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during the medical check up of …

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during the medical check up of 35 students of a class there with were recorded as follows weight in kg less than 30 80 less than 43 less than 40 25 + 144 + 9 + 14614 less than 4828 less than 50 32 less than 5235
  • 1 answers

Sia ? 6 years, 7 months ago

More than method: cumulative frequency

Weight No. of students Weight more than Cumulative frequency
38-40 3 38 35
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

On X-axis plot lower class limits.On Y-axis plot cumulative frequency. 
We plot the points (38,35),(40,32),(42,30),(44,26),(46,26),(48,7),(50,3).
Less than method :

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 14
46-48 14 28
48-50 4 32
50-0 3 35

On x-axis plot upper class limits.On Y-axis plot cumulative frequency
We plot the points (38,0),(40,3),(42,5),(44,9),(46,4),(48,28),(50,32),(52,35).

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
Now, N = 35
{tex}\frac { N } { 2 } = 17.5{/tex}
The cumulative frequency just greater than {tex}\frac {N}{2}{/tex} is 28 and the corresponding class is 46 - 48.
Thus 46 - 48 is the median class such that
L = 46, f = 14, C= 14 h = 2
Median {tex}= L + \frac { \frac { N } { 2 } - c _ { 1 } } { f } \times h{/tex}
{tex}= 46 + \frac { 17.5 - 14 } { 14 } \times 2{/tex}
{tex}= 46 + \frac {7}{14}{/tex}
= 46.5
Median = 46.5 kg
Hence verified.

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