 34

atu
162.5-171.5 4 38

171.5-180.5 2 40

yL
Now, n = 40
So, =n

2
= 20
40

This observation lies in the class 144.5 - 153.5.

So, 144.5 - 153.5 is the median class.
Therefore,
em
l = 144.5
h=9
cf = 17
ad

f = 12
n
−cf
20−17
Median = l + ( h = 144.5 + ( 9
2
∴ )× )×
f 12
Ac

= 144.5 + 2.25 = 146.75 mm

Hence, the median length of the leaves is 146.75 mm
46. Let f1 and f2 be the missing frequencies.
10 + f1 + 25 + 30 + f2 +10 = 100
V²

⇒ f1 + f2 = 25

Median is 32, which lies in 30-40. So, the median class is 30-40.
∴ l = 30, h = 10, f = 30, N = 100 and cf = 10 + f1 + 25 = f1 + 35.
M

N
( −cf )

Now, Median, M =l + {h ×
2
}
f

50−( f1 +35)}
⇒ 30 + [10 × ] = 32
30

(15− f )
1
⇒ 30 + = 32
3

⇒ (15 − f1 ) = 6

⇒ f1 = 9

f2 = 25 - 9 = 16
Hence, f1 = 9 and f2 = 16.
47. First, we will convert the graph given into tabular form as shown below:
Class interval Frequency (fi) Mid value (xi) fixi Cumulative Frequency

1–4 6 2.5 15 6

4–7 30 5.5 165 36

15 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 7 – 10 40 8.5 340 76

10 – 13 16 11.5 184 92

13 – 16 4 14.5 58 96

16 – 19 4 17.5 70 100

N = ∑ fi = 100 Σfi xi = 832

i. N = 100
Σfi xi
Mean = N
=
832

100
= 8.32
ii. N

2
=
100

2
= 50
The cumulative frequency just greater than N

2
is 76, then the median class is 7 - 10 such that
l = 7, h = 10 - 7 = 3, f = 40, F = 36
N
−F

Median = l + 2

f
× h
50−36
=7+ × 3

r
40
42
= 7 + 1.05 = 8.05

atu
= 7 +
40

iii. Mode = 3 Median - 2 Mean

= 3 × 8.05 - 2 × 8.32 = 7.51

Class Interval Frequency fi Cumulative frequency

yL
48.
60.5 - 70.5 5 5

70.5 - 80.5 15 20

80.5 - 90.5 20 40
em
90.5 - 100.5 30 70

100.5 - 110.5 20 90
ad

110.5 - 120.5 8 98
N = 98⇒ N

2
= 49

The cumulative frequency just greater than 49 is 70 and the corresponding class is 90.5 - 100.5.
Ac

Median class is 90.5 - 100.5

∴ l = 90.5, h = 10, f = 30, c. f . = 40
N
( −cf )

Now, Median = l + {h ×
2
}
f
V²

49−40
= 90.5 + [10 × ]
30

= 90.5 + 3 = 93.50
M

49. Salary (in thousand ₹) No. of persons (f) cf

5-10 49 49

10-15 133 182

15-20 63 245

20-25 15 260

25-30 6 266

30-35 7 273

35-40 4 277

40-45 2 279

45-50 1 N = 280
N

2
= 280

2
= 140
140 lie in the interval 10-15

16 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 So, Median class is interval 10-15
Median = l + ( - C)h

f
N

Here l = 10, h = 5, f = 133, c = 49

5
So, Median = 10 + 133
(140 - 49)
= 10 + 5×91

133

= 10 + 3.421
= 13.421
Medium salary is ₹ 13.421 thousand, i.e., ₹ 13420

Percentage of marks Number of students

Cummulative frequency
(Class Interval) (fi)​
50.
30-35 16 16

35-40 14 30

40-45 18 48

tur
45-50 20 68

50-55 18 86

La
55-60 12 98

60-65 2 100
Now, N= 100 So, N

2
= 50 my
∴ Median class is 45-50
So, l = 45,
cf (cummulative frequency of preceding class) = 48
f (frequency of median class) = 20
de
h=5
n
−cf

∴ Median = l + ( 2

f
) × h
ca

50−48
Median = 45 + ( 20
) × 5

Median = 45 + ( 2

4
)
A

Median = 45 + 0.5
Median = 45.5
Class Interval Frequency Cumulative frequency (Cf)
51.
V²

0-10 6 6

10-20 3 9
M

20-30 x 9+x

30-40 12 21 + x

40 - 50 19 40 + x

Total ∑ fi = 40 + x
Total frequency = 40 + x
i.e., N = 40 + x
N 40+x
∴ =
2 2

Median = 35
∴ Median class = 30 - 40

Thus, l = 30, Cf = 9 + x, f = 12 and h = 10

N
− Cf

Median = l + (
2
) × h
f

17 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 40+x
{ −(9+x)}

35 = 30 + ...[Given, median = 35]

2
⇒ × 10
12

(22−x)×10
⇒ 35 - 30 = 2×12
(22−x)×5
⇒ 5= 12

⇒ 12 = 22 - x
⇒ x = 10

52. Class interval Mid value (x) Frequency (f) fx Cumulative frequency

0 - 20 10 6 60 6

20 - 40 30 8 240 17

40 - 60 50 10 500 24

60 - 80 70 12 840 36

80 - 100 90 6 540 42

tur
100 - 120 110 5 550 47

120 - 140 130 3 390 50

N = 50 Σf x = 3120

La
Σf x
Mean = N
=
3120

50
= 62.4
We have,
N = 50 my
Then, N
=
2
50

2
= 25
The cumulative frequency just greater than N

2
is 36, then the median class is 60 - 80 such that
l = 60, h = 80 - 60 = 20, f = 12, F = 24
N
−F

Median = l + 2
× h
de
f

25−24
= 60 + 12
× 20
20
= 60 + 12
ca

= 60 + 1.67
= 61.67
Here the maximum frequency is 12, then the corresponding class 60 - 80 is the modal class
A

l = 60, h = 80 - 60 = 20, f = 12, f1 = 10, f2 = 6

f −f
Mode = l + 2f − f1 − f2
1
× h
V²

12−10
= 60 + 2×12−10−6
× 20
= 60 + 40

= 65
M

53. We have,
Class Intervals Frequency (f) C.F

Below 140 4 4

140-145 7 11

145-150 18 29

150-155 11 40

155-160 6 46

160-165 5 51

N = ∑ f = 51

Here, N

2
=
51

2
= 25.5 which is in the class 145-150
Here, l1 = 145, h = 5, N = 51, C = 11, F = 18

18 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 N
−C

∴ Median = l 1 +
2

f
× h

25.5−11
= 145 + × 5
18
72.5
= 145 + ⇒ 149.03
18

∴ Median height of the girls = 149.03

Class Frequency (fi) Cumulative Frequency (cf)
54.
15 - 20 8 8

20 - 25 13 21

25 - 30 21 42

30 - 35 12 54

35 - 40 5 59

40 - 45 4 63

tur
N = Σf = 63
Now, N = 63
⇒
N
=
63
= 31.5

La
2 2

The cumulative frequency just greater than 31.5 is 42 and the corresponding class is 25-30.
Thus, the median class is 25-30.
∴ l = 25, h = 5, f = 21, cf = 21, = 31.5 N

Now,
N
−cf
my
Median, M = l + {h × ( 2

f
)}

31.5−21
= 25 + {5 × ( 21
)}
de
= 25 + 2.5
= 27.5
Hence, the median = 27.5
ca

Marks Number of Students (fi) Cf

55.
0 - 10 3 3
A

10 - 20 8 11

20 - 30 15 26
V²

30 - 40 10 36

40 - 50 8 44
M

Total 44

So. N = 44
⇒
N

2
=
44

2
= 22
The cumulative frequency just greater than ( N

2
= 22) is 26, so the corresponding median class is 20 - 30 and accordingly we get
Cf = 11 (cumulative frequency before the median class).
Now, since median class is 20 - 30.
∴ l = 20, h = 10 f = 15, = 22 and Cf = 11
N

Median is given by,

N
− Cf

Median = l + (
2
) × h
f

22−11
⇒ Median = 20 + ( 15
) × 10

= 20 + 7.33

19 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 = 27.33
Thus, median marks is 27.33

56. Class Interval Frequency Cumulative frequency

0-10 6 6

10-20 9 15

20-30 10 25

30-40 8 33

40-50 x 33 + x

∑ fi = 33 + x
Total frequency = 33 + x
i.e., N = 33 + x
N 33+x
∴ =

tur
2 2

Median = 25
So, class corresponding to this 20-30,
So, l = 20, Cf = 15, f = 10 and h = 10

La
N
− cf

Median = l + (
2
) × h
f

33+x
−15

⇒ 25 = 20 + ( 2

10
) × 10 my
33+x−30
⇒ 25 = 20 + 2
..[Given, Median = 25]
3+x
⇒ 25 - 20 = 2

⇒ 10 = 3 + x
de
⇒ x=7
57. To calculate the median age, we need to find the class intervals and their corresponding frequencies.
It is shown below:
ca

Class interval Frequency Cumulative Frequency

Below 20 2 2
A

20-25 4 6

25-30 18 24
V²

30-35 21 45

35-40 33 78

40-45 11 89
M

45-50 3 92

50-55 6 98

55-60 2 100
Now, n = 100
So, = n

2
= 50
100

This observation lies in class 35 - 40.

So, 35 - 40 is the median class.
Therefore,
l = 35
h=5
cf = 45
f = 33

20 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 n
−cf
50−45
Median = l + ( h = 35 + ( 5
2
∴ )× )×
f 33

25
= 35 + 33
= 35 + 0.76 = 35.76 years
Hence, the median age is 35.76 years.
58. First, let us convert the graphical distribution in the form of a table as shown below:
Weight No. of Students

40-44 4

44-48 6

48-52 10

52-56 14

56-60 10

60-64 8

tur
64-68 6

68-72 2

i. The modal class is the class with the highest frequency. As the maximum frequency is 14 and the class corresponding to it is

La
52-56, so
Modal class = 52-56
f −f
ii. Mode = l + ( 1 0

2f1 − f0 − f2
) × h my
where l is lower limit of the modal class,
h is the size of the class interval,
f1 is the frequency of the modal class,
f0 is frequency of the class preceding the modal class,
de

f2 is frequency of the class succeeding the modal class

So, here l = 52, f1 = 14, f0 = 10, f2 = 10, h = 4
ca

14−10
Thus, Mode = 52 + 28−10−10
× 4 = 54
Hence, mode weight of the students is 54 kg.
iii. Given: Mean = 55.2 kg
A

Now, Mode = 3 Median - 2 Mean

1 1
Thus, Median = 3
(Mode + 2 Mean) = 3
(54 + 2× 55.2) = 54.8
Hence, median weight of the students is 54.8 kg.
V²

59. Class interval Frequency Cumulative frequency

20-30 p p
M

30-40 15 p + 15

40-50 25 40 + p

50-60 20 60 + p

60-70 q 60 + p + q

70-80 8 68 + p +q

80-90 10 78 + p + q

Total = 90
sum of cumulative frequencies = 90
78 + p + q = 90
⇒ p + q = 12 ...(i)

Given , Median = 50
Hence, median class is 50 - 60.

21 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 ∴ l = 50, h = 10, f = 20, c.f.= 40 + p, N = 90
N
−c.f .

Now, Median = l + h
2
×
f

45−(40+p)
⇒ 50 = 50 + 20
× 10
5−p
⇒ 0= 2

⇒ 5-p=0
⇒ p = 5

substitute p in Eq(i), we get

⇒ q = 12 - 5 = 7

Hence, p = 5 and q = 7
60. First, we will convert the graph into the tabular form as shown below:
Marks obtained 0 - 20 20 - 40 40 - 60 60 - 80 80 -100

Number of students 15 18 21 29 17

i. Modal class is the class having maximum number of frequency.

tur
Here, maximum frequency is 29 and it belongs to class 60-80, so Modal class = 60-80
f1 − f0
ii. Mode = = l + × h
2f1 − f0 − f2

Here, l = 60, f1 = 29, f0 = 21, f2 = 17 and h = 20

La
29−21
Mode = 60 + 2×29−21−17
× 20

= 60 +
8

58−38
× 20 = 68
iii. Mode = 3 median - 2 mean
Mode = 68 and mean = 53 (given)
my
∴ 3 median = mode + 2 mean

3 median = 68 + 2 × 53
174
Median = 3
= 58
de
Hence, Median = 58
61. Calculation of median
xi fi cf
ca

5 1 1

6 5 6
A

7 11 17

8 14 31
V²

9 16 47

10 13 60
M

11 10 70

12 70 140

13 4 144

15 1 145

18 1 146

20 1 147

N = Σf = 147
i

We have
147
N = 147 ⇒ N

2
=
2
= 73.5
N
The cumulative frequency just greater than 2
is 140 and the corresponding value of variable x is 12.
Hence, median = 12. This means that for about half the number of days, more than 12 students were absent.

62. variable frequency c.f

22 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 15-25 8 8

25-35 10 18

35-45 X 18+x

45-55 25 43+x

55-65 40 83+x

65-75 Y 83+x+y

75-85 15 98+x+y

85-95 7 105+x+y
median = 58
∴ median class = 55 - 65
∴ l = 55

∑ f = 140

tur
i

105 + x + y = 140
x + y = 35 ...(i)
N
−c⋅f

median = l + (

La
2
)h
f

70−(43+x)
58 = 55 + ( 40
) 10
70−43−x
58 = 55 + ( 40
)× 10 my
27−x
58 = 55 + ( 40
)× 10
27−x
58 = 55 + ( 4
)

27−x
3= 4
de
12 = 27 - x
x = 27 - 12
x = 15
ca

from (i)
x + y = 35
15 + y = 35
A

y = 35 - 15
y = 20

63. Marks No. of students class interval Frequency Cumulative frequency

V²

less than 10 0 0-10 0 0

less than 30 10 10-30 10 10

M

less than 50 25 30-50 15 25

less than 70 43 50-70 18 43(F)

less than 90 65 70-90 22(f) 65

less than 110 87 90-110 22 87

less than 130 96 110-130 9 96

less than 150 100 130-150 4 100

N = 100
We have
N = 100
∴
N

2
=
100

2
= 50
The cumulative frequency just greater than N

2
is 65 then median class is 70 - 90 such that
l = 70, f = 22, F = 43, h = 90 - 70 =20

23 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 N
−F

∴ Median = l + 2

f
× h
50−43
= 70 + 22
× 20
7×20
= 70 + 22

= 70 + 6.36
= 76.36

64. Class interval Frequency Cumulative frequency

0-10 5 5

10-20 25 30

20-30 x 30 + x

30-40 18 48 + x

40-50 7 55 + x

N = 55 + x

tur
Let the missing frequency be x
Given, Median = 24 ...(1)
From table, total frequency N = 55 + x Or, ( N
= 27.5 + ( x

La
) )
2 2

Hence, c.f. just greater than ( N

2
) is (30 + x), which corresponding class is 20 - 30.
Then, median class = 20 - 30
∴ l = 20, h = 30 - 20 =10, f =x, F = 30 my
N
−F

∴ Median= l + 2
× h
f
55+x
−30

⇒ 24 = 20 + 2

x
× 10
55+x
−30

24 - 20 = 2
10
de
⇒ ×
x

55+x
⇒ 4x = ( 2
− 30) × 10
⇒ 4x = 5 (55 + x) - 300
ca

⇒ 4x - 5x = -25

⇒ -x = -25

⇒ x = 25
A

∴ Missing frequency = 25

65. Class Frequency (f) c.f.

V²

1400 - 1550 6 8

1550 - 1700 13 21

1700 - 1850 25 46
M

1850 - 2000 10 56

∑f = n = 54
Here, n = 54
54
⇒
n

2
=
2
= 27
The cumulative frequency greater than or equal to 27 is 46 or 27th term lies in the class interval 1700-1850.
∴ Median class is 1700-1850.

66. Here, the frequency table is given in inclusive form. So, we first transform it into exclusive form by subtracting and adding h

2
to
the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit
of the previous class.
Here, h = 1 So, = 0.5 h

Transforming the above table into exclusive form and preparing the cumulative frequency table, we get:-
Weekly wages (in ₹) No of workers Cumulative frequency

59.5-69.5 5 5

24 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 69.5-79.5 15 20

79.5-89.5 20 40

89.5-99.5 30 70

99.5-109.5 20 90

109.5-119.5 8 98

N = Σf = 98
i

We have, N(Total frequency) = 98 Or, h

2
= 49
The cumulative frequency just greater than h

2
is 70 and the corresponding class is 89.5-99.5. So, 89.5-99.5 is the median class.
Now,
l = 89.5 (lower limit of median class),
h = 10 (length of interval of median class),
f = 30 (frequency of median class)

tur
F = 40 (cumulative frequency of the class just preceding the median class)
Now, Median is given by:-
N
−f

=l+ 2

F
× h

La
49−40
= 89.5 + 30
× 10
= 89.5 + 3 = 92.5

67. Class Frequency (f)

my Cumulative frequency (cf)

0-10 5 5

10-20 x 5+x

20-30 20 25 + x
de
30-40 15 40 + x

40-50 y 40 + x + y
ca

50-60 5 45 + x + y

N = Σf = 60
A

N = 60
45 + x + y = 60
⇒ x + y =15 ...(i)
V²

Median = 28.5
Clearly, it lies in the class interval 20 - 30.
So, 20 - 30 is the median class.
M

∴ l = 20, h = 10, f = 20, F = 5 + x and N = 60

we know that,
N
−F

Median = l + 2

f
× h
30−(5+x)
⇒ 28.5 = 20 + 20
× 10
25−x
⇒ 28.5 = 20 + 2
25−x
⇒ 8.5 = 2
⇒ 25 - x = 17 ⇒ x = 8
Putting x = 8 in Eq(i), we get
8 + y = 15
y=7
Hence, x = 8 and y = 7

68. Class interval Frequency Cumulative frequency

85-100 10 10

100-115 4 14

25 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 115-130 7 21

130-145 9 30
Here, N = 30 ⇒ N

2
= 15

The cumulative frequency just greater than 15 is 21.

Hence, median class is 115-130.
∴ l = 115, h = 15, f = 7, cf = cf of preceding class = 14
N
( −cf )

Now, Median
2
= 1 + {h × }
f

(15−14)
= 115 + {15 × }
7

1
= 115 + {15 × }
7

= 115 + 2.1
= 117.1
Thus, the median bowling speed is 117.1 km/hr.

tur
69. Table:
Marks Frequency c.f.

15 8 8

La
20 10 18

25 12 30

30 5 35
my
35 4 39

40 7 46

45 6 52
de

60 4 56

70 8 64
ca

80 6 70

90 8 78
A

N = 78

Median = average of 39th and 40th observation

V²

35+40
Median = = 37.5
2

70. First, we will convert the graph into tabular form given below:
Monthly
Number of Class Cumulative
M

consumption di = xi - 135 ui =
xi −135
fiui
consumers (fi) mark (xi) 5
Frequency
(in units)

65-85 4 75 –60 –3 –12 4

85-105 5 95 –40 –2 –10 9
105-125 13 115 –20 –1 –13 22
125-145 20 135 0 0 0 42
145-165 14 155 20 1 14 56
165-185 8 175 40 2 16 64
185-205 4 195 60 3 12 68

Total ∑ fi = 68 ∑ fi ui = 7

i. Let a = 135.
Now, h = 20
Using the step-deviation method,

26 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 ∑ f ui
¯¯
¯ i 7
M ean, x = a + ( ) × h = 135 + ( ) × 20
∑ fi 68

= 135 +
35

17
= 135 + 2.05 = 137.05
ii. Now, N = 68
So,N
=
2
= 34
68

This observation lies in class 125-145.

Therefore, 125-145 is the median class.
So, l = 125, CF = 22, f = 20
N
−CF
2
∴ M edian = l + ( ) × h
f

34−22
= 125 + (
20
) × 20 = 125 + 12 = 137
iii. Mode = 3 Median - 2 Mean
= 3× 137 - 2× 137.05 = 136.9

71. Class interval Frequency Cumulative frequency

tur
0-6 4 4

6-12 x 4+x

12-18 5 9+x

La
18-24 y 9+x+y

24-30 my 1 10 + x + y

Total =20
10 + x + y = 20
⇒ x + y =10 ...(i)

Median = 14.4
de
Hence, median class is 12 - 18.
∴ l = 12, h = 6, f = 5, F = cumulative frequency of preceding class = 4 + x, N = 20
N
−F

Median= 1 + h
ca

2
×
f

10−(4+x)
⇒ 14.4 = 12 + 5
× 6
6−x
⇒ 2.4 = × 6
A

5
6−x
⇒ 0.4 = 5

⇒ x=4
substitute x value in (i), we get
V²

⇒ y = 10 - 4 = 6

72. C.I. f c.f.

M

20 - 40 12 12

40 - 60 18 30

60 - 80 23 53

80 - 100 15 68

100 - 120 12 80

120 - 140 12 92

140 - 160 8 100

∑ fi = 100

n
n = 100 ⇒ = 50
2

Median Class= 60 − 80
l = 60, c. f . = 30, f = 23, h = 20

27 / 29
Founder Director:Mr. VRS Sir ||| Mob. 9373033537
 n
−cf

we know that, Median = l +

2
× h
f

50−30
= 60 + × 20
23

= 77.39

73. Class interval Mid value (x) Frequency (f) fx Cumulative frequency

0 – 50 25 2 50 2

50 – 100 75 3 225 5

100 – 150 125 5 625 10

150 – 200 175 6 1050 16

200 – 250 225 5 1127 21

250 – 300 275 3 825 24

300 – 350 325 1 325 25

tur
N = 25 Σf x = 4225
Σf x
Mean = N
=
4225

25
= 169
We have,

La
N = 25
N 25
Then, 2
=
2
= 12.5
The cumulative frequency just greater than N

2
is 16, then the median class is 150 - 200 such that
l = 150, h = 200 – 150 = 50, f = 6, F = 10
N
−F
my
Median = l + h
2
×
f

12.5−10
= 150 + 6
× 50
= 150 + 125
de
6

= 150 + 20.83
= 170.83
Here the maximum frequency is 6, then the corresponding class 150 - 200 is the modal class
ca

l = 150, h = 200 – 150 = 50, f = 6, f1 = 5, f2 = 5

f −f1
Mode = l + × h
A

2f − f − f
1 2
6−5
= 150 + 2×6−5−5
× 50
50
= 150 + 2

= 150 + 25
V²

= 175
74. The given series is in inclusive form. Converting it to exclusive form and preparing cumulative frequency table, we get
M

Class Frequency Cumulative Frequency

0.5 - 5.5 13 13

5.5 - 11.5 10 23

11.5 - 17.5 15 38

17.5 - 23.5 8 46

23.5 - 29.5 11 57
Here, N = 57 ⇒ N

2
= 28.5

The cumulative frequency just greater than 28.5 is 38,

Hence, the median class is 11.5 - 17.5.
Upper limit of the median class = 17.5
75. According to the question,
x f cumulative frequency

1 8 8

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 2 10 18

3 11 29

4 16 45

5 20 65

6 25 90

7 15 105

8 9 114

9 6 120

N = 120
N = 120 ⇒ N

2
= 60.
The cumulative frequency just greater than N
i.e., 60 is 65 and

tur
2

the value of x corresponding to 65 is = 5.

∴ Median = 5

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my
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V²
M

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