Comparing random variables with a normal distribution

[Powertravel]

Homework Statement

You have 7 apples whose weight (in gram) is independent of each other and normally distributed, N($\mu$= 150, $\sigma$² = 20²).
You also have a cabbage whose weight is independent of the apples and N(1000, 50²)

What is the probability that the seven apples will weigh more than the cabbage?

Homework Equations

Let X represent the weight of the seven apples combined, and Y the weight of the cabbage.
X~N(1050, 2800)
Y ~N(1000, 50²)

The Attempt at a Solution

I have an easy time calculating the probability that a random variable will yield a number within a specific interval. For example I know how to get the probability that the 7 apples will weigh more that 1000g,
p(X > 1000) =
$\varphi$(X > (1000 - $\mu$_X)/$\sigma$_x) =
$\Phi$(0.94) = 0.8264, which I got from a chart for $\Phi$(x).

I am completely lost however on how to calculate the probability that a certain random variable will yield a bigger number that another random variable, both normally distributed but with different parameters: p(X > Y).

Thank you.

 

[micromass]

Let $X_i$ be the distribution of the apples. And let Y be the distribution of the cabbage.

You know that $X_i\sim N(150,20^2)$ and $Y\sim N(1000,50^2)$.

Can you find out the distribution of -Y??
Can you use this to find out the distribution of $X:=X_1+X_2+X_3+X_4+X_5+X_6+X_7-Y$??
Can you use this to find $P(X\geq 0)$??

 

[Powertravel]

Thank you :)
I like how stuff is obvious when someone tells you.

-y ~ N(-1000, 50²)

$X$_i ~ N(1050 - 1000, 2800 + 2500)

P( $X$$\geq$0) = $\Phi$($\stackrel{50}{\sqrt{5300}}$) = $\Phi$(0.69) = 0.7549

Thanks again :D