The class marks of a frequency distribution are given as follows:
15, 20, 25, …… The class corresponding to the class mark 20 is ………………..
Solution:
Class size = 20 – 15 = 5
Consider a as the lower limit of the required class
Upper limit can be written as a + 5
(a + (a + 5))/ 2 = 20
2a + 5 = 40
2a = 35
a = 17.5
Upper limit a + 5 = 17.5 + 5 = 22.5
Hence, the required class is 17.5-22.5
In the class intervals 10 – 20, 20 – 30, the number 20 is included in
(a) 10-20
(b) 20 – 30
……………………
Solution:
(d) none of these
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