This is the problem I have to solve for a job at school. Can anyone help me, what kind of distributions approximations do I use and how to calculate the requested probabilities?
In one factory there are two M1 and M2 machines that produce screws. The length in millimeters of the screws' and is usually distributed with mean 40 and standard deviation 1 for those produced by M1, and with media 41 and standard deviation 2 for those produced by M2.
From a box containing 60 screws produced by M1 and 40 screws produced by M2 a screw is drawn at random.
(i) Determine the probability of this screw having a length greater than 40 mm.
(ii) Assuming now that you have verified that the screw removed from the carton has lower length at 40 mm, determine the probability of this screw being produced by machine M2.
For (i), the probability is $$P(screw>40) = P(screw>40|M_1)P(M_1) + P(screw>40|M_2)P(M_2)$$ We are given that $P(M_1) = 0.6$ and $P(M_2) = 0.4$. Since $M_1$'s screw lengths are distributed $\mathcal N(40,1)$, $P(screw > 40|M_1) = 0.5$. Since $M_2$'s screw lengths are distributed $\mathcal N(41,2)$, $P(screw>40|M_2) = 1 - \Phi(-0.5)$, where $\Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-z^2/2}\;dz$ is the CDF of the standard normal. So, we get: \begin{align} P(screw > 40) &= P(screw>40|M_1)P(M_1) + P(screw>40|M_2)P(M_2)\\ &= 0.5\cdot0.6 + (1-\Phi(-0.5))\cdot 0.4\ \\ &= 0.5765... \end{align}
For (ii), I'm assuming you want $P(M_2|screw <40)$. Using Bayes' theorem, we can recycle some of our work from (i): \begin{align} P(M_2|screw<40) &= \frac{P(screw<40|M_2)P(M_2)}{P(screw<40)} \\ &=\frac{\Phi(-0.5)\cdot0.4}{1-0.5765...} \\ &=0.2914... \end{align}