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

Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to equip students with the skills to apply statistical methods to real-world problems enhancing their data interpretation abilities.

Statistics

Statistics involves methods for summarizing and analyzing data to extract meaningful insights. It includes descriptive statistics which summarize data sets through measures like mean, median, and mode, and inferential statistics which make predictions or inferences about a population based on the sample data. Organizing and analyzing data statistics helps in making informed decisions and understanding trends.

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

Solution:

No. of rooms	No. of houses	Cumulative Frequency
Less than or equal to 1	4	4
Less than or equal to 2	9	13
Less than or equal to 3	22	35
Less than or equal to 4	28	63
Less than or equal to 5	24	87
Less than or equal to 6	12	99
Less than or equal to 7	8	107
Less than or equal to 8	6	113
Less than or equal to 9	5	118
Less than or equal to 10	2	120

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.

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

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

Solution:

Marks	No. of Students	Marks Less than	Cumulative Frequency
600 – 640	16	640	16
640 – 680	45	680	61
680 – 720	156	720	217
720 – 760	284	760	501
760 – 800	172	800	673
800 – 840	59	840	732
840 – 880	18	880	750

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

Solution:

Converting the given frequency distribution into continuous frequency distribution:

Class-interval	No. of Students	Less than	Cumulative frequency
0.5 – 4.5	2	4.5	2
4.5 – 9.5	6	9.5	8
9.5 – 14.5	10	14.5	18
14.5 – 19.5	5	19.5	23
19.5 – 24.5	3	24.5	26

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.

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

Draw the frequency polygon for it.

Solution:

Now, computing the following data, we have,

Profit per shop	Mid-value	No of shops:
Less than 0	0	0
Less than 0 – 50	25	12
Less than 50 – 100	75	18
Less than 100 – 150	125	27
Less than 150 – 200	175	20
Less than 200 – 250	225	17
Less than 250 – 300	275	6
Above 300	300	0

Now, the frequency polygon can be computed as follows :

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

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 income	Cumulative frequency
Less than 120	12
Less than 140	26
Less than 160	34
Less than 180	40
Less than 200	50

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. 

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

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 55	50-55	2	2
Less than 60	55-60	8	10
Less than 65	60-65	12	22
Less than 70	65-70	24	46
Less than 75	70-75	38	84
Less than 80	75-80	16	100

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 yield	c.f.	Class
More than or equal to 50	100	50-55
More than or equal to 55	84	55-60
More than or equal to 60	46	60-65
More than or equal to 65	22	65-70
More than or equal to 70	10	70-75
More than or equal to 75	2	75-80
More than or equal to 80	0	80-85

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:

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Question7. During the medical check-up of 35 students of a class, their weights were recorded as follows :

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	Class	c.f.
Less than 38	0	36-38	0
Less than 40	3	38-40	3
Less than 42	5	40-42	2
Less than 44	9	42-44	4
Less than 46	14	44-46	5
Less than 48	28	46-48	14
Less than 50	32	48-50	4
Less than 52	35	50-52	3

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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

∴  = 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 = 

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

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.

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∵ Total number of days (n) = 66

Now,  = 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:

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

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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:

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:

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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

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