RD Sharma Solutions Class 10 Statistics Exercise 7.6

RD Sharma Solutions Class 10 Chapter 7 Exercise 7.6

RD Sharma Class 10 Solutions Chapter 7 Ex 7.6 PDF Download

Exercise 7.6

 

Q.1 Draw an ogive by less than the method for the following data:

 

No. of rooms No. of houses
1

2

3

4

5

6

7

8

9

10

4

9

22

28

24

12

8

6

5

2

 

Soln:

 

No. of rooms No. of houses Cumulative Frequency
Less than or equal to 1

Less than or equal to 2

Less than or equal to 3

Less than or equal to 4

Less than or equal to 5

Less than or equal to 6

Less than or equal to 7

Less than or equal to 8

Less than or equal to 9

Less than or equal to 10

4

9

22

28

24

12

8

6

5

2

4

13

35

63

87

97

107

113

118

120

We need to plot the points (1,4) , (2,3) , (3,35) , (4,63) , (5,87) , (6,99) , (7,107) , (8,113) , (9,118) , (10,120), by taking upper class limit over the x-axis and cumulative frequency over the y-axis.

 

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Q.2): The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

 

Marks No. of Students
600-640

640-680

680-720

720-760

760-800

800-840

840-880

16

45

156

284

172

59

18

 

Soln:

 

Marks No. of Students Marks less than Cumulative Frequency
600-640

640-680

680-720

720-760

760-800

800-840

840-880

16

45

156

284

172

59

18

640

680

720

760

800

840

880

16

61

217

501

693

732

750

Plot the points (640-16), (680, 61), (720,217), (760,501), (800,673), (840,732), (880,750) by taking upper class limit over the x-axis and cumulative frequency over the y-axis.

 

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Q.3) Draw an ogive to represent the following frequency distribution:

Class-interval 0-4 5-9 10-14 15-19 20-24
No. of students 2 6 10 5 3

 

Soln:

The given frequency distribution is not continuous, so we will first make it continuous and then prepare the cumulative frequency:

 

Class-interval No. of Students Less than Cumulative frequency
0.5-4.5

4.5-9.5

9.5-14.5

14.5-19.5

19.5-24.5

2

6

10

5

3

4.5

9.5

14.5

19.5

24.5

2

8

18

23

26

 

Plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5,26) by taking the upper class limit over the x-axis and cumulative frequency over the y-axis.

 

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Q.4) The monthly profits (in Rs) of 100 shops are distributed as follows:

 

Profits per shop: 0-50 50-100 100-150 150-200 200-250 250-300
No of shops: 12           18           27           20           17           6

 

Draw the frequency polygon for it

 

Soln:

We have

 

Profit per shop Mid-value No. of shops
Less than 0

0-60

60-120

120-180

180-240

240-300

300-360

Above 300

0

25

75

125

175

225

275

300

0

12

18

27

20

17

6

0

4

 

Q.5) The following distribution gives the daily income of 50 workers of a factory:

 

Daily income (in Rs): 100-120     120-140     140-160      160-180     180-200
No of workers: 12                14                  8                   6                          10

 

Convert the above distribution to a ‘less than’ type cumulative frequency distribution and draw its ogive.

 

Soln:

We first prepare the cumulative frequency table by less than method as given below

Daily income Cumulative frequency
<120 12
<140 26
<160 34
<180 40
<200 50

Now we mark on x-axis upper class limit, y-axis cumulative frequencies.

Thus we plot the point (120,12)(140,26)(160,34)(180,40)(200,50).

 

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Q.6) The following table gives production yield per hectare of wheat of 100 farms of a village:

 

Production yield: 50-55     55-60     60-65     65-70     70-75     75-80 in kg per hectare
No of farms: 2              8              12           24           38           16

 

Draw ‘less than’ ogive and ‘more than’ ogive

 

Soln:

Less than method:

Cumulative frequency table by less than method

Production yield Number of farms Production yield more than Cumulative frequency
50-55 2 55 2
55-60 8 60 10
60-65 12 65 22
65-70 24 70 46
70-80 38 75 84
75-80 16 80 100

Now we mark on x-axis upper class limit, y-axis cumulative frequencies.

We plot the point (50,100) (55, 98) (60, 90) (65, 78) (70, 54) (75, 16)

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Q.7)During the medical check-up of 35 students of a class, their weight recorded as follows:

Weight (in kg) No of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

 

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the verify the result my using the formula.

 

Soln: Less than method

It is given that

On x-axis upper class limits. Y-axis cumulative frequency

We plot the points (38,0) (40,3)(42,5)(44,9)(46,4)(48,28)(50,32)(52,35)

More than method: cumulative frequency

Weight No. of students Weight more than Cumulative frequency
38-40 3 38 34
40-42 2 40 32
42-44 4 42 30
44-46 5 44 26
46-48 14 46 21
48-50 4 48 7
50-52 3 50 3

X-axis lower class limits on y-axis cf

We plot the points (38,35)(40,32)(42,30)(44,26)(46,26)(48,7)(50,3)

 

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We find the two types of curves intersect at a point P. From point P perpendicular PM is draw on x-axis

The verification,

We have

Weight (in kg) No. of students Cumulative frequency
36-38 0 0
38-40 3 3
40-42 2 5
42-44 4 9
44-46 5 28
46-48 14 32
48-50 4 32
50-0 3 35

Now, N = 35

\(\frac{N}{2}=17.5\)

The cumulative frequency just greater than \(\frac{N}{2}\) is 28 and the corresponding class is 46 – 48

Thus 46 – 48 is the median class such that

L = 46, f = 14, \(C_{1}\)=14 h = 2

Median = \(L+\frac{\frac{N}{2}-c_{1}}{f}\times h\)

= \(46+\frac{17.5-14}{14}\times 2\)

= 46 +7/14

46.5

Median = 46.5 kg

Hence verified

 

 

Q.9) The following table shows the height of trees:

Height No. of trees
Less than  7

Less than  14

Less than  21

Less than  28

Less than  35

Less than  42

Less than  49

Less than  56

26

57

92

134

216

287

341

360

Draw ‘less than ‘ogive and ‘more than ‘ogive

 

Soln:

By less than method

Height No. of trees
Less than  7

Less than  14

Less than  21

Less than  28

Less than  35

Less than  42

Less than  49

Less than  56

26

57

92

134

216

287

341

360

Plot the points (7,26) , (14,57) , (21,92) , (28,134) , (35,216) , (42,287) , (49,341) , (56,360) by taking upper class limit over the x-axis and cumulative frequency over the y-axis.

By more than method:

 

Height Frequency Height more than C.F.
0-7

7-14

14-21

21-28

28-35

35-42

24-49

49-56

26

31

35

42

82

71

54

19

0

5

10

15

20

25

30

35

360

334

303

268

226

144

73

19

 

Take lower class limit over the x-axis and CF over the y-axis and plot (0,360) , (7,334) , (14,303) , (21,268)  (28,226) , (35,144) , (42,73) , (49,19).

 

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Q.10) The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Profit (In lakhs In Rs) Number of shops (frequency)
More than or equal to 5 30
More than or equal to 10 28
More than or equal to 15 16
More than or equal to 20 14
More than or equal to 25 10
More than or equal to 30 7
More than or equal to 35 3

Draw both ogive for the above data and hence obtain the median.

 

Soln:

More than method

Profit (In lakhs in Rs) Number of shops (frequency)
More than or equal to 5 30
More than or equal to 10 28
More than or equal to 15 16
More than or equal to 20 14
More than or equal to 25 10
More than or equal to 30 7
More than or equal to 35 3

 

Now, we mark on x-axis lower class limits, y-axis cumulative frequency

Thus, we plot the points (5,30)(10,28)(15,16)(20,14)(25,10)(30,7) and (35,3)

Less than method

Profit in lakhs No of shops Profits less than C.F
0-10 2 10 2
10-15 12 15 14
15-20 2 20 16
20-25 4 25 20
25-30 3 30 23
30-35 4 35 27
35-40 3 40 30

Now we mark the upper class limits along x-axis and cumulative frequency along y-axis.

Thus we plot the points (10,2)(15,14)(20,16)(25,20)(30,23)(35,27)(40,30)

We find that the two types of curves intersect of P from point L it is drawn on x-axis

The value of a profit corresponding to M is 17.5. Hence median is 17.5 lakh

 

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