A machine fastens plastic screw-on caps onto containers of motor oil. If the mac

Question

A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The capping machine torque has the normal distribution with mean 6.9 inch-pounds and standard deviation 0.99 inch-pounds. The cap strength (the torque that would break the cap) has the normal distribution with mean 10.4 inch-pounds and standard deviation 1.3 inch-pounds.

(b) What is the probability that a cap will break while being fastened by the capping machine?

Explanation / Answer

b)

The mean difference is

u(X-Y) = 6.9 - 10.4 = -3.5

and standard deviation of difference is

sigma(X-Y) = sqrt(Var(x) +Var(y)) = sqrt(0.99^2+1.3^2) = 1.634044063

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
C = critical value =    0      
u = mean =    -3.5      
          
s = standard deviation =    1.634044063      
          
Thus,          
          
z = (C - u) / s =    2.141925104      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.141925104   ) =    0.016099755 [ANSWER]

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