Skip to content
Related Articles

Related Articles

Improve Article

Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.6

  • Last Updated : 18 Mar, 2021
Geek Week

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

No. of rooms12345678910
No. of houses49222824128652

Solution:

No. of roomsNo. of housesCumulative Frequency
Less than or equal to 144
Less than or equal to 2913
Less than or equal to 32235
Less than or equal to 42863
Less than or equal to 52487
Less than or equal to 61299
Less than or equal to 78107
Less than or equal to 86113
Less than or equal to 95118
Less than or equal to 102120

We plot the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118), (10, 120) respectively by taking the upper-class limit over the x-axis and cumulative frequency over the y-axis of the graph.

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

MarksNo. of Students
600 – 64016
640 – 68045
680 – 720156
720 – 760284
760 – 800172
800 – 84059
840 – 88018

Prepare a cumulative frequency distribution table by less than method and draw an ogive.

Solution:

MarksNo. of StudentsMarks Less thanCumulative Frequency
600 – 6401664016
640 – 6804568061
680 – 720156720217
720 – 760284760501
760 – 800172800673
800 – 84059840732
840 – 88018880750



Question 3. Draw an Ogive to represent the following frequency distribution:

Class-interval0 – 45 – 910 – 1415 – 1920 – 24
No. of students261053

Solution:

Converting the given frequency distribution into continuous frequency distribution:

Class-intervalNo. of StudentsLess thanCumulative frequency
0.5 – 4.524.52
4.5 – 9.569.58
9.5 – 14.51014.518
14.5 – 19.5519.523
19.5 – 24.5324.526

We plot the specified points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5, 26) on a graph by taking the upper class limit over the x-axis and cumulative frequency over the y-axis respectively.

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

Profit per shopNo of shops:
0 – 5012
50 – 10018
100 – 15027
150 – 20020
200 – 25017
250 – 3006

Draw the frequency polygon for it.

Solution:

Now, computing the following data, we have,

Profit per shopMid-valueNo of shops:
Less than 000
Less than 0 – 502512
Less than 50 – 1007518
Less than 100 – 15012527
Less than 150 – 20017520
Less than 200 – 25022517
Less than 250 – 3002756
Above 3003000

Now, the frequency polygon can be computed as follows :



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

Daily income (in Rs):No of workers:
100 – 12012
120 – 14014
140 – 1608
160 – 1806
180 – 20010

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

Solution:

Using the less than method, the following distribution can be converted to a continuous distribution, as,

Daily incomeCumulative frequency
Less than 12012
Less than 14026
Less than 16034
Less than 18040
Less than 20050

Mark the point (120, 12), (140, 26), (160, 34), (180, 40), (200, 50), taking upper class limit on the x-axis and cumulative frequencies on y-axis respectively. 

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

Production yield50-5555-6060-6565-7070-7575-80 in kg per hectare
Number of farms2812243816

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

Solution:

(i) Computing the Less than ogive

Production yield

 in kg 1 hectare

Class

No. of farms.

(f)



c.f.
Less than 5550-5522
Less than 6055-60810
Less than 6560-651222
Less than 7065-702446
Less than 75 70-753884
Less than 8075-8016100

We plot the specified points (55, 2), (60, 10), (65, 22), (70, 46), (75, 84) and (80, 100) and connect them to form an ogive. 

(ii) More than
 

Production yieldc.f.Class
More than or equal to 5010050-55
More than or equal to 558455-60
More than or equal to 604660-65
More than or equal to 652265-70
More than or equal to 701070-75
More than or equal to 75275-80
More than or equal to 80080-85

We plot the specified points (50, 100), (55, 84), (60, 46), (65, 22), (70, 10), (75, 2) and (80, 0) and connect to form a more than ogive as shown below:
 

Question 7. During the medical check-up of 35 students of a class, their weights were recorded as follows :

Weight (in Kg)

Number 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 result by using the formula. (C.B.S.E. 2009)

Solution: 

Weight (in Kg)

Number of students

Classc.f.

Less than 38

0

36-380

Less than 40

3

38-403

Less than 42



5

40-422

Less than 44

9

42-444

Less than 46

14

44-465

Less than 48

28

46-4814

Less than 50

32

48-504

Less than 52

35

50-523

Plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) on the graph and join them in free hand to get an ogive as shown.

Here N = 35 which is odd

∴ \frac{N}{2}=\frac{25}{2}    = 252 = 17.5

From 17.5 on y-axis draw a line parallel to x-axis meeting the curve at P. From P, draw PM ⊥ x-axis

∴ Median which is 46.5 (approx)

Now N = 17.5 lies in the class 46 – 48 (as 14 < 17.5 < 28)

∴ 46-48 is the median class

Here l= 46, h = 2,f= 14, F= 14

∴ Median = l+\frac{\frac{N}{2}-F}{f}\times h=46+\frac{17.5-14}{14}\times2=46+\frac{3.5\times2}{14}=46+0.5=46.5     

Question 8. The annual rainfall record of a city for 66 days is given in the following tab
 

Rainfall (in cm):

0-10

10-20

20-30

30-40

40-50

50-60

Number of days:

22



10

8

15

5

6

Calculate the median rainfall using ogives of more than type and less than type. [NCERT Example]

Solution:

We observe that, the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10. So,

the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20, 30, 40, 50 and 60.Also, we observe that annual rainfall

record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 0-10. So, annual rainfall record for 66 – 22 = 44 days is more than or equal to 10 cm. Continuing in this

manner we will get remaining more than or equal to 20, 30 , 40, 50, and 60.

Now, we construct a table for less than and more than type.

(i) Less than type

(ii) More than type

Rainfall (in cm)

Number of days

Rainfall (in cm)

Number of days

Less than 0

0

More than or equal to 0

66

Less than 10

0+22=22

More than or equal to 10

66-22=44

Less than 20

22+10=32

More than or equal to 20

44-10=34

Less than 30



32+8=40

More than or equal to 30

34-8=26

Less than 40

40+15=55

More than or equal to 40

26-15=11

Less than 50

55+5=60

More than or equal to 50

11-5=6

Less than 60

60+6=66

More than or equal to 60

6-6=0

To draw less than type ogive we plot the points (0,0), (10,22), (20,32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.

To draw the more than type ogive we plot the points (0, 66), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the graph paper and join them by free hand.

∵ Total number of days (n) = 66

Now, \frac{n}{2}    = 33

Firstly, we plot a line parallel to X-axis at intersection point of both ogives, which further intersect at (0, 33) on Y- axis. Now, we draw a line perpendicular to X-axis at intersection point of both ogives,

which further intersect at (21.25, 0) on X-axis. Which is the required median using ogives.

Hence, median rainfall = 21.25 cm.

Question 9. The following table gives the height of trees:

Height

Number of trees

Less than 7

26

Less than 14

57

Less than 21



92

Less than 28

134

Less than 35

216

Less than 42

287

Less than 49

341

Less than 56

360

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

Solution:

(i) First we prepare less than frequency table as given below:

Height

Class interval

Frequency

c.f.

Less than 7

0-7

26

26

Less than 14

7-14

31

57

Less than 21

14-21

35

92

Less than 28

21-28

42

134

Less than 35

28-35

82

216

Less than 42

35-42

71

287

Less than 49

42-49

54

341

Less than 56

49-56

19

360

Now we plot the points (7, 26), (14, 57), (21, 92), (28, 134), (35, 216), (42, 287), (49, 341), (56, 360) on the graph and join then in a frequency curve which is ‘less than ogive’

(ii) More than ogive:

First we prepare ‘more than’ frequency table as shown given below:

More than

Class interval

c.f.

Frequency

More than 0

0-7

360



19

More than 7

7-14

341

54

More than 14

14-21

287

71

More than 21

21-28

216

82

More than 28

28-35

134

42

More than 35

35-42

92

35

More than 42

42-49

57

31

More than 49

49-56

26

26

More than 56

56-

0

0

Now we plot the points (0, 360), (7, 341), (14, 287), (21, 216), (28, 134), (35, 92), (42, 57), (49, 26), (56, 0) on the graph and join them in free hand curve to get more than ogive.

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

Profit (in lakhs ₹) 

Class intervals

Number of shops (c.f.)

Frequency

More than or equal to 5

5-10

30

2

More than or equal to 10

10-15

28

12

More than or equal to 15

15-20

16

2

More than or equal to 20

20-25

14

4

More than or equal to 25

25-30

10

3

More than or equal to 30

30-35



7

4

More than or equal to 35

35-40

3

0

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

Solution:

Class interval

Frequency

c.f.

5-10

3

3

10-15

4

7

15-20

3

10

20-25

4

14

25-30

2

16

30-35

12

28

35-40

2

30

Now plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) on the graph and join them to get a more than curve.

Less than curve:

Now plot the points (10, 3), (15, 7), (20, 10), (25, 14), (30, 16), (35, 28) and (40, 30) on the graph and join them to get a less them ogive. The two curved intersect at P. From P, draw PM 1 x-axis, M is the

median which is 22.5

∴ Median = Rs. 22.5 lakh

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :