14. In a certain examination there were 100 candidates of whom 21 failed, 6 secured
distinction, 12 were placed in the first division, 18 in the second division and
43 were placed in the third division. It is known that at least 75 percent marks
required for distinction, 40 percent for passing, 50 percent for second division
and 60 percent for first division. Firstly convert above data in class intervals
then calculate the median of the distribution of marks.
Answers
The median of the distribution of marks is 46. 74
Explanation:
Classifying the data in class intervals as:
Marks in Number of Cumulative Frequency (Cf)
Percentage Candidates (f)
0 - 40 21 21
40 - 50 43 64
50 - 60 18 82
60 - 75 12 94
75 - 100 6 100
Total (N) 100
Computing the median as:
Median (M) = Size of (N/2)th item
M = Size of (100/2)th item
M = 50th item
So, it will lie in the class interval of 40-50
The formula for computing the same:
M = L1 + L2 - L2 / f (m - c)
where
L1 is lower class of interval which is 40
L2 is upper class of interval which is 50
f is number of candidates is 43
m is mid value which is 50
c is above class interval cf which is 21
Putting the values above:
M = 40 + (50 - 40) / 43 (50 -21)
M = 40 + (10 + 29) / 43
M = 40 + 6.74
M = 46.74
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