Probability of a difference in a normal distribution

Question

I have a normal distribution of cabbage with $\sigma=0.7$ and $\mu=2.4$ kg. The question is; what is the probability that $2$ randomly chosen cabbage has a weight-difference of more than $1$ kg?

I don't know where to start, other than $P(|X-Y|>1)$

Answer 1

Hint:

Good start.

Now realize that $Z:=X-Y$ again has normal distribution (find $\sigma_Z$).

This evidently with $\mu_Z=\mu_X-\mu_Y=0$ so that: $$Z\stackrel{d}{=}-Z$$ leading to: $$P(|Z|>1)=P(Z>1)+P(Z<-1)=P(Z>1)+P(-Z>1)=2P(Z>1)$$

Finally you can make use of $Z=\sigma_ZU$ where $U$ has standard normal distribution, leading to: $$P(Z>1)=P(U>\sigma_Z^{-1})=1-\Phi(\sigma_Z^{-1})$$where $\Phi$ denotes the CDF of standard normal distribution.

Answer 2

The sum of normal r.v.'s $Z=X+Y$ is normal with $\mu_Z = \mu_X+\mu_Y$ and $\sigma_Z^2=\sigma_X^2+\sigma_Y^2$. Now think what are $\mu$ and $\sigma^2$ for $Z=X-Y$ and compute weight of the tails of its distribution.

Answer 3

If $X\sim N(\mu_X,\,\sigma_X^2)$ is independent of $Y\sim N(\mu_Y,\,\sigma_Y^2)$,$$aX+bY\sim N(a\mu_X+b\mu_Y,\,a^2\sigma_X^2+b^2\sigma_Y^2)$$for constants $a,\,b$. In particular, if$$\mu_X=\mu_Y=2.4,\,\sigma_X=\sigma_Y=0.7,$$then $X-Y\sim N(0,\,2\times 0.7^2)$. So you want to calculate$$\begin{align}1-P(-1\le X-Y\le 1)&=1-\left(\Phi\left(\frac{1}{0.7\sqrt{2}}\right)-\Phi\left(\frac{-1}{0.7\sqrt{2}}\right)\right)\\&=2\left(1-\Phi\left(\frac{1}{0.7\sqrt{2}}\right)\right),\end{align}$$where $\Phi$ is the $N(0,\,1)$ CDF.