14. In a certain examination there were 100 candidates of whom 21 fail

1.	14. In a certain examination there were 100 candidates of whom 21 failed, 6 secureddistinction, 12 were placed in the first division, 18 in the second division and43 were placed in the third division. It is known that at least 75 percent marksrequired for distinction, 40 percent for passing, 50 percent for second divisionand 60 percent for first division. Firstly convert above data in class intervalsthen calculate the median of the distribution of marks.
Answer» 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 You can LEARN more from here about median: brainly.in/question/2666187 You can learn more from here about median: brainly.in/question/7415136

14. In a certain examination there were 100 candidates of whom 21 failed, 6 secureddistinction, 12 were placed in the first division, 18 in the second division and43 were placed in the third division. It is known that at least 75 percent marksrequired for distinction, 40 percent for passing, 50 percent for second divisionand 60 percent for first division. Firstly convert above data in class intervalsthen calculate the median of the distribution of marks.

Answer»

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

You can LEARN more from here about median:

You can learn more from here about median: