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 7 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength (the torque that would break the cap) has the normal distribution with mean 9.8 inch-pounds and standard deviation 1.22 inch-pounds.

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

Explanation / Answer

Let

A = torque
B = strength

Then

u(A - B) = u(A) - u(B) = 7 - 9.8 = -2.8

sigma(A - B) = sqrt(sigma^2(A) + sigma^2(B)) = sqrt(0.9^2 + 1.22^2) = 1.516047493

So now, we get P(A - B > 0), where it will break.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    -2.8      
          
s = standard deviation =    1.516047493      
          
Thus,          
          
z = (x - u) / s =    1.84690784      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.84690784   ) =    0.032380249 [ANSWER]

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