Math, asked by lavanyanivathitha15, 3 months ago

In an engineering examination a student is considered to have failed, secured second less than 45%, between 45% and 60%, between 60% and 75% and above 75% respectively. In a particular yesr 10% of the students failed in the examination and 5% got distinction. Find the percentage of students who have got second class.(ude Normal distribution)

Answers

Answered by amitnrw
4

Given : In an engineering examination a student is considered

failed  less than 45%,

second class between 45% and 60%

1st class between 60% and 75%

Distinction above 75%

10% of the students failed in the examination

5% got distinction

To Find : the percentage of students who have got second class

Solution:

Let say Mean Marks = M %

and Standard deviation = S %

Z score = ( Value - Mean)/SD

Failed 10 % for Less than 45 %

Value = 45 %

Z score for 10 % =  -1.281

-1.281 = (45 - M)/S

=> -1.281S = 45 - M

5% got distinction.

=> Z score for 100 - 5 = 95 % = 1.645

Value =  75 %

1.645 = ( 75 - M)/S

=> 1,645S = 75 - M

=> 2.926S = 30

=> S = 10.253

  M = 58.134

second class between 45% and 60%

Z score for 45 % = ( 45 -  58.134)/10.253 = -1281   =  10 %

Z score for 60 % = ( 60 -  58.134)/10.253 =  0.182   =  57.2 %

percentage of students who have got second class = 57.2 -  10  = 47.2 %

47.2  percentage of students  got second class

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Answered by Anonymous
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47.2

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47.2

Answered by Anonymous
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47.2

Answered by Anonymous
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47.2

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