Thread: Probabilities involving the sum or difference of independent Normal random variable

Applying torque A machine fastens plastic screw-on caps onto containers of motor oil. If the machineapplies more torque than the cap ean withstand, the capwill break. Both the torque applied and the strength ofthe caps vary. The capping-machine torque T follows aNormal distribution with mean 7 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength C (thetorque that would break the cap) follows a Normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds.





a.) What is the probability that a cap will break whilebeing fastened by the machine?
b.) Explain why it is reasonable to assume that the capstrength and the torque applied by the machineare independent.



have no idea how to solve this... please help

They give you both distributions for the torque and cap strength. You are concerned that T<C. Thus a new distribution must be created for the DIFFERENCE Y=C-T. You wish to know P(Y < 0). I'll leave you to fill in details. As for why you can assume independence; it's highly unlikely this company manufactures both the machine that applies the cap AND the cap; thus it's fair to assume that the strength of the caps are manufactured independent of knowledge of that specific machine. If that wasn't true (say the company buys caps from a distributor that has knowledge about the machines they use), it's unlikely the two r.v.'s would be independent.