Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

The document Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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III. LONG ANSWER TYPE QUESTIONS

Q1. The median of the following frequency distribution is 35. Find the value of x.

 

Class Interval

Frequency

Cumulative

frequency

0-20

7

7

20-40

8

15

40-60

12

27

60-80

10

37

80-100

8

45

100-120

5

50

Total

50

 

 

Also find the modal class.

Sol. Let us prepare the cumulative frequency table:

Class intervals

f

cf

0-10

2

2 + 0 = 2

10-20

3

2 + 3 = 5

20-30

x

5 + x = (5 + x)

30-40

6

(5 + x) + 6 = 11 + x

40-50

5

(11 + x) + 5 = 16 + x

50-60

3

(16 + x) + 3 = 19 + x

60-70

2

(19 + x) + 2 = 21 + x

Total

(21 + x)

 

 

Here,  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Obviously,  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev lies in the class interval 30−40.


l = 30, Cf = x + 5, f = 6 and h = 10 

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

⇒ 5 × 12 = 110 − 10x
⇒ 10x = 110 − 60
⇒ 10x =50 ⇒ x = 5 

(i) ∴ The required value of x is 5.

(ii) ∵ The maximum frequency is 6
∴ The modal class is 30−40.

Q2. Draw the cumulative frequency curve (ogive) by more than method from the following data:

 

Income (in Rs)

Number of persons

0-1000

7

1000-2000

15

2000-3000

35

3000-4000

28

4000-5000

10

5000-6000

5

 

Sol. First, we prepare the cumulative frequency table as follows:

Income (in Rs)

No. of persons

Income

more than

Cumulative

frequency

0-1000

7

more than 0

100 - 0 = 100

1000-2000

15

more than 1000

100 - 7 = 93

2000-3000

35

more than 2000

93 - 15 = 78

3000-4000

28

more than 3000

78 - 35 = 43

4000-5000

10

more than 4000

43 - 28 = 15

5000-6000

5

more than 5000

15 - 10 = 5

 

Thus, we plot the points (0, 100), (1000, 93), (2000, 78), (3000, 43), (4000, 15), (5000, 5) and (6000, 0).

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Q3. The mean of the following data is 53, find the missing frequencies.

Age (in years)

0-20

20-40

40-60

60-80

80-100

Total

Number of people

15

f

21

f2

17

100


Sol. 

 

Age (in years)

Number of People (fi)

Mid value (xi)

fi × xi

0-20

15

10

15 x 10 = 150

20-40

fa

30

f1 x 30 = 30 f1

40-60

21

50

21 x 50 = 1050

60-80

f2

70

h x 70 = 70 f2

80-100

17

90

17 x 90 = 1530

Total

100

 

2730 + 30 f1 + 70 f2

 

Since, 15 + f1 + 21 + f2 + 17 = 100
∴ 53 + (f1 + f2) = 100

⇒ f1 + f2 = 100 − 53 = 47              ...(1)

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

⇒ 2730 + 30 f1 + 70 f2 = 5300
⇒ 30 f1 + 70 f2 = 5300 − 2730 = 2570
⇒ 3 f1 + 7 f2 = 257
⇒ 3 (47 − f2) + 7 f2 = 257     |∵ f1 = 47 − f2
⇒ 141 − 3 f2 + 7 f2 = 257
⇒ 4 f2 = 116 or f2 = 29
∴ f1 = 41 − 29 = 18
Thus, f1 = 18 and f2 = 29.

Q4. The mean of the following distribution is 18.

 

Class interval

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequencies

3

6

9

13

f

5

4


Find f

Sol. Let the assumed mean (a) = 18

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

∴ We have the following table:

Class intervalxifiLong Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRevfi ui

11-13

12

3

-3

3 x (-3) = -9

13-15

14

6

-2

6 x (-2) = -12

15-17

16

9

-1

9 x (-1) = -9

17-19

18

13

0

13 x 0 = 0

19-21

20

f

1

f x 1 = f

21-23

22

5

2

5 x 2 = 10

23-25

24

4

3

3 x 3 = 9

 

 

∑ fi = 40 + f

 

∑fi ui = f− 8

 

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

⇒  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Thus, the required frequency = 8 

Q5. The percentage of marks obtained by 100 students in an examination are given below:

Marks

30 - 35

35 - 40

40 - 45

45 - 50

50 -55

55 - 60

60-65

Frequency

14

16

18

23

18

8

3

 

Find the median of the above data.

Sol. 

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Here n = 100 ⇒ n/2 = 50, which lies in the class 45 – 50, where

l1 (lower limit of the median – class) = 45
c (The cumulative frequency of the class preceding the median class) = 48
f (The frequency of the median class) = 23
h (The class size) = 5

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

∴ The median percentage of marks is 45.4.

Q6. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

 

Class interval

0 - 6

6 -12

12 -18

18 - 24

24 - 30

Frequency

4

x

5

y

1


Sol.

 

Class interval

Frequency

Cumulative

Frequency

0-6

4

4 + 0 = 4

6-12

x

4 + x = (4 + x)

12-18

5

5 + (4 + x) = 9 + x

18-24

y

y + (9 + x) = 9 + x + y

24-30

1

1 + (9 + x + y) = 10 + x + y

 

Since, n = 20
∴ 10 + x + y = 20
⇒ x + y = 20 − 10
⇒ x + y = 10.      ...(1)

Also, we have
Median = 14.4,
which lies in the class interval 12−18.
∴ The median class is 12−18, such that

l = 12, f = 5, Cf = 4 + x and h = 6

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

⇒ 24 + 6x = (9.6) × 5
⇒ 24 + 6x =48
⇒ 6 x = 48 − 24 = 24

⇒ x = 24/6 = 4

Now, from (1), we have:

x + y = 10
⇒ 4 + y = 10
⇒ y = 10 − 4 = 6
Thus, x =4 and y = 6

Q7. The following table gives production yield per hectare of wheat of 100 farms of a village:

 

Production yield (kg/hectare)

Number of farms

40-45

4

45-50

6

50-55

16

55-60

20

60-65

30

65-70

24


Change the distribution to a ‘more than type’ distribution, and draw its ogive.

Sol. The given distribution is converted to a “more than type” as:

Production yield (kg/hectare)

Number of farms (cumulative frequencies)

More than or equal to 40

100 - 0 = 100

More than or equal to 45

100 - 4 = 96

More than or equal to 50

96 - 6 = 90

More than or equal to 55

90 - 16 = 74

More than or equal to 60

74 - 20 = 54

More than or equal to 65

54 - 30 = 24

 

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Q8. The distribution below gives the weights of 30 students of a class. Find the mean and the median weight of the students:

 

Weight (in kg)

Number of students

40-45

2

45-50

3

50-55

8

55-60

6

60-65

6

65-70

3

70-75

2

 

Sol. Let the assumed mean, a = 57.5
∴ h = 5

∴  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

We have the following table:

Weight of students 
 (in kg)

Class

mark 
(xi)

Frequency

fi

Cumulative

frequency 
 Cf

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

fi ui

 

 

 

 

 

 

40-45

42.5

2

2 + 0 = 2

-3

-6

45-50

47.5

3

2 + 3 = 5

-2

-6

50-55

52.5

8

5 + 8 = 13

-1

-8

55-60

57.5

6

13 + 6 = 19

0

0

60-65

62.5

6

19 + 6 = 25

1

6

65-70

67.5

3

25 + 3 = 28

2

6

70-75

72.5

2

28 + 2 = 30

3

6

Total

 

∑ fi = 30

 

 

∑fiui = - 2

 

∴ The mean of the given data  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

For finding the median:

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

And it lies in the class 55–60.

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Q9. The lengths of 40 leaves of a plant are measured correct upto the nearest millimetre and the data is as under:

 

Length (in mm)

Numbers of Leaves

118-126

4

126-134

5

134-142

10

142-150

12

150-158

4

158-166

5


Find the mean and median length of the leaves.

Sol. For finding the mean:

Let the assumed mean a = 146
h = 8

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Now, we have the following table:

Length
 (in mm)
Class
 mark
 (xi)

 
Frequency
 (fi)
Cumulative
 frequency
 Cf
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRevfi ui

118-126

122

4

4 + 0

= 4

 

-3

-12

126-134

130

5

4 + 5

= 9

 

-2

-10

134-142

138

10

9 + 10

= 19

 

-1

-10

142-150

146

12

19 + 12

= 31

 

0

0

150-158

154

4

31 + 4

= 35

 

1

4

158-166

162

5

35 + 5

= 40

 

2

10

Total

 

∑ fi = 40

 

 

∑ fi ui =-18

 

∴ Mean  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
= 142.4 mm

Median: Since  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev  [Median class is 142-150]

l = 142
cf = 19
f = 12 and h = 8

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev
Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Q10. The table below shows the daily expenditure on food of 30 households in a locality:

 

Daily expenditure (in Rs)

Numbers of households

100-150

6

150-200

7

200-250

12

250-300

3

300-350

2


Find the mean and median daily expenditure on food.

Sol. For finding the mean

Let the assumed mean a = 225.

h = 50

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

We have the following table:

Daily
 expenditure
xific.f.Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRevfi ui

100-150

125

6

6 + 0 = 6

-2

(-2) x 6 = -12

150-200

175

7

6 + 7 = 13

-1

(-1) x 7 = -7

200-250

225

12

13 + 12 = 25

0

(0) x 12 = 0

250-300

275

3

25 + 3 = 28

1

(1) x 3 = 3

300-350

325

2

28 + 2 = 30

2

(2) x 2 = 4

Total

 

∑ fi = 30

 

 

∑ fiui =-12

 

∴ Mean  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

⇒  Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

To find median:

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

And 15 lies in the class 200−250.
∴ Median class is 200−250.

∴ l = 200
cf = 13
f = 12 and h = 50

Thus, Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

Long Answer Type Questions(Part- 1)- Statistics Class 10 Notes | EduRev

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