Question

In a course, there are 2 exams consisting of 1 midexam and 1 final. In calculation of overall grades, 50 percent of midexam and final grades are taken. The grades of the midexam and the final are independent and normally distributed with means of 60 and 40, and with standard deviations of 18 and 24, respectively.

a) What is the probability a randomly selected student will get an overall grade that is larger than 80?

b) If the instructor wants 84.13 percent of students passing the course, what should be the value of passing grade (minimum overall grade which a student should score to pass the course)?

Answer 1

The values provided in the above questions as below

The grades of the midexam and the final are independent and normally distributed with means of 60 and 40, and with standard deviations of 18 and 24, respectively.

Consider,

X = the grades of the midexam

X ~ N(60, 182)

Here, x = 60 and x = 18

Y = the grades of the final

Y = N(40, 242)

Here. Y = 40, Y = 24

We calculate the overall grade using the 50 percent of midexam and final grades are taken.

Let Z = the overall grade using the 50 percent of midexam and final grades are taken

z = mean of Z = 50 percent of the means of the midexam and final grades = 60 + 40 = 100 & it's 50 percent = 50

mean of Z = z = 50

z = standard deviation of Z = 50 percent of the standard deviations of the midexam and final grades

= 18 + 24 = 42

& it's 50 percent = 21

standard deviation of Z = z = 21

Z ~N(z, z2) that is

Z~ N(50, 212)

z = 50, z = 21

a) We find the probability a randomly selected student will get an overall grade that is larger than 80.

consider

= overall grade that is larger than 80

-------------(1)

We convert the above into z using following formula

--------------(2)

Using equation (2) in equation (1) we get

We find the above value using table of Standard Normal Curve Areas

We find the probability whose corresponding = 1.43

The probability a randomly selected student will get an overall grade that is larger than 80 is 0.0764

b) If the instructor wants 84.13 percent of students passing the course, what should be the value of passing grade

Here, probability = 84.13 percent = 0.8413

We find Z value of whose corresponding probability = 0.8413 using table of Standard Normal Curve Areas

value = 1

Consider, X = the value of passing grade if the instructor wants 84.13 percent of students passing the course.

We find value of X using following formula

Using all values we get

If the instructor wants 84.13 percent of students passing the course then the value of passing grade is 71.

Summary :-

a) The probability a randomly selected student will get an overall grade that is larger than 80 is 0.0764

b) If the instructor wants 84.13 percent of students passing the course then the value of passing grade is 71.