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EX 14.4(2) During the medical check-up of 35 students of a class, their weight were recorded as, follows :
Numbers of
students
Less than
38
Less than
40
Less than
42
Less than
44
Less than
46
0
3
5
9
14
Less than
48
Less than
50
Less than
52
28
32
35
Draw a less than type ogive for the given data.Hence obtain the median weight from graph and verify the result by using the formula.
obtain the median
weight from graph
Draw a less than type ogive
Weight
(in Kg)
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Weight
(in Kg)
Numbers of students
Cumulative Frequency
36 - 38
38 - 40
40 - 42
42 - 44
44 - 46
0
3 – 0 = 3
5 – 3 = 2
9 – 4 = 5
14 – 9 = 6
0
3
5
9
14
(38,0)
(40,3)
(42,5)
(44,9)
(46,14)
46 - 48
48 - 50
50 - 52
28 – 14 = 14
32 – 28 = 4
35 – 32 = 3
28
32
35
(48,28)
(50,32)
(52,35)
Point to be plotted
Median from graph is 46.5 kg
Now, =
N
2
=
35
2
=
17.5.
Which lies in the class 46 - 48
∴
Median class is 46 – 48.
∴ L =
46,
f =
14,
cf =
14
and h =
2
Exercise 14.4 – Q.2
x'
x
y'
y
0
Weight in Kg.
No. of students
5
10
15
20
25
30
35
38
40
42
44
46
48
50
52
Scale : X-axis, 1cm = 1 Kg
Y-axis, 1cm = 5 students
(38,0)
(40,3)
(42,5)
(44,9)
(46,14)
(48,28)
(50,32)
(52,35)
Soln.
Now let us plot the points on a graph
2
2
2
2
2
2
2
38
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Points to
be plotted
(38,0)
(40,3)
(42,5)
(44,9)
(46,14)
(48,28)
(50,32)
(52,35)
Leaving 2 cm from bottom we draw horizontal X-axis and leaving 2cm from left we draw vertical Y-axis
On X-axis if classes are not starting from ‘0’ leaving 2cm from X-axis we start putting limits .
Looking at the biggest Y-co ordinate we select the scale on Y - axis
We plot the points one after the other and write the co-ordinate
Join all points with smooth curve
The difference between origin and lower limit of first class is not same as the width of classes hence we put a Krink mark
(46.5)
Exercise 14.4 – Q.2
∴ Median =
L +
N
2
- c.f
f
× h
=
46 +
- 14
14
× 2
17.5
=
46 +
3.5
7
=
46 +
0.5
∴ Median =
46.5
Hence the median is same as obtained from graph.
7
0.5
Exercise 14.4 – Q.2