Question

On stat your assessment is based on: Final Exam 47% Learn based on‐line assessment 34% Assignments 19%

Consider three random variables X, Y and Z which respectively represent the exam, on‐line assessment total and assignment scores (out of 100%) of a randomly chosen student. Assume that X, Y and Z areindependent (this is clearly not true, but the answers may be a reasonable approximation).

Suppose that past experience suggests the following properties of these assessment items (each out of 100%): E(X) = 61, sd(X) = 20, E(Y) = 72, sd(Y) = 22 and E(Z) =65, sd(Z) = 24.

a) Find the distribution parameters, E(T) and Var(T), for the total mark, T, where:T = 0.47X+0.34Y+0.19Z.

b) Assume the on‐line assessment, exam and assignment scores are Normally distributed. If the pass mark is 50%, calculate the probability that a randomly selected student will pass.

c) Find the expected number of A+ grades (90% and above) to be awarded in July if there are 840 students on the course this semester.

d) If a random sample of 16 stat students is selected, what is the probability that their average grade is at least a B (that is, on average they get 70% or more in total)?

How are b, c and d answered using excel?

Answer 1

X, Y and Z which respectively represent the exam, on‐line assessment total and assignment scores (out of 100%) of a randomly chosen student.

E(X) = 61, sd(X) = 20, E(Y) = 72, sd(Y) = 22 and E(Z) =65, sd(Z) = 24

a) If T = 0.47X+0.34Y+0.19Z

then the expectation of T is

The variance of T is

b) Sine X, Y and Z are normally distributed, we can say that T = 0.47X+0.34Y+0.19Z is also normally distributed with mean

and standard deviation

the probability that a randomly selected student will pass is same as the probability that a randomly selected student will have T>50

We will use the excel function to get P(X<x)

=NORM.DIST(x,mean,standard deviation, TRUE)

the required probability P(T>50) is

=1-NORM.DIST(50,65.5,12.8493,TRUE)

which is 0.8861

ans: the probability that a randomly selected student will pass is 0.8861

c) the probability that a randomly selected student will get A+ grade is same as the the probability that a randomly selected student will score T>90

The required probability is

Using the Excel function

=1-NORM.DIST(90,65.5,12.8493,TRUE)

we get

P(T>90) = 0.0283

ans: the expected number of A+ grades (90% and above) to be awarded in July if there are 840 students on the course this semester is P(T>90)* 840 = 23.8

d) Let be the average total marks of a random sample of n=16 students. We know that has a normal distribution with

mean and

standard deviation (also called standard error of mean)

The probability that their average grade of a sample of 16 students is at least a B (that is, on average they get 70% or more in total) is

We will use the excel function

=1-NORM.DIST(70,65.5,3.2123,TRUE)

and get

0.0806

ans: The probability that their average grade of a sample of 16 students is at least a B is 0.0806

and get