# Homework Help: Normal Distribution Problem

1. Dec 2, 2006

### 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?

2. Dec 2, 2006

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 $$cov(X,Y) = 0$$

Last edited: Dec 2, 2006
3. Dec 3, 2006

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.

Last edited: Dec 3, 2006
4. Dec 4, 2006

### Harmony

Thanks, I understand it now.