Comparing random variables with a normal distribution

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SUMMARY

The discussion focuses on calculating the probability that the combined weight of seven apples, modeled as a normal distribution N(1050, 2800), exceeds the weight of a cabbage, modeled as N(1000, 502). The solution involves transforming the cabbage's distribution to -Y ~ N(-1000, 502) and combining it with the apples' distribution. The final probability calculation yields P(X ≥ 0) = Φ(0.69) = 0.7549, indicating a 75.49% chance that the apples weigh more than the cabbage.

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Homework Statement



You have 7 apples whose weight (in gram) is independent of each other and normally distributed, N(\mu= 150, \sigma2 = 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 - \muX)/\sigmax) =
\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.
 
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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)??
 
Thank you :)
I like how stuff is obvious when someone tells you.

-y ~ N(-1000, 502)

Xi ~ N(1050 - 1000, 2800 + 2500)

P( X\geq0) = \Phi(\stackrel{50}{\sqrt{5300}}) = \Phi(0.69) = 0.7549

Thanks again :D
 

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