some students of class X donated for welfare of old age persons. Their contributions are shown in the following frequency distribution table: Amount=0-20,20-40,40-60,60-80,80-100. Number of students= 5,8,12,11,4 . Find the median and mode of their contributions
Answers
Solution :-
Amount(in Rs.) No. of students c.f.
0-20 5 5
20-40 8 13
40-60 12 25
60-80 11 36
80-100 4 40
Total. ⅀fi = 40.
we have :-
N = 40
(N/2) = 20 .
Median class = 40 - 60 .
F(frequency of median class) = 12
L(Lower limit of median class) = 40 .
c.f.(cumulative frequency of median class) = 13
h(size of median class) = 60 - 40 = 20.
Median :-
L + [ {(N/2) - c.f} / F ] * h
Putting all values we get :-
→ Median = 40 + [(20 - 13) / 12] * 20
→ Median = 40 + (7/12) * 20
→ Median = 40 + (35/3)
→ Median = (120 + 35)/3
→ Median = (155/3)
→ Median = 51.67 (Ans.)
Now, for Mode we have :-
Modal class = 40 - 60 .
f1(frequency of modal class) = 12 .
f0(freq. previous than modal class) = 8.
f2(freq. next to modal class) = 11.
h(size of modal class) = 60 - 40 = 20.
L(Lower limit of modal class) = 40 .
Mode :-
L + [ (f1 - f0) / (2f1 - f0 - f2) ] * h
Putting all values we get :-
→ Mode = 40 + [(12 - 8) / (2*12 - 8 - 11)] * 20
→ Mode = 40 + [4 /(24 - 19)] * 20
→ Mode = 40 + (4/5) * 20
→ Mode = 40 + 4 * 4
→ Mode = 40 + 16
→ Mode = 56 (Ans.)
Hence, Median and mode for their contribution is 51.67 and 56.
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