HELP Sums of Random Variables problem: Statistics

[geno678]

HELP!Sums of Random Variables problem: Statistics

Homework Statement

3. Assume that Y = 3 X1+5 X2+4 X3+6 X4 and X1, X2, X3 and X4 are random variables that represent the dice rolls of a 6 sided, 8 sided, 10 sided and 12 sided dice, respectively.

a. If all four dice rolls yield a 3, what is the sum of these random variables?

b. What is the expected value of the sum of these random variables?

Homework Equations

=11+22+⋯+
1,2,… represent coefficients
1,2,…, represent variables
If =21+82
1 = First dice roll ; 2 = Second dice roll
Lets say we roll 3, then 5
Y = 2(3) + 8(5) = 46

for part (a)

for part(b)

for expected value
==11+22+⋯+
this part i don't understand.

The Attempt at a Solution

Ok so I attempted part (a). I have no clue if I'm correct though.

So I was given Y = 3(X1) + 5 (X2) + 4(X3) + 6(X4)

Here's where I'm confused. The problems says that all four dice yield a 3.

I made X1 = 1/6, X2 = 1/8, X3 = 1/10, X4 = 1/12

I plugged these values into my equation.

Y = 3(1/6) + 5 (1/8) + 4 (1/10) + 6(1/12) = 2.025

However, in the example that my professor gave me I'm thinking that

X1, X2, X3, and X4 = 3.

Is the answer really Y = 3(3) + 5 (3) + 4(3) + 6(3) = 54

 

[vela]

The answer for part (a) is 54, as you suspected. X_i is the value of a die, not the probability of getting a particular roll.

Part of your post is unreadable since a bunch of characters in the relevant equations section show up as empty squares.

 

[geno678]

Seriously. That's weird I can see them. I'll type them in next time.
Ok awesome. So my hunch was correct.

For part (b) I have to find the expected value. The formula they gave me was Uy= E(y) =
a1 * E(X1) + a2 * E(X2) + ... an * E(Xn).

How would I approach this problem?

 

[vela]

You should know by now how to calculate an expected value of a single random variable. If not, look it up in your textbook. Start there and calculate the expected value for each die.

 

[geno678]

Oh ok. I'll do it, and then show you my results.

 

[geno678]

Ok so E(X1) = (1+2+3+4+5+6)/6 = 3.5
E(X2) = (1+2+3+4+5+6+7+8)/8 = 4.5
E(X3) = (1+2+3+4+5+6+7+8+9+10)/10 = 5.5
E(X4) = (1+2+3+4+5+6+7+8+9+10+11+12)/12 = 6.5

E(y) = (1)3.5 + (1)4.5 + (1)5.5 + (1)6.5 = 20

 

[vela]

Almost! The coefficients aren't equal to 1, right?

 

[geno678]

No, the previous equation was Y = 3 X1+5 X2+4 X3+6 X4.

But in my example, they used one for the coefficients. So I'm confused on that part.

 

[geno678]

They said when 2 dices are rolled, and the expected value is

E(y) = 1(3.5) + 1(3.5) = 7

 

[geno678]

Unless it's E(y) = 3(3.5) + 5(4.5) + 4(5.5) + 6(6.5) =

or unless my coefficients are 1 through 4, idk my examples aren't very good

 

[vela]

This is how it works: if you have

$$Y = a_1 X_1 + a_2 X_2 + \cdots + a_n X_n$$

its expected value is given by

$$\begin{align*}<br /> E(Y) & = E(a_1 X_1 + a_2 X_2 + \cdots + a_n X_n) \\<br /> & = a_1 E(X_1) + a_2 E(X_2) + \cdots + a_n E(X_n)<br /> \end{align*}$$

 

[geno678]

Then that means the answer is E(y) = 3(3.5) + 5(4.5) + 4(5.5) + 6(6.5) = 94

Y = 3 X1+5 X2+4 X3+6 X4.

 

[vela]

Right!

 

[geno678]

Awesome. Thank you for your help!