What is the probability that 5 large loaves are heavier than 10 small loaves?

[Harmony]

The weight of a large loaf of bread is a normal variable with mean 420g and standard deviation 30g. The weight of a small loaf is a normal variable with mean 220g and standard deviation 10g.

1) Find the probability that 5 large loaves of bread are heavier than 10 small loaves.

My Working:
Let X be the weight of a large loaf, Y be the weight of a small loaf

X~N(420, 900) , Y~N(220,100)

5X>10Y
5X-10Y>0
X-2Y>0

E(X-2Y)=E(X) - 2E(Y)=420-440=-20
Var(X-2Y)=Var(X)+4Var(Y)=900+400=1300

P(X-2Y>0)=P(Z>20/(1300)^1/2)=P(Z>0.5547)=0.2896

But the answer given is totally different. Is there anything I miss in my working?

 

[courtrigrad]

Are $$X,Y$$ normally distributed random variables? If so, then:

$$Z_{X} = \frac{X-420}{30}$$, and $$Z_{Y} = \frac{Y-220}{10}$$

Also $$var(aX+bY) = a^{2}var(X) + b^{2}var(Y) + 2abcov(X,Y)$$

Since $$X,Y$$ are independent, then $$cov(X,Y) = 0$$

 

[pizzasky]

You are working with different loaves of bread here, with EACH of their weights being normally distributed. It would not be correct to consider the case where 5X > 10Y, because this would imply you are thinking about the scenario where 5 times the weight of ONE large loaf is greater than 10 times the weight of ONE small loaf.

Instead you should think about the distribution of $$X_{1} + X_{2} + X_{3} + X_{4} + X_{5}$$.

Observe that $$5X \sim N(2100,22500)$$ but $$X_{1} + X_{2} + X_{3} + X_{4} + X_{5} \sim N(2100,4500)$$, so these 2 distributions are indeed different.

 

[Harmony]

Thanks, I understand it now.