(You are not allowed to ask multiple questions in the same post. Here is a response to 3 of the questions posted)
On STA, the assessment is based on scores in the Final Exam(X), Learn based on-line assessment (Y) and Assignments (Z). The 3 variables X, Y and Z are independent.
It is known from past experience that E(X) = 61, sd(X) = 20, E(Y) = 72, sd(Y) = 22, E(Z) = 65 and sd(Z) = 24.
The total marks are derived from X, Y and Z using the formula T = 0.5*X + 0.3*Y + 0.2*Z
E(T) = 0.5*E(X) + 0.3*E(Y) + 0.2*E(Z)
= 0.5*61 + 0.3*72 + 0.2*65
= 65.1
Var(T) = 0.5*Var(X) + 0.3*Var(Y) + 0.2*Var(Z)
= 0.5*20^2 + 0.3*22^2 + 0.2*24^2
= 460.4
sd(T) = 21.457
If the marks require to pass is 50%, to determine the probability that a randomly selected student will pass, first find the z-score.
z = (50 - 65.1)/21.457
= -0.7037
Using the normal distribution table, the cumulative probability of a score less than 50% is 0.24080 or 24.08%
The z-score corresponding to 90% is (90 - 65.1)/21.457 = 1.16046
This gives the probability of students getting 90% and above as 1 - 0.87707 = 12.293%
As there are 610 students, the number of students getting an A+ is 75.
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