Question

Question* On STAT your assessment is based on: Final Exam Learn based online assessment Assignments 4790 +' 3490 19% Consider three random variables X, Y and Z which respectively represent the exam, online assessment total and assignment scores (out of 100%) of a randomly chosen student. Assume that X, Y and Z are independent (this is clearly not true, but the answers may be a reasonableapproximation).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, ← (3)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. " (4)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)?

Answer 1

Let T denote the total score obtained. Then

T = 0.47X + 0.34y +0.19Z

Now,

E(T)-E(0.47x +0.34Y +0.19Z) 0.47E(X)+0.34E(Y)+0.19E(Z) 0.47(61) 0.34(72) 0.19(65)65.5

and,

34Y+0.19Z) = 0.472V(X)+0.342 V(Y)+0.192V (Z) = V(T) = V(0.47X+0. 0.472 (202) 0.34 (222) + 0.192(242) -165.104 0.342 (222) + 0.192 (242) = 165.104

So, we can say

T ~ N(μ-65.5, σ2-165.104)

3)

90 65.5 P(Z > 1.91) = 1-P(Z < 1.91) P(Z ) /165·104 90) 1-0.97193-0.02807

Required expected number of students = 0.02807(840) = 23.5788 $\approx$ 24

4)

Using Central Limit theorem, we know,

165.104 X ~ N(μ, 띄 ~ N(65.5, ) 7n

Required probability =

70 65.5 10.319 ) = P(Z > 1.40) = 1-P(Z < 1.40) P(X > 70) = P(Z > 10.919240.08076