Math, asked by RdShelar5483, 1 year ago

During the medical check-up of 35 students of a class their weights were recorded as follows:
Weight (in kg)
38 − 40
40 − 42
42 − 44
44 − 46
46 − 48
48 − 50
50 − 52
Number of students
3
2
4
5
14
4
3
Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph

Answers

Answered by ishitamogha21
97
hope this answer will help you.
Attachments:

ishitamogha21: it is right
rockbunty2003pe0rbe: say me how 14 will come twice
rockbunty2003pe0rbe: In median and cf
Answered by amitnrw
6

Given :  medical check-up of 35 students of a class

their weights were recorded

To Find : Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph

Solution:

Weight (in kg)    f           Less  than        More than

38 − 40              3               3                     35

40 − 42              2                5                    32

42 − 44              4                9                    30

44 − 46               5              14                   26

46 − 48               14             28                  21

48 − 50               4              32                     7        

50 − 52               3               35                   3

Median is 46.5 kg  from graph

from formula :

46 +  2 * (17.5 - 14)/14

= 46 + 0.5

= 46.5

Less than ogive

The points with the upper limits of the class are plotted on x axis  and the corresponding less than cumulative frequencies on y axis.

The points are joined by free hand smooth curve to give less than cumulative frequency curve or the less than Ogive

More than ogive

The points with the Lower limits of the class are plotted on x axis  and the corresponding more than cumulative frequencies on y axis.

The points are joined by free hand smooth curve to give more than cumulative frequency curve or the More than Ogive

Learn More:

Draw a less than and more than ogive in the same graph and find ...

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Draw more than ogive for following frequency distribution and hence ...

brainly.in/question/12809143

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