Question

There are 10000 students participate in an exam and the exam score approximately follow a normal distribution. Given that 359 students get scores above 90 in the exam and 1151 students get lower than 60. If there are 2500 students pass the exam, find out the lowest score to pass

Answer 1

Given :

n =number of student = 10000

p1 = Proportion of students get scores more than 90 = 359 /10000 =0.0359

p2 = Proportion of students get scores less than 60 = 1151/1000 = 0.1151

Let the random variable X is defined as

X : exam score of students

From the given condition

-----------------(I)

From normal probabiltiy table

--------------(II)

From (I) and (II)

-------------------(A)

------------------(III)

From normal probabiltiy table

--------------(IV)

From (III) and (IV)

-------------------(B)

Solving equation ( A) and (B) simultaneously we get values of mu and sigma are

Since 2500 students pass in the examination.

p = Proportion of students pass in the examination. = 2500 /10000 = 0.25

Let a be the minimum required score to pass the exam.

Hence

---------------------( V)

From normal probability table

P( Z> 0.6745) = 0.25 ------------------(VI)

From (V) and (VI)

Hence minimum score is 79