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
Answers
Less Than Cumulative Frequency distribution table & Graph Of Less Than Ogive Is In The Attachment.
From the table : Plot these points (38,0) (40,3)(42,5)(44,9)(46,4)(48,28)(50,32)(52,35)by taking upper class limit over the x-axis and cumulative frequency over the y-axis.
★★ LESS THAN TYPE OGIVE:
It is the graph drawn between upper limits and cumulative frequencies of a distribution. Here, we mark the points with upper limit and x- coordinate and corresponding cumulative frequency as y- coordinate and join them by freehand smooth curve. This type of graph is cumulated upward.
For Median :
Here ,n = 35
n/2 = 17.5
From (0,17.5) draw a line parallel to the x-axis cutting the curve at a point M having x - coordinate as 46.5 .
Hence, the Median is 46.5 kg .
VERIFICATION :
Here ,n = 35
n/2 = 17.5
Since, the Cumulative frequency just greater than 17.5 is 28 and the corresponding class is 46 - 48 . Therefore 46 - 48 is the median class.
Here, l = 46 , f = 14 , c.f = 14, h = 2
MEDIAN = l + [(n/2 - cf )/f ] ×h
= 46 + [(17.5 - 14)/14] × 2
= 46 + [(3.5 × 2)/14]
= 46 + 7 /14
= 46 + 1/2
= 46 + 0.5
MEDIAN = 46.5 kg
Both the methods give same answer.
Hence , verified .
★★ For Finding the MEDIAN from Cumulative frequency curve either less than ogive or more than ogive, locate n/2 on the Y-axis , where n is the total number of observations. From this point ,draw a line parallel to the x-axis cutting the curve at a point. From this point, draw perpendicular to the x-axis. The abscissa of point of intersection of the perpendicular with the x-axis is the median of the data.
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