Find the probability of an event affected by two variables of normal distribution. > A company is producing juice bottles of 2L. One machine is fulling half of the bottle with concentrate juice and an other machine is fulling the other half with water. Let X be the quantity of concentrate of the first machine be a random variable that obey to N(0.98;0.0009) and let Y be the quantity of water the the second machine is giving be a random variable that obey to N(1.02;0.0016). What is the probability of a bottle to have more than 1.98L of juice? I am very not sure how to start that problem. Do I add a third variable that is representing the two normal variables? Thank you.

The total amount of fluid in the bottle is given by the sum of the amount added by the first machine and the amount added by the second machine.

Now, the sum of two normal distributions is again normal: $$Z:= X + Y \sim \mathcal{N}(0.98 + 1.02, 0.0009 + 0.0016) = \mathcal{N}(2,0.0025).$$

Here the variances have been added because we assume that the two machines inject liquid independently.

Now you need to compute $$\mathbb{P}(Z \ge 1.98) = \mathbb{P}(\frac{Z-2}{\sqrt{0.0025}} \ge \frac{1.98-2}{\sqrt{0.0025}}) = 1-\Phi(\frac{1.98-2}{\sqrt{0.0025}})$$