In a certain examination,there were 100 candidates of whom 21 failed,6 secured distinctions,12 were placed in the first division,43 were - Economics - Measures of Dispersion

Question

In a certain examination,there were 100 candidates of whom 21
failed,6 secured distinctions,12 were placed in the first
division,43 were placed in the third division and 18 in the second
division It is known that at least 75% marks are required for
distinction,40% for passing,50% for second division and 60% for
first division Organise the date and calculate median,first and the
third quartiles

Prince Verma · Accepted Answer

Class
Frequency
Cf

0 - 40
21
21

40 - 50
43
64

50 - 60
18
82

60 - 75
12
94

75 - 100
6
100

Total
100

M = N2 = 1002 = 50Median class is 40-50Median = L + N2- Cff × i= 40 + 50 - 2143×10 = 46
74First Quartile-Q1 = Size of N4th item = 1004th item = 25th itemThe class corresponsing to Q1 is 40-50Q1 = l + N4- Cff × i = 40 + 25 - 2143×10    = 40
94 Third quartile-Q3 = Size of 3(N4)th item = 3(1004)th item = 75th itemThe class corresponsing to Q3 is 50-60Q3 = l + 3(N4)- Cff × i = 50 + 75 - 6443×10    = 52
5

Class	Frequency	Cf
0 - 40	21	21
40 - 50	43	64
50 - 60	18	82
60 - 75	12	94
75 - 100	6	100
Total	100