ESE 2018

Question

In a sample of 100 students, the mean of the marks (only integers) obtained by them in a test is 14 with its standard deviation of 2.5(marks obtained can be fitted with a normal distribution ).the percentage of students scoring 16 marks is

a)36

b)23

c)12

d)10

(Area under standard normal curve between z=0 and z=0.6 is 0.2257 ; and between z=0 and z=1.0 is 0.3413)

Comments

can you send me the question paper? i think there is some mistake in question

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getting z as 2/2.5 =>0.8 but the given values are not useful.

Taken z value as 0.6 (should not be taken like this but taken because during the final value some approx is done) then the answer will be option B

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i dont have question paper.

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It's 2018 ese question...no error

Answer 1

the answer will be option B

 

Comments

P(x=16) = (area between z=0 and z=1.0 is 0.3413 ) – (area between z=0 and z=0.6 is 0.2257)

P(X=16) = 0.12 and in percentage it will be 12%.

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@alokmehta07  Why subtract the areas?