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## Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.6

1. 1. Draw an given by less than method for the following data:
No.of rooms: 1  2  3  4  5  6  7  8  9  10
No. of houses: 4  9  22  28  24  12  8  6  5  2

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

Solution 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.

Solution 5. The  following distribution the daily income of 50 workers  of  a factory :
Daily income in : 100-120  120-140  140-160  160-180  180-200
Number of workers : 12  14   8   6  10
Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive .

Solution 6. The following table gives production yield per hectare of wheat of 100 farms of a village:
Production yield in kg per hectare: 50-55  55-60  60-65  65-70  70-75  75-80
Number of farms: 2   8  12   24   38   16
Draw ‘less than’ ogive and ‘more than’ ogive.

Solution 7. 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 .

Solution 8. The annual rainfall record of a city for 66 days is given in the following table : Calculate the median rainfall using ogives of more than type and less than type .

Solution

Prepare a table for less than type .
Now, plot the less than ogive using suitable points . Here, N = 66
N/2 = 33
In order to find the median rainfall , we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting to curve at Q. From this point Q, draw a line parallel to the y - axis and meeting the x-axis at the point R. The x - coordinate of R is 21.25 . Thus, median rainfall is is 21.25 cm .  Here , N = 66
N/2 =  33
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis.Let the point be P.From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinates of R is 21.25.

9. The following table gives the height of tress . Draw ' less than 'ogive and more than 'ogive'

Solution

Consider the following table . Now draw less than ogive using suitable points . Now, prepare the cumulative frequency table for men than series . Now draw more than ogive using suitable points . 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

Firstly , we prepare the cumulative frequency table for less than type . Again, prepare the cumulative frequency table for more than type . Now, ''more than ogive'' and ''less than ogive'' can be drawn as follows : The x - coordinate of the point of intersection of the '' more than ogive'' and less- than ogive'' gives the median of the given distribution..
So, the corresponding median is 17.5 lakh .
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