4. 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
the distribution of marks to be normal).
Answers
Answer:
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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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