Question 1174946: Please help me solve this problem
The number of home runs scored by a certain team in one baseball game is a random variable
with the distribution
x 0 1 2
P(x) 0.4 0.4 0.2
The team plays 2 games. The number of home runs scored in one game is independent of
the number of home runs in the other game. Let Y be the total number of home runs. Find
E(Y ) and Var(Y ).
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
X = number of home runs in one game
Y = n*X = number of home runs in n games
n is some positive whole number
If n = 2, then,
Y = n*X
Y = 2X
This represents how many home runs there are after 2 games.
Recall that if Y = a+bX, then,
E(Y) = expected value of Y = mean of Y
E(Y) = E(a+bX)
E(Y) = E(a)+E(bX)
E(Y) = E(a)+b*E(X)
E(Y) = a+b*E(X)
I'm using a linear transformation.
We'll set this equation aside for now.
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Let's compute E(X)
The given table is
But I find it's easier to have the table oriented like this
Let's build a third column that consists of multiplying each x and P(x) paired value
| X | P(X) | X*P(X) | | 0 | 0.4 | 0*0.4 = 0 | | 1 | 0.4 | 1*0.4 = 0.4 | | 2 | 0.2 | 2*0.2 = 0.4 |
Cleaning up that third column has us left with this
| X | P(X) | X*P(X) | | 0 | 0.4 | 0 | | 1 | 0.4 | 0.4 | | 2 | 0.2 | 0.4 |
Add up everything in this new column to get
0+0.4+0.4 = 0.8
This is exactly the value of E(X), or the mean of X.
For one game, we expect on average about 0.8 home runs.
We can use a non-integer decimal value to represent averages even if it doesn't make much sense to have some fractional value of a home run.
It's like saying that we have the following number of homeruns: {0, 1, 0, 2}
Each value representing a single-game home run count.
They average to (0+1+0+2)/4 = 3/4 = 0.75, which isn't a whole number.
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Now we'll go back to
E(Y) = a+b*E(X)
In the case of Y = 2X, we see that a = 0 and b = 2
Compare Y = 2X to Y = a+bX.
So,
E(Y) = a+b*E(X)
E(Y) = 0+2*E(X)
E(Y) = 2*E(X)
E(Y) = 2*0.8
E(Y) = 1.6
Over the course of the two games, we expect to get about 1.6 home runs on average.
In short, if we expect about 0.8 home runs in one game, then we expect about 1.6 home runs over the course of 2 games. These values are averages.
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Now onto the variance. The goal is to compute Var(Y), but first we'll need to compute Var(X)
One way to do this is to use this formula
Var(X) = E(X^2) - [E(X)]^2
which I find easier to remember since it has a strange sort of "symmetry" to it in a way. The first term is the expected value of the X^2 terms, while the second part is the square of E(X).
Computing E(X^2) isn't too bad. We return to the table and build out a X^2 column and a X^2*P(X) column like so
| X | P(X) | X^2 | X^2*P(X) | | 0 | 0.4 | 0 | 0 | | 1 | 0.4 | 1 | 0.4 | | 2 | 0.2 | 4 | 0.8 |
Adding up everything in the X^2*P(X) column will get us E(X^2)
E(X^2) = 0+0.4+0.8 = 1.2
Which then tells us,
Var(X) = E(X^2) - [E(X)]^2
Var(X) = 1.2 - [0.8]^2
Var(X) = 1.2 - 0.64
Var(X) = 0.56
Now we apply this rule
If Y = a+bX, then Var(Y) = b^2*Var(X)
We use a = 0 and b = 2 again
So,
Var(Y) = b^2*Var(X)
Var(Y) = 2^2*Var(X)
Var(Y) = 4*Var(X)
Var(Y) = 4*0.56
Var(Y) = 2.24
This tells us how spread out the Y distribution is. The larger the variance, the more spread out the values will be.
Side note: when computing the variance, it's often better to use technology of some kind to make life easier.
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Answers:
E(Y) = 1.6
Var(Y) = 2.24
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