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audia4
July 24th 2005, 12:09 PM
1.
WE have a distribution in which P(X=0) and P(X=1) = 0.6. From this distribution we draw two samples, X1 and X2, and we let their averages be Y. Of course, Y is a random variable; what is Var(Y)

a. 0.6, b. 0.174, c. 0.16, d 0.36 e. none of the above.

2. X1, ..., X5 are independent variables with E(X_i) = u and var(xi) =io^2, where u = theoretical mean, o^2 = variance.
for i = 1,...5, what is the 50th percentile of (xbar - 3u)/o^2(xbar is the sample mean)

a. 0 b. 1.645 c. 3.84 d. 11.07 e. 19.21

3. X and Y both have uniform distributions over (0,1). Evaluate P(X^2>=Y^2)
Hint(think conditioning and double integral)

a. 1/3 b. 2/5 c. 3/5 d 2/3 e 1

Thank you

.Godspeed.
July 24th 2005, 05:13 PM
1.
WE have a distribution in which P(X=0) and P(X=1) = 0.6. From this distribution we draw two samples, X1 and X2, and we let their averages be Y. Of course, Y is a random variable; what is Var(Y)

a. 0.6, b. 0.174, c. 0.16, d 0.36 e. none of the above.

I'm not entirely sure, but it sounds like you are asking what Var(Y)=Var((X1+X2)/2) is.

Var((X1+X2)/2) = Var(X1+X2)/4 =

(because of independence ) (Var(X1) + Var(X2))/4 = 2Var(X)/4 = Var(X)/2

E[X] = 0*.4 + 1*.6 = .6, E[X^2] = 0^2*.4 + 1^2*.6 = .6
Var[X] = E[X^2] - E[X]^2 = .6 - .6^2 = .24

--> Var(X)/2 = .24/2 = .12, e.

This seems right, but I'm always incredibly leery about getting the "none of the above" answer choice. If someone sees something I messed up, please show us because I don't want to mess audia up. Thanks.

Alice2005
July 24th 2005, 09:57 PM
'm not entirely sure, but it sounds like you are asking what Var(Y)=Var((X1+X2)/2) is.

Var((X1+X2)/2) = Var(X1+X2)/4 =

(because of independence ) (Var(X1) + Var(X2))/4 = 2Var(X)/4 = Var(X)/2

E[X] = 0*.4 + 1*.6 = .6, E[X^2] = 0^2*.4 + 1^2*.6 = .6
Var[X] = E[X^2] - E[X]^2 = .6 - .6^2 = .24

--> Var(X)/2 = .24/2 = .12, e.

________________________________________________

Just want to point out two things.

E[X] = 0*.6 + 1*.6 = .6,

since it states in the question that (x=1) and (x=2) are both 0.6.
but that doesn't matter since its mutilplied by 0 anyway.

The second thing is, I don't know why you are assuming they are independent.

.Godspeed.
July 24th 2005, 10:46 PM
'm not entirely sure, but it sounds like you are asking what Var(Y)=Var((X1+X2)/2) is.

Var((X1+X2)/2) = Var(X1+X2)/4 =

(because of independence ) (Var(X1) + Var(X2))/4 = 2Var(X)/4 = Var(X)/2

E[X] = 0*.4 + 1*.6 = .6, E[X^2] = 0^2*.4 + 1^2*.6 = .6
Var[X] = E[X^2] - E[X]^2 = .6 - .6^2 = .24

--> Var(X)/2 = .24/2 = .12, e.

________________________________________________

Just want to point out two things.

E[X] = 0*.6 + 1*.6 = .6,

since it states in the question that (x=1) and (x=2) are both 0.6.
but that doesn't matter since its mutilplied by 0 anyway.

The second thing is, I don't know why you are assuming they are independent.

Oh, I thought that what was meant was that P(X=0)=.4 and P(X=1)=.6. And I thought the two draws from the sample were drawn independent of each other; maybe I was wrong, though.

wat
July 25th 2005, 04:08 AM
1.
WE have a distribution in which P(X=0) and P(X=1) = 0.6. From this distribution we draw two samples, X1 and X2, and we let their averages be Y. Of course, Y is a random variable; what is Var(Y)

a. 0.6, b. 0.174, c. 0.16, d 0.36 e. none of the above.

2. X1, ..., X5 are independent variables with E(X_i) = u and var(xi) =io^2, where u = theoretical mean, o^2 = variance.
for i = 1,...5, what is the 50th percentile of (xbar - 3u)/o^2(xbar is the sample mean)

a. 0 b. 1.645 c. 3.84 d. 11.07 e. 19.21

3. X and Y both have uniform distributions over (0,1). Evaluate P(X^2>=Y^2)
Hint(think conditioning and double integral)

a. 1/3 b. 2/5 c. 3/5 d 2/3 e 1

Thank you

Please check #1. If the problem is stating that P(X=0) = P(X=1) = 0.6, this is not a valid probability distribution.

For #3, draw the square that represents the unit square (i.e., the square with coordinates (0,0), (0,1), (1,0), (1,1). Then, within that square, draw two graphs: y=x^2 and x = y^2 (or y=sqrt(x)). Figure out which area represents "x^2 => y^2". Then integrate.

audia4
July 25th 2005, 01:46 PM
To Wat:

I followed your suggestion, but when the two graphs are drawn, it is clear that x^2<=y^2, so P(x^2>=y^2) = 0, so I am stuck once again.

THank you.

wat
July 25th 2005, 02:40 PM
To Wat:

I followed your suggestion, but when the two graphs are drawn, it is clear that x^2<=y^2, so P(x^2>=y^2) = 0, so I am stuck once again.

THank you.

Then the graph is drawn incorrectly. Consider the point (0.5, 0.5). It is well within the boundaries in the square, and it certainly satisfies "x^2 => y^2" because they're equal. So, there exists a point in the set {(x,y) | x^2 => y^2, 0<=x,y<=1}, which more or less indicates that the probability is not 0. This line of reasoning happens for all x=y, 0<=x,y<=1. So, you know at least the y=x line in the square counts for the probability.

Graph it again - use a graphing calculator or Excel if you have to. There is an area where X^2=>Y^2. It should look like an oval from the lower left corner to the upper right corner with sharp points at (0,0) and (1,1).

wat
July 25th 2005, 04:05 PM
2. X1, ..., X5 are independent variables with E(X_i) = u and var(xi) =io^2, where u = theoretical mean, o^2 = variance.
for i = 1,...5, what is the 50th percentile of (xbar - 3u)/o^2(xbar is the sample mean)

a. 0 b. 1.645 c. 3.84 d. 11.07 e. 19.21

Just to be clear about #2: Is Var(X_i) = i * o^2? As in, if the variance of the first variable is 4, the variance of the 3rd variable is 12?

audia4
July 26th 2005, 07:33 AM
to wat: Thank you for the reply. REgarding the last question, yes, if the variance of X_i is 4, then Var(X_3) = 3*4 = 12

Thanks