A machine fills tins with paint. The mass of a tin is normally distributed with a mean of 0.3 kg and a standard deviation of 0.02 kg. The mass of the paint that goes into the tin is normally distributed with a mean of 10 kg and a standard deviation of 0.05 kg.
The Quality Control department recommends that the average mass of 4 randomly selected filled tins must lie between 10.20 and 10.40 kg with 99.9% probability.
Question : What is the probability that the mass of a single filled tin lies between 10.20 and 10.40 kg?
Answer : This is what I have done,
X~N(0.3, 0.0004), where X represents the mass of tin
Y~N(10, 0.0025), where Y represents the mass of paint
Let U = X+Y. Since both are independent, U~(10.3, 0.0029)
By applying the Central Limit Theorem,
P(10.2 < U < 10.4) = P( $\left(\frac{10.2 - 4(10.3)}{\sqrt{4*0.0029}}\right)$<$\left(\frac{U - 4(10.3)}{\sqrt{4*0.0029}}\right)$<$\left(\frac{10.4 - 4(10.3)}{\sqrt{4*0.0029}}\right)$ )
However, this would result in large values of CDF where I wouldn't be able to use the table for my answer.
Can anyone tell me where I went wrong? Thanks~