Probability of Random Variable Question | Probability Error I had this question in my past paper:

Bags of sugar are packed in boxes, each box containing 20 bags. The masses of the boxes, when
empty, are normally distributed with mean 0.4 kg and standard deviation 0.01 kg. The masses of the
bags are normally distributed with mean 1.02 kg and standard deviation 0.03 kg.
i) Two full boxes are chosen at random. Find the probability that they differ in mass by less than 0.02 kg

I did the calculations and I got my probability as 2 x 0.4582 however in the marking scheme they've deducted the final answer from 1. Why is that so?
 A: I do not know how you arrived at your result but I would do in the following manner:
Set X= boxes' mass and Y=bags' mass.
The gross mass is
$$Z=(X+Y_1+Y_2+\dots +Y_{20})\sim N(20.8;0.0181)$$
and
$$W=(Z_1-Z_2)\sim N(0;0.0362)$$
you are requested to calculate
$$\mathbb{P}[|W|<0.02]=\mathbb{P}[-0.02<W<0.02]=$$
$$=\Phi\left(\frac{0.02}{\sqrt{0.0362}}  \right)-\Phi\left(\frac{-0.02}{\sqrt{0.0362}}  \right)=0.5419-0.4581=0.0837$$
A: What you did is probably:

*

*Variance of the weight of a full box is $0.01^2+20\times 0.03^2=0.0181$

*Variance of the difference between two full boxes is $2\times 0.0181=0.0362$  (and mean is $0$)

*Probability first full box was more than $0.02$kg bigger than the second full box is probability standard normal exceeds $\frac{0.02}{\sqrt{0.0362}}\approx 0.1051177$ which is about $0.4582$

*Probability either box is more than $0.02$kg bigger than the other is double that, about $0.9163$
and what you needed to do was the next step

*

*Probability that they differ in mass by less than $0.02$kg is $1$ minus that, about $0.0837$
