In a certain examination the percentage of passes and distinctions were 46 and 9
respectively. Estimate the average marks obtained by the candidates and their standard
deviation, the minimum pass and distinction marks being 40 and 75 respectively. assume normal distribution
Answers
Given : In a certain examination the percentage of pa sses and distinctions were 46 and 9 respectively.
the minimum pa ss and distinction marks being 40 and 75 respectively. assume normal distribution
To Find : Estimate the average marks obtained by the candidates and their standard deviation,
Solution:
z score = ( value - Mean ) /SD
Mean = x
SD = s
46 % pa ss and 40 marks
=> -0.1 z score for 46 % and value = 40
=> - 0.1 = ( 40 - x) S
=> -0.1S = 40 - x
9 % distinctions and 75 marks
=> 91 % before 75 marks
z score = 1.34 and Value = 75
=> 1.34 = ( 75 - x) /S
=> 1.34S = 75 - x
1.44S = 35
=> S = 24.3
-0.1S = 40 - x
=> x = 42.43
Mean = 42.43
Standard deviation = 24.3
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