In an examination, minimum 40 marks for passing and 75 marks for distinction are
required. In this examination 45% students passed and 9% obtained distinction. Find
average marks and standard deviation of this distribution of marks.
[P(z=0.125)=0.05 and P(z=1.34)=0.41]
Answers
Given : minimum 40 marks for pa ssing and 75 marks for distinction are
required. In this examination 45% students pa ssed and 9% obtained distinction.
To Find : average marks and standard deviation of this distribution of marks.
Solution:
45% of Students pa ssed
at least 40 marks in order to pa ss
=> for more than 40 marks percentage = 45 % => 55% got upto 40 marks
z score = 0.125
9% of Students pa ssed
at least 75 marks in order to pa ss with distinction
for more than 75 marks percentage = 9 % => 91% got upto 75 marks
z score = 1.34
Z score = ( value - Mean)/SD
=> 0.125 = (40 - Mean)/SD
=>0.125SD = 40 - mean
1.34 = (75 - Mean)/SD
=> 1.34SD = 75 - Mean
=>1.215SD = 35
=> SD = 28.8
=> mean = 36.4
Mean = 36.4 and SD = 28.8
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