Question

in an examination it is laid down that a student passes if he secured 30% or marks his place in the first second and third according as he secure 60% or more marks between 45% and 60% marks and between 30 and 45% respectively he gets a destination in case he secure80 percent or more marks.it is notice from the result that 10% of the student fail in examination where is 5% of them were obtained assassination calculate the percentage of student place in the second division is normal distribution of marks ​

Answer 1

Given : in an examination it is laid down that a student passes if he secured 30% or marks his place

To find : the percentage of student place in the second division is normal distribution of marks

Solution:

student passes if he secured 30% or more marks

10% of the student fail in examination

Z score for 10%  =  -1.281

Let say Mean Marks = M  & SD  = S

Z score = ( Value - Mean)/SD

=> -1.281  = (30 - M )/S

=> -1.281S = 30 - M

80 percent or more  Distinction

5% of them   obtained Distinction (=> 95 % Did not obtained)

Z score for 95 %  = 1.645

=> 1.645 = (80 - M)/S

=> 1.645S = 80 - M

=> S = 17.1

M = 51.9

marks between 45% and 60% for second division

Z score for 45%  = ( 45 - 51.9)/17.1  =  -0.403 =>  34.4 %

Z score for 60%  = ( 60 - 51.9)/17.1  =   0.473 =>  68.2 %

percentage of student place in the second division  = 68.2 - 34.4  = 33.8 %

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