Using expectation/variance algebra in normal distribution A shop sells apples and pears. The masses, in grams, of the apples may be assumed to have a A~$N(180, 12^2)$ distribution and the masses of the pears, in grams, may be assumed to have a P~$N(100, 10^2)$ distribution. Find the probability that the mass of a randomly chosen apple is more than double the mass of a randomly chosen pear.
Although this question appeared in the linear combination of normal variables section, I wasn't quite sure how to incorporate it into this particular problem. I thought that it could just simply be 
$P(A>2P)$, where 2P can be found by multiplying the mean, 100, by 2. So it would be 
$P(A>200)$. But I don't think this approach is correct. How should I solve this? The answer is 0.196.
 A: Guide:
If assume independence, then we have $A-2P$ is normal distribution with mean $$\mathbb{E}[A-2P]=\mathbb{E}[A]-2\mathbb{E}[P]=180-2(100)$$
and 
$$Var[A-2P]=Var[A]+4Va[P]=12^2+4(10^2)$$
A: Indeed, $$\Pr[A > 2P] \ne \Pr[A > 200].$$  This would be true if $P = 100$ almost surely, but it is not; $P$ is itself random and you must account for this in your probability calculation.
Recall that if $A$ and $P$ are independent normal random variables, a linear combination of them is also normally distributed.  Specifically, if $X \sim \operatorname{Normal}(\mu_X, \sigma^2_X)$ and $Y \sim \operatorname{Normal}(\mu_Y, \sigma^2_Y)$, and $X$ and $Y$ are independent, we have $$aX + bY \sim \operatorname{Normal}(a\mu_X + b\mu_Y, a^2 \sigma^2_X + b^2 \sigma^2_Y).$$  Note that $a$ and $b$ need not be positive.
With this in mind, what can we say about $$\Pr[A > 2P] = \Pr[A - 2P > 0]?$$  What is the distribution of the random variable $A - 2P$?
A: You need to consider the sampling distribution of difference between means. Note: $D=A-2P \sim N(\mu_D;\sigma_{D}^2)$. Note:
$$\mu_D=E(A-2P)=E(A)-2E(P)=180-2(100)=-20;\\
\sigma_{D}^2=Var(A-2P)=Var(A)+4Var(P)=12^2+4(10^2)=544.$$
Hence:
$$P(D>0)=\Phi\left(z>\frac{0-\mu_D}{\sigma_{D}}\right)=\Phi\left(z>\frac{0-(-20)}{\sqrt{544}}\right)=\Phi(z>0.8574)=0.196.$$
