Solution

STATISTICS

Class 10 - Mathematics

1. (a) 10
Explanation:
Arithmetic mean of 7, 8, x, 11, 14 is x
7+8+x+11+14
⇒
5
=x
40+x
⇒
5
= x ⇒ 40 + x = 5x
⇒ 5x - x = 40 ⇒ 4x = 40
⇒ x = = 10
40

2.
(d) 315

n
Explanation:
150-155 is the modal class

i
Height in cm Number of students (f) Cumulative Frequency (CF)

es.
150-155 15 15

155-160 13 28

160-165

165-170
10

8 tut 38

46
sti
170-175 9 55

175-180 5 60
ein

Total 60
Here, N

2
= 30, the commulative frequency just above 30 is 38 and the corresponding class is
160-165 which is the median class.
hence the required sum = 115 + 165 = 315
ac

3.
w.

(c) 470
Explanation:
n
−c

Median = l +
ww

2
× h
f

44
−8

= 400 + 100
2
×
20

= 400 + 14

20
× 100
= 400 + 14 × 5
= 400 + 70
= 470

4.
(d) 26
Explanation:
mode = 3 median - 2 mean
= 3(30) - 2(32)
= 90 - 64
= 26

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 5.
(c) decreases by 2
Explanation:
decreased by 2.

6.
(c) Mean
Explanation:
Mode is the value with the maximum frequency. Thus, it can be determined from the graph.
Median is the middle value of the data. Thus, it can be determined from the graph.
Mean is the ratio of sum of all data values and the total number of values. Thus, it cannot be determined by graphically.

7. (a) 80
Explanation:
In the given data, Maximum frequency is 15.

n
Therefore, the modal class is 80 - 90.

i
The lower limit of the modal class is 80.

es.
8. (a) n
Explanation:
We know that the mean or average of observations, is the sum of the values of all the observations divided by the total number
of observations.

tut
and, we have first n odd natural numbers as
1, 3, ...., 2n - 1
sti
Clearly the above series is an AP(Arithmetic progression) with first term, a = 1 and common difference, d = 2
And no of terms is clearly n.
And last term is (2n - 1)
ein

We know, sum of terms of an AP if first and last terms are known is:
Sn = (a + an)
n

Putting the values in above equation we have sum of series i.e.

= n2 ...(i)
n(2n)
1 + 2 + 3 + ... + n = n
(1 + 2n - 1) =
ac

2 2

As,
Sum of all terms
Mean =
w.

no of terms
n2
⇒ Mean = n
=n
9.
ww

(d) assumed mean

Explanation:
∑ f ui
In the formula, x̄ ¯
¯
= a + h(
i
) ‘a’ stands for Assumed Mean i.e., assumed mean value is the mid value (xi) of class
∑ fi

intervals of a set of grouped data.

10.
(c) mean = mode = median
Explanation:
For a symmetrical distribution,
we have Mean = mode = median

11. Class marks are 12, 14, 16.......

Class size= 2
Class Intervals are 11 − 13, 13 − 15, 15 − 17, 17 − 19, 19 − 21, 21 − 23, . . . . . so on.

12. C.I. Frequency C.f.

0-5 2 2

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 5 - 10 9 11

10 - 15 6 17

15 - 20 7 24

20 - 25 1 25
25
Here, N

2
=
2
= 12.5
17 is just greater than 12.5
∴ Median class is (10 - 15)

Upper limit of median class = 15

Now, modal class = class with maximum frequency
∴ Modal class is (5 - 10)

Lower limit of modal class = 5

∴ Sum of lower limit of modal class and upper limit of median class
= 5 + 15
= 20

n
Class Interval Frequency(fi) Class mark xi (f )

i
× xi
13. i

es.
1-3 12 2 24

3-5 22 4 88

5-7 27 6 162

7-9
tut 19

Σfi = 80
8 152

Σ (fi × xi ) = 426
sti
Σ( fi × xi )
∴ mean= Σf
=
426

80
= 5.325
i

14. Class Interval Frequency C.F

ein

0 - 10 5 5

10 - 20 8 13

20 - 30 7 20
ac

30 - 40 12 32
w.

40 - 50 28 60

50 - 60 20 80
ww

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

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

Σf x
⇒ 50 = 100

⇒ Σf x = 5000
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

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16. ∵ the highest frequency is 14
∴ Modal class- 35 - 40

So, l = 35, h = 40 - 35 = 5,
f0 = 5, f1 = 14, f2 = 8
Now,
f1 − f0
Mode = l + ( 2f1 − f0 − f2
) × h

14−5
Mode = 35 + ( 28−5−8
) × 5

Mode = 35 + ( 9

3
)

Mode = 35 + 3
Mode = 37
17. We first, find the class mark xi of each class and then proceed as follows

Marks Class marks (xi) frequency (fi) fixi

10-20 15 2 30

n
20-30 25 4 100

i
es.
30-40 35 7 245

40-50 45 6 270

50-60 55 1 55

Therefore, mean x̄ = ¯
¯
Σ f×x
=
tut
700

20
= 35
∑ fi = 20 ∑ fi xi = 700
sti
Σ fi

Hence, the mean of scores of 20 students in mathematics test 35.

18. Modal class = 5 - 7
Here, l = 5,f1 = 80, f0= 45, h = 2,f2 = 55
ein

(f − f )
Mode = l + 1 0

2f1 − f0 − f2
× h

(80−45)
=5+ × 2
2(80)−45−55

80−45
=5+
ac

× 2
160−45−55
35
=5+ 160−100
× 2

=5+ 35×2
w.

60

=5+ 70

60

= 5 + 1.17
= 6.17
ww

Class Interval Frequency fi Mid-Value xi fixi

19.
25-35 6 30 180

35-45 10 40 400

45-55 8 50 400

55-65 12 60 720

65-75 4 70 280

∑ fi = 40 ∑ fi xi = 1980
∑ fi xi
Mean =
∑ fi

1980
=
40

Mean = 49.5

Marks Obtained Number of students (fi) xi fixi

20.
0 - 10 12 5 60

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 10 - 20 23 15 345

20 - 30 34 25 850

30 - 40 25 35 875

40 - 50 6 45 270

Total 100 2400

2400
Mean = 100

= 24
xi fi fi xi
21.
2 3 6

4 2 8

6 3 18

n
10 1 10

i
p+5 2 2p + 10

es.
∑ fi = 11 ∑ fi xi = 2p + 52

Σfi = 11,

Σfi xi = 2p + 52

⇒ 6 =
Mean
2p+52

11
=
Σfi xi

∑ fi

tut
sti
⇒ 66 = 2p + 52

⇒ 2p = 14

⇒ p = 7

22. Total number of observations = N = 19 (odd)

ein

th

observation = Value of 10th observation

N +1
⇒ Median = Value of ( 2
)

given, median = 30
10th observation is 30.
ac

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.
w.

Hence, median is not affected.

⇒ Median = 30

23. Given, mode = 36 , median = 43

ww

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

3 median = mode + 2 mean.......(1)
Or, -2 mean = mode - 3 median
Or, -2 mean = 36 - 3 × 43​
Or, - 2 mean = 36 - 129
Or, - 2 mean = - 93
Thus, Mean = 46.5
24. Mode = 1000
Median = 1250
we know that ,
Mode = 3 Median - 2 Mean
3 Median - Mode
⇒ Mean =
2
3×1250−1000
=
2
3750−1000
=
2
2750
=
2

= 1375

∴ mean = 1375

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 Number of days Number of students (fi) Class mark (xi) fixi
25.
0-6 11 3 33

6-10 10 8 80

10-14 7 12 84

14-20 4 17 68

20-28 4 24 96

28-38 3 33 99

38-40 1 39 39

Total ∑ fi = 40 ∑ fixi = 499

Using the direct method,

∑ f xi
¯¯
x̄ =
i
=
499
= 12.475

n
∑ fi 40

Hence, the mean number of days a student was absent is 12.48

i
26. We shall first convert the given data to continuous classes. Then, the data become

es.
Length (in mm) Number of leaves Cumulative frequency

117.5-126.5 3 3

tut
126.5-135.5 5 8

135.5-144.5 9 17

144.5-153.5 12 29
sti
153.5-162.5 5 34

162.5-171.5 4 38
ein

171.5-180.5 2 40
Now, n = 40
n 40
So, = = 20
ac

2 2

This observation lies in the class 144.5 - 153.5.

So, 144.5 - 153.5 is the median class.
Therefore,
w.

l = 144.5
h=9
ww

cf = 17
f = 12
n
−cf
20−17
∴ Median = l + ( 2

f
)× h = 144.5 + ( 12
)× 9

= 144.5 + 2.25 = 146.75 mm

Hence, the median length of the leaves is 146.75 mm

27. Class interval 10-15 15–20 20-25 25-30 30-35 35-40

Frequency 30 45 75 35 25 15
Here the maximum frequency is 75, then the corresponding class 20 – 25 is the modal class.
l = 20, h = 25 - 20 = 5, f = 75 , f1 = 45, f2 = 35
f −f1
We know that, Mode = l + 2f − f1 − f2
× h
75−45
= 20 + 2×75−45−35
× 5
= 20 + 150

70

= 20 + 2.14
= 22.14
28. Calculation of mean:

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 xi −50
Class Mid-values (xi) Frequency (fi) di = xi -50 ui = fiui
20

0-20 10 7 -40 -2 -14

20-40 30 p -20 -1 -p

40-60 50 10 0 0 0

60-80 70 9 20 1 9

80-100 90 13 40 2 26

N = ∑ f = 39 + p
i ∑ fi ui = 21 - p
According to the question, mean of the given data is 54.
From table,
A = 50,​​N = 39 + p, h = 20, Σf u = 21 - p i i

We know that, Mean = A + h { 1

N
Σfi ui }

21−p
54 = 50 + 20 × {

n
⇒ } ​​​​​​​
39+p

21−p
⇒ 4 = 20 × { } ​​​​​​​

i
39+p

es.
21−p
⇒ 1=5( 39+p
)

⇒ 39 + p = 105 - 5p
⇒ 6p = 66

tut
⇒ p = 11

29. Calculation of mean by using step-deviation method.

xi −A
ui =
Class Interval Frequency(fi) Mid value xi h
(f )
sti
i × ui
xi −203.5
=
1

200 - 201 13 200.5 -3 -39

ein

201 - 202 27 201.5 -2 -54

202 - 203 18 202.5 -1 -18

203 - 204 10 203.5 = A 0 0

ac

204 - 205 1 204.5 1 1

205 - 206 1 205.5 2 2

w.

Σfi = 70 Σ (fi × ui ) = -108

A = 203.5, h = 1
ww

∑ fi ui
Mean = A + {h × }
∑ fi

−108
= 203.5 + {1 × }
70

= 203.5 − 1.54 = 201.96

mean weight of packets is 201.96

30. Class interval Frequency Cumulative frequency

20-30 p p

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

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

Given , Median = 50
Hence, median class is 50 - 60.
∴ l = 50, h = 10, f = 20, c.f.= 40 + p, N = 90
N
−c.f .

Now, Median = l + 2

f
× h
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

n
Percentage of marks Number of students
Cummulative frequency

i
(Class Interval) (fi)​
31.

es.
30-35 16 16

35-40 14 30

40-45

45-50
18

20 tut 48

68
sti
50-55 18 86

55-60 12 98

60-65 2 100
ein

Now, N= 100 So, N

2
= 50
∴ Median class is 45-50
So, l = 45,
ac

cf (cummulative frequency of preceding class) = 48

f (frequency of median class) = 20
h=5
w.

n
−cf

∴ Median = l + ( 2
) × h
f

50−48
Median = 45 + (
ww

) × 5
20

Median = 45 + ( 2

4
)

Median = 45 + 0.5
Median = 45.5
32. Table:
Marks Frequency

0 - 10 3

10 - 20 5

20 - 30 7

30 - 40 10

40 - 50 12

50 - 60 15

60 - 70 12

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 70 - 80 6

80 - 90 2

90 - 100 8

Marks Number of Students (fi) Cf

33.
0 - 10 3 3

10 - 20 8 11

20 - 30 15 26

30 - 40 10 36

40 - 50 8 44

Total 44

n
So. N = 44
= 22

i
N 44
⇒ =
2 2

es.
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.

tut
∴ l = 20, h = 10 f = 15, = 22 and Cf = 11
N

Median is given by,

N
− Cf

Median = l + ( 2
) × h
sti
f

22−11
⇒ Median = 20 + ( 15
) × 10

= 20 + 7.33
ein

= 27.33
Thus, median marks is 27.33
34. For step - deviation method, we prepare the following table:
ac

xi −A
ui =
Class Interval Frequency fi Mid value xi xi −99
h
fiui
=
6
w.

84 - 90 15 87 -2 -30

90 - 96 22 93 -1 -22
ww

96 - 102 20 99 = A 0 0

102 - 108 18 105 1 18

108 - 114 20 111 2 40

114 - 120 25 117 3 75

∑ fi = 120 ∑ fi ui = 81

Thus, A = 99, h = 6, ∑ f i = 120 and ∑ fi ui = 81

∑ fi ui
Mean = A + {h × }
∑ fi

81
= 99 + {6 × }
120

= 99 + 4.05
= 103.05
35. The given series is an inclusive series.
Making it an exclusive series, we get
Class interval Frequency fi Mid-value xi ui =
xi −A
fiui
h
xi −500.5
=
200

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 0.5-200.5 14 100.5 -2 -28

200.5-400.5 15 300.5 -1 -15

400.5-600.5 14 500.5 =A 0 0

600.5-800.5 7 700.5 1 7

∑ fi = 50 ∑ fiui = -36

Thus, A = 500.5, h = 200, ∑ fi = 50 and ∑ fiui = -36

∑ f ui
Mean= A +{h × ∑f
i
}
i

−36
= 500.5 +{200 × 50
}

= 500.5 - 144
= 356.5
Thus, the mean daily income of employees is ₹ 356.50
36. Let us first construct the table for di × fi,

n
where di = xi - A (Assumed mean), as shown below:

i
Class Marks (xi) Frequency (fi) di = xi - A di × fi

es.
Class

0-5 2.5 8 -10 -80

5-10 7.5 7 -5 -35

10-15

15-20
12.5 = A

17.5
tut 10

13
0

5
0

65
sti
20-25 22.5 12 10 120

∑ fi = 50 ∑ fi di = 70
ein

Let A = 12.5,
Then ∑ f d = 70
i i

Now, the required mean

∑ fi di
ac

(x̄) = A +
∑ fi

= 12.5 + 70

50

= 12.5 + 1.4
w.

= 13.9

37. Class interval Frequency Cumulative frequency

ww

85-100 10 10

100-115 4 14

115-130 7 21

130-145 9 30
N
Here, N = 30 ⇒ 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.

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38. Class interval (Indusive) Class interval (Exclusive) Frequency Cumulative frequency

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

169-171 168.5-171.5 127 402

172-174 171.5-174.5 18 420

N = 420
N = 420
∴
N

2
=
420

2
= 210
The cumulative frequency just greater than N

2
is 275.
so, 165.5-168.5 is the median class.
l = 165.5, f = 142, F = 133 ,h = 168.5 - 165.5 = 3

n
N
−F

∴ Median = l + 2

f
× h

i
210−133
= 165.5 + × 3

es.
142
77×3
= 165.5 + 142

= 165.5 + 1.63
= 167.13

39.
Class Interval

0 - 10 3tut
Frequency(fi) Mid Value xi

5
(fi × xi

15
)
sti
10 - 20 4 15 60

20 - 30 p 25 25p
ein

30 - 40 3 35 105

40 - 50 2 45 90

Σfi = 12 + p Σ (fi × xi ) = 270 + 25p

ac

Σ( f × xi )
Now, mean= i

Σfi

270+25p
⇒ 24 =
w.

12+p

⇒ 24(12 + p) = 270 + 25p

⇒ 288 + 24p = 270 + 25p

ww

⇒ p = 18

C.I. Frequency(fi) Mid value xi ui= x −25

i
(f × ui )
40. 10
i

0 - 10 7 5 -2 -14

10 - 20 10 15 -1 -10

20 - 30 15 25 = A 0 0

30 - 40 8 35 1 8

40 - 50 10 45 2 20

Σfi = 50 Σ (fi × ui ) = 4

A = 25, h = 10

∑ fi ui
we know that, Mean = A + {h × }
∑ fi

4
= 25 + {10 × }
50

= 25 + 0.8 = 25.8

No of heads per toss No of tosses fixi

41.

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 0 38 0

1 144 144

2 342 684

3 287 861

4 164 656

5 25 125

∑ fi = 1000 ∑ fixi = 2470

∑ fi xi
Mean number of heads per toss = =
2470

1000
= 2.47
∑ fi

Therefore, Mean = 2.47

Deviation
Class Interval Frequency(fi) Class mark xi (f × di )
di= (xi - A) = (xi - 57) i

42.

n
50 - 52 18 51 -6 -108

i
52 - 54 21 53 -4 -84

es.
54 - 56 17 55 -2 -34

56 - 58 28 57 = A 0 0

58 - 60

60 - 62

62 - 64
16

35

15
tut 59

61

63
2

6
32

140

90
sti
Σfi = 150 Σ (fi × di ) = 36

Let the assumed mean A = 57

ein

Σ( f × di )
¯¯
¯ i
∴ x = A+
Σfi

36
= (57 + )
150

= 57 + 0.24
ac

= 57.24
Hence, mean weight = 57.24 kg
w.

xi −A
ui =
Class Interval Frequency(fi) Mid value xi h
xi −550
(f
i × ui )
=
43. 20
ww

500 - 520 14 510 -2 -28

520 - 540 9 530 -1 -9

540 - 560 5 550 = A 0 0

560 - 580 4 570 1 4

580- 600 3 590 2 6

600 - 620 5 610 3 15

Σfi = 40 Σ (fi × ui ) = -12

Let assumed mean (A) = 550, h = 20 (class-interval).....(1)
from the above table,
∑ f = 40 and ∑ f u = −12 .....(2)
i i i

Now,
∑ fi ui
Mean = A + {h × ∑f
}
i

= 550 + (20 ×
−12

40
) [ from (1) & (2) ]

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 = 550 - 6
= 544
Hence, the mean of the frequency distribution is 544.
44. Calculation of Mean:
xi −a

C.I. Class mark(xi) fi ui = fi ui

h

0 - 10 5 4 -3 -12

10 - 20 15 4 -2 -8

20 - 30 25 7 -1 -7

30 - 40 35 10 0 0

40 - 50 45 12 1 12

50 - 60 55 8 2 16

60 - 70 65 5 3 15

n
Total Σ fi = 50 Σ fi ui = 16

i
es.
a = 35, h = 10

we know that,
Σfi ui
Mean = a + Σfi
× h

tut
16
= 35+ × 10
50

= 35 + 3.2 = 38.2

45. First we prepare the following cumulative table to compute the median
sti
Class Frequency Cumulative frequency

5-10 5 5
ein

10-15 6 11

15-20 15 26

20-25 10 36
ac

25-30 5 41

30-35 4 45
w.

35-40 2 47

40-45 2 49
ww

N = 49
We have, N=49
49
∴
N
=
2
= 24.5
2

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
l = 15, f = 15, F = 11 and h = 5
N
−F
24.5−11
∴ Median = l +
2

f
× h = 15 +
15
× 5 = 15 +
13.5

3
= 19.5
46. We may find class marks for each interval by using the relation
upperlimit+ lowerclasslimit
x =
2

Class size of this data = 0.04

Concentration of
Frequency fi Class interval xi di = xi – 0.14 ui fiui
SO2

0.00 – 0.04 4 0.02 -0.12 -3 -12

0.04 – 0.08 9 0.06 -0.08 -2 -18

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 0.08 – 0.12 9 0.10 -0.04 -1 -9

0.12 – 0.16 2 0.14 0 0 0

0.16 – 0.20 4 0.18 0.04 1 4

0.20 – 0.24 2 0.22 0.08 2 4

Total ∑ fi = 30 ∑ fi ui = -31
let a = 0.14
∑ f ui
¯¯
¯ i
Mean x = a + ( ) × h
∑f
i

−31
= 0.14 + (0.04) ( 30
)

= 0.099 ppm
47. Let the missing frequency be f the assumed mean be A = 47 and h = 3.
Calculation of Mean
Class-Intervals mid-values xi fi di = xi - 47.5 ui =
xi −47.5
fiui
3

n
40-43 41.5 31 -6 -2 -62

i
43-46 44.5 58 -3 -1 -58

es.
46-49 47.5 60 0 0 0

49-52 50.5 f 3 1 f

52-55

tut
53.5

N = ∑ f = 176 + f
i
27 6 2 54

Σfi ui = f - 66
sti
We have,
¯¯¯
X
¯
= 47.2 , A = 47.5 and h = 3
∴ X
¯¯¯
¯
=A+h{ 1
Σfi ui }
ein

f −66
⇒ 47.2 = 47.5 + 3 × { 176+f
} ​​​​​​​

f −66
- 0.3 = 3 × { 176+f
}

−1 f −66
- 176 - f = 10 f - 660 ⇒ 11f = 484 ⇒ f = 44
ac

⇒ = ⇒
10 176+f

Hence, the missing frequency is 44

CI fi xi di ui fiui
w.

48.
0 - 10 5 5 -30 -3 -15

10 - 20 10 15 -20 -2 -20
ww

20 - 30 18 25 -10 -1 -18

30 - 40 30 35 0 0 0

40 - 50 25 45 10 1 20

50 - 60 12 55 20 2 24

60 - 70 5 65 30 3 15

Total 100 6
Σfi ui
mean = A + Σfi
× h

= 35 + 100
6
× 10

= 356

10
or 35.6

Marks (C.I) No of Students (fi) xi Fixi

49. i.
0-10 8 5 40

10-20 16 15 240

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 20-30 36 25 900

30-40 34 35 1190

40-50 6 45 270
∑ fi = 100
∑ fi x = 2640
i

∑ fi xi
mean(x̄) =
∑ fi

= 2640

100

= 264

10

= 26.4

Marks (C.I) No of Students (fi) c.f

ii.
0-10 8 8

10-20 16 24

n
20-30 36 60

i
30-40 34 95

es.
40-50 6 100
N = 100
N 100
= 50

tut
=
2 2

c.f Just greater than N

2
i.e 50 is 60
∴ median class = 20 - 30
lower limit (l) of median class = 20
sti
N
−cf

median = l + (
2
) × h
f
ein

50−24
= 20 + ( 36
)× 10
26
= 20 + 36
× 10
= 27.22
ac

iii. maxm class frequency is 36, belonging to class interval 20-30

∴ modal class = 20 - 30

l = 20
w.

f −f
mode = l + ( 1

2f − f − f
0
) × h
1 0 2

36−16
= 20 + ( 2×36−16−34
)× 10
ww

20×10
= 20 + 72−50

= 20 + 200

22

= 29.09
50. Let the missing frequencies are a and b.
Class Interval Frequency fi Cumulative frequency

0-5 12 12

5 - 10 a 12 + a

10 - 15 12 24 + a

15 - 20 15 39 + a

20 - 25 b 39 + a + b

25 - 30 6 45 + a + b

30 - 35 6 51 + a + b

35 - 40 4 55 + a + b = 70

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 Then, 55 + a + b = 70
a + b = 15 ......(1)
Median is 16, which lies in 15 - 20
So, The median class is 15 - 20
Therefore, l = 15, h = 5, N = 70, f = 15 and cf = 24 + a
Median is 16, which lies in the class 15 - 20. Hence, median class is 15 - 20.
∴ l = 15, h = 5, f = 15, c. f . = 24 + a
N
( −cf )

Now, Median = l + {h × 2
}
f

(35−24−a)
∴ 16 = 15 + {5 × }
15

11−a
⇒ 16 = 15 + { }
3

11−a
⇒ 1 =
3

⇒3 = 11 − a

⇒a = 8

n
Now, 55 + a + b = 70

i
⇒55 + 8 + b = 70

es.
⇒63 + b = 70

⇒ b = 7

Hence, the missing frequencies are a = 8 and b = 7.

Monthly consumption Number of consumers cf

tut
51.
65 - 85 7 7

85 - 105 8 15
sti
105 - 125 7 22

125 - 145 20 42
ein

145 - 165 14 56

165 - 185 9 65

185 - 205 5 70
ac

N 70
= = 35
2 2

l = 125, c.f = 22, f = 20, h = 20

w.

N
−cf

Median = I + ( 2

f
) × h
ww

(35−22)
= 125 + × 20
20

= 125 + 13 = 138

Note: No marks to be deducted in case student substitutes the values correctly in the formula without writing values of l, cf, etc.
52. We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 interval.
Class size (h) = 500
f −f1
Mode = l + 2f − f − f
× h
1 2

Lower limit (l)of modal class = 1500

Frequency (f) of modal class = 40
Frequency (f1) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
40−24
mode = 1500 + 2×40−24−33
× 500
16
= 1500 + 80−57
× 500
= 1500 + 347.826
= 1847.826 ≈ 1847.83

Expenditure (in ₹.) Number of families fi xi di = xi - 2750 ui uifi

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 1000-1500 24 1250 -1500 -3 -72

1500-2000 40 1750 -1000 -2 -80

2000-2500 33 2250 -500 -1 -33

2500-3000 28 2750=a 0 0 0

3000-3500 30 3250 500 1 30

3500-4000 22 3750 1000 2 44

4000-4500 16 4250 1500 3 48

4500-5000 7 4750 2000 4 28

Σfi = 200 Σfi di = - 35

Σfi di
Mean x̄ = a +
¯
¯
Σf
× h
i
−35
¯¯
x̄ = 2750 + 200
× 500

n
¯
¯
x̄ = 2750 - 87.5
¯¯
x̄ = 2662.5

i
53. Modal Class: 45 - 60

es.
Mode = 55
15−a
55 = 45 + × 15
30−(a+10)

a=5

tut
⇒

6 + 7 + a + 15 + 10 + b = 51
⇒ a + b = 13

⇒ b = 13 - 5 = 8
sti
54. From the given data we have
Class internal Frequency fi Mid-value xi ui =
xi −A
=
x1 −162.5
fi × ui
h 5
ein

150-155 15 152.5 -2 -30

155-160 8 157.5 -1 -8

160-165 20 162.5 = A 0 0
ac

165-170 12 167.5 1 12

170-175 5 172.5 2 10
w.

∑ fi = 60 ∑ fi ui = −16

As the class 160-165 has the maximum frequency.

ww

so it is the modal class.

∴ xk = 160, h = 5, fk = 20, fk-1 = 8, fk+1 = 12
(f −f )

Now, Mode = x
k k−1

k + h{ }
(2f −f −f )
k k−1 k+1

20−8
= 160 + 5 { }
2(20)−8−12

12
= 160 + 5 ×
20

= 160 + 3
= 163
Thus, the modal height is 163 cm, which means the maximum number of students has height 163 cm.
Thus, A = 162.5, h = 5, ∑ fi = 60 and ∑ fiui = -16
∑ fi ui
Mean = A + {h × }
∑ fi

−16
= 162.5 + {5 × }
60

= 162.5 - 1.33
= 161.17
Thus, the mean height is 161.17 cm, which means on an average, the height of a student in a class is 161.17 cm.

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55. Since value of number of mangoes and number of boxes are large numerically. So we use step-deviation method
xi −a
True Class Interval No. of boxes(fi) Class mark(xi) ui = fiui
h

49.5-52.5 15 51 -2 -30

52.5-55.5 110 54 -1 -110

55.5-58.5 135 57 0 0

58.5-61.5 115 60 1 115

61.5-64.5 25 63 2 50

∑ fi = 400 ∑ fiui = 25

Let assumed mean (a) = 57,

h=3,
∑ fi ui 25
¯¯
∴ u =
¯

∑ fi
=
400
= 0.0625 (approx.)

n
Using formula, Mean (x) = a + hu ¯¯
¯ ¯¯
¯

= 57 + 3 (0.0625

i
= 57 + 0.1875

es.
= 57.1875
= 57.19 (approx)
Therefore, the mean number of mangoes is 57.19

56. Class interval

0-10 tut
Frequency

10
Cumulative frequency

10
sti
10-20 20 30

20-30 f1 30 + f1(F)
ein

30-40 40 (f) 70 + f1

40-50 f2 70 + f1 + f2

50-60 25 95 + f1 + f2
ac

60-70 15 110 + f1 + f2
w.

N = 170
According to the question ,
Given,
ww

Median = 35
then median class = 30 - 40
∴ l = 30, h = 40 - 30 = 10, f = 40, F = 30 + f1

N
−F

∴ Median = 1 + 2

f
× h
85−(30+ f )
⇒ 35 = 30 + 40
1
× 10
85−30−f1
⇒ 35 - 30 = 40
× 10
55−f
1
⇒ 5 =
4

⇒ 20 = 55 - f1
⇒ f1 = 55 - 20 = 35
Given
Sum of frequencies = 170
⇒ 10 + 20 + f1 + 40 + f2 + 25 + 15 = 170
⇒ 10 + 20 + 35 + 40 + f2 + 25 + 15 = 170
⇒ f2 = 170 - 10 - 20 - 35 - 40 - 25 - 15

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 ⇒ f2 = 25
∴ f1 = 35 and f2 = 25

Age (in years) No. of patients (fi) Mid point (xi) xifi
57.
5 - 15 6 10 60

15 - 25 11 20 220

25 - 35 21 30 630

35 - 45 23 40 920

45 - 55 14 50 700

55 - 65 5 60 300

Total 80 2830
⇒ Mean =
2830

80

n
= 35.375
Modal class = (35 - 45)

i
es.
23−21
⇒ Mode = 35 + (
2×23−21−14
) × h

= 36.81
Therefore, mode and mean of given data are 36.81 years and 35.375 years respectively.

tut
58. Table:
Class Interval Frequency fi Mid value xi fixi Cumulative Frequency

0 -10 6 5 30 6
sti
10 - 20 8 15 120 14

20 - 30 10 25 250 24
ein

30 - 40 15 35 525 39

40 - 50 5 45 225 44
ac

50 - 60 4 55 220 48

60 - 70 2 65 130 50
w.

∑ fi = 50 ∑ fi ui = 1500

∑ fi xi 1500
i. Mean = ∑f
=
50
= 30
i
ww

ii. Median
N = 50⇒ N

2
= 25

median class is 30 - 40.

∴l = 30, h = 10, f = 15, c. f . = 24
N
( −cf )

Median = l + { h ×
2
}
f

25−24
= 30 + {10 × }
15

= 30 + 0.67 = 30.67

iii. Mode
Maximum frequency = 15
Hence, modal class is 30 - 40.
(f −f )

Now, Mode = x
k k−1

k + h{ }
(2f −f −f )
k k−1 k+1

15−10
= 30 + 10 { }
2(15)−10−5

= 30 + 3.33 = 33.33

Class Interval Mid - value di = xi -15 ui =

( xi −15)
Frequency fi fiui
59. 6

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 0–6 3 -12 -2 -1 -14

6 – 12 9 -6 -1 5 -5

12 – 18 15 0 0 10 0

18 – 24 21 6 1 12 12

24 – 30 27 12 2 6 12

N = 40 ∑ fi ui =5
Let the assumed mean be (A) = 15
h=6
∑ f ui
Mean = A + h N
i

5
= 15 + 6( 40
)

= 15 + 0.75
= 15.75

n
60. i. Class intereval Frequency Cumulative frequency

i
10-20 12 12

es.
20-30 30 42

30-40 x 42 + x(F)

40-50

50-60 tut
65(f)

y
107 + x

107 + x + y
sti
60-70 25 132 + x + y

70-80 18 150 + x + y
ein

N = 230
let the unknown frequencies are 'x' and 'y'.
Median = 46
Then, median class = 40 - 50
ac

N
−F

∴ Median = l + 2

f
× n
Here,
w.

L = Lower limit of median class

cf = Cumulative frequency of class prior to median class.
f = Frequency of median class.
ww

h = Class size.
∴ l = 40, h = 50 - 40 = 10, f = 65, F = 42 + x

115−(42+x)
⇒ 46 = 40 + 65
× 10
115−(42+x)
⇒ 46 - 40 = 65
× 10
⇒
6×65

10
= 73 - x
⇒ x = 73 - 39 = 34
Given
N = 230
⇒ 12 + 30 + 34 + 65 + y + 25 + 18 = 230

⇒ y = 230 - 184 = 46

Class interval Mid value Frequency fx

ii.
10-20 15 12 180

20-30 25 30 750

30-40 35 34 1190

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 40-50 45 65 2925

50-60 55 46 2530

60-70 65 25 1625

70-80 75 18 1350

N = 230 Σfx = 10550

∑ fx
∴ Mean = N
​​​​​​​

=
10550

230
= 45.87
61. i. Modal class is 19.5 - 24.5
Lowe limit = 19.5
14.5 - 19.5 - 24.5 - 29.5 - 34.5 - 39.5 - 44.5 - 49.5 -
Age (in years)
ii. a. 19.5 24.5 29.5 34.5 39.5 44.5 49.5 54.5
Number of
62 132 96 37 13 11 10 4

n
participants

i
cf 62 194 290 327 340 351 361 365

es.
n

2
=
365

2
= 182.5
median class = 19.5 - 24.5
OR
b. 62 + 132 + 96 + 37 + 13 + 11 + 10 = 361
iii. 3 median = mode + 2 mean
tut
62. i. The maximum class frequency is 15 belonging to class interval 150-155
sti
∴ 150 - 155 is the modal class

lower limit (l) of modal class =150

Height (in cm) frequency C.F
ein

ii.
135-140 2 2

140-145 8 10
ac

145-150 10 20

150-155 15 35
w.

155-160 6 41

160-165 5 46
ww

165-170 4 50

∑ fi = 50

∑ fi = 2 + 8 + 10 + 15 + 6 + 5 + 4 =50 =N
N

2
=
50

2
= 25
c.f just greater that N

2
i.e, 25 is 35
∴ Median class 150-155
Height (in cm) frequency (fi) xi
iii.
135-140 2 137.5

140-145 8 142.5

145-150 10 147.5

150-155 15 152.5

155-160 6 157.5

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 160-165 5 162.5

165-170 4 167.5
lower limit+upper limit
xi = 2

middle term of xi,is the assumean mean

Hence, Assumed Mean = 152.5
OR
n
−c.f

Median = l ( 2
) × h
f

25−20
= 150 + ( 15
)× 5

= 150 + 5

15
× 5

= 150 + 5

= 150 + 1.67
= 151.67

n
63. Number announced 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75

i
Number of times (f) 8 9 10 12 9

es.
cf 8 17 27 39 48 = N

i. N

2
= 24
∴ median class is 30 - 45

ii. P(picking up an even number) =

iii. a. Median = 30 +
(
48

2
−17)

× 15
37

75

tut
sti
10

= 40.5
OR
b. Modal class is 45 - 60
ein

12−10
Mode = 45 + 2×12−10−9
× 15

= 51
64. i. Median class : 100 - 110
ac

ii. No. of leaves equal to or more than 10cm(100 mm) = 23

a. C.I f cf
w.

70-80 3 3

80-90 5 8
ww

90-100 9 17

100-110 12 29

110-120 5 34

120-130 4 38

130-140 2 40 = N
Median = 100 + 10

12
(20 − 17) = 102 ⋅ 5

OR
b. Modal class is 100 - 110
Mode = 100 + 10 × 12−9

24−9−5
= 103

65. i. Women having heart beat in range 68 - 77

= 4 + 3 + 8 = 15
ii. Median class = 74 - 77
f1 − f0
iii. a. Mode = l + ( 2f1 − f0 − f2
) × h

l = 74, f1 = 8, f0 = 3, f2 = 7, h = 3

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 8−3
∴ Modal value = 74 + ( 16−3−7
) × 3

= 76.5
OR

b. No. of heart beats f cf

65 - 68 2 2

68 - 71 4 6

71 - 74 3 9

74 - 77 8 17

77 - 80 7 24

80 - 83 4 28

83 - 86 2 30

n
N
−Cf

Median = I + 2
× h
f

i
(15−9)
= 74 + × 3

es.
8

= 76.25
66.
(d) A is false but R is true.
Explanation:
A is false but R is true.
tut
sti
67.
(c) A is true but R is false.
Explanation:
ein

Median = (mode + 2mean)

1

= 1

3
(60 + 2 × 66) = 64
ac

68.
(d) A is false but R is true.
Explanation:
w.

A is false but R is true.

ww

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