Question

Normal distribution of Calculus 1 course in Chemical Engineering Department
According to the grade point average is 50 and the standard deviation is 3. Since the passing grade is 45:
a) What is the percentage of probability of Chemical Engineering students from the course?
b) What is the probability that Chemical Engineering students will pass the course?
c) What is the grade of the student who gets a grade better than 90% of the class from Calculus 1 course?
d) What is the grade of the student who gets a grade worse than 25% of the class from Calculus 1 course?

Answer 1

Solution :

Let we define; X = the grade points of Calculus 1 course student in Chemical Engeneering Department.

We have X is normally distributed with average 50 and the standard deviation 3. The passing grade is 45.

The probability that Chemical Engineering students will pass the course is given by; P( X > 45 ).

The Z score for 45 is given by; Z0 = (45 - mean) / sd = (45 - 50) / 3 = -1.67.

From the Z table; we get P(Z < -1.67) = 0.0475. This gives us P(Z > -1.67) = 1 - P(Z < -1.67) = 1 - 0.0475 = 0.9525.

And hence P( X > 45 ) = P( Z > -1.67 ) = 0.9525.

Let grade of the student who gets a grade better than (lowest) 90% of the class from Calculus 1 course be "x1".

That means P( X < x1 ) = 0.90. But from the Z table we get P( Z < 1.28) = 0.90.

This gives us; 1.28 = (x1 - 50) / 3, hence solving the equation for x1 we get, x1 = 50 + 1.28*3 = 53.84.

Let the grade of the student who gets a grade worse than (top) 25% of the class from Calculus 1 course be "x2".

That means P( X > x1 ) = 0.25 or P(X < x1) = 0.75. But from the Z table we get P( Z < 0.67) = 0.75.

This gives us; 0.67 = (x2 - 50) / 3, hence solving the equation for x2 we get, x2 = 50 + 0.67*3 = 52.01.