14. The following table gives the distribution of marks secured by some students in a certain examination :
Marks:
0-20 21-30 31-40 41-50 51-60 61-70
No. of Students :42
38
120
84
48
36
71-80
31
Find: (1) Median marks.
(ii) The percentage of failure if minimum for a pass is 35 marks.
ARS. (1) Md=40.46 (ii) 31.58%.
Answers
Given : distribution of marks secured by some students in a certain examination :
To Find : Median marks.
Solution:
Marks f cf
0-20 42 42
21-30 38 80
31-40 120 200
41-50 84 284
51-60 48 332
61-70 36 368
71-80 31 399
399/2 = 199.5
Median = 30.5 + 10 * ( 199.5 - 80)/120
= 30.5 + 9.958
= 40.458
= 40.46
Median Marks = 40.46
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