# 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 Free 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.

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.

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.

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

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).

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)

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)

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).

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

#### Practise This Question

x=4 is a solution to the equation 2x+8=0