ECON106 TPS 4 1 Die & Coin Y=0 Y=1 P(X=x) X-1 1/12 1/12 1/6 X-2 1/12 1/12 1/6 x=

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ECON106 TPS 4 1 Die & Coin Y=0 Y=1 P(X=x) X-1 1/12 1/12 1/6 X-2 1/12 1/12 1/6 x=3 1/12 1/12 1/6 X-4 1/12 1/12 1/6 x-5 1/121/121/6 X-6 1/12 1/12 1/6 P(Y-y) 1/2 1/2 1 In the above joint and marginal probability table, X repr the flip of a fair coin (heads=0, tails=1). esents the roll of a six-sided die and Y represents a) create a PDF and CDF of the variable W-x-Y. b) find E(W) c) find Var(W) d) determine if X and Y are independent e) determine if X and W are independent 2 Semester Grade In a certain chemistry class, the semester grade is calculated using homework, 2 midterms, and a final exam, all graded out of 100%. The table below describes the weighting. % of semester grade 40 20 15 25 task a ore Ouore Final 65 16 Midterm 2 80 15 Midterm 1 70 12 HW 89 10 a) What's the expected semester grade for a random student in the class? b) What other information would you need in order to calculate the variance of the semester grade for the class? Once you knew that information, how would you use it to find the variance?

Explanation / Answer

Question 1

Here first we will make a table of X and Y and their differences and we get the following table

Now we can see W has values from 0 to 6

so here probability distribution of W is

f(W) = 1/12 ; W = 0

= 2/12 ; W = 1

= 2/12 ; W = 2

= 2/12 ; W = 3

= 2/12 ; W = 4

= 2/12 ; W = 5

= 1/12 ; W = 6

(b) Here

E[W] = 0 * 1/12 + 1 * 2/12 + 2 * 2/12 + 3 * 2/12 + 4 * 2/12 + 5 * 2/12 + 6 * 1/12 = 3

(c) Var [W] = (0 - 3)2 * 1/12 + (1 - 3)2 * 2/12 + (2 - 3)2 * 2/12 + (3 - 3)2 * 2/12 + (4 - 3)2 * 2/12 + (5 - 3)2 * 2/12 + (6 - 3)2 * 1/12 = 3.1667

(d) TO check X and Y are independent

f(x,y) = f(x) f(y)

so at x = 1 and y = 1

f(x,y) = 1/12

f(x = 1) = 1/6 ; f(y = 1) = 1/2

f(x,y) = f(x) f(y)

soo yeas events are independent

(e) for X and W to be indepndent

f(X,W) = f(X) f(W)

for X = 1 and W = 1

f(X =1 , W = 1) = 1/12

f(X = 1) = 1/6

f(W = 1) = 1/6

f(X= 1) * f(W = 1) = 1/6 * 1/6 = 1/36

which is not equal to 1.12

so we can say that X and W are not independent.

X Y W P(X,Y) 1 0 1 1/12 2 0 2 1/12 3 0 3 1/12 4 0 4 1/12 5 0 5 1/12 6 0 6 1/12 1 1 0 1/12 2 1 1 1/12 3 1 2 1/12 4 1 3 1/12 5 1 4 1/12 6 1 5 1/12

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