A snap-on lid must fit on a cup. The metal cup outside diameter follows a normal

Question

A snap-on lid must fit on a cup. The metal cup outside diameter follows a normal distribution with a mean of 3.000 in. and a standard deviation of 0.005 in. The inside diameter of the plastic lid also follows a normal distribution with a mean of 2.999 in. and a standard deviation of 0.006 in. Estimate the percentage of cups that will fail to function properly if a cup-lid clearance of more than 0.003 will leak and an interference of less than - 0.010 will not fit (the plastic lid will stretch to some extent).

Explanation / Answer

difference in mean = 3.000 -2.999 = 0.001

difference in diameters should be between : 0.003 tp 0.010

max outer dia = 3.000 +y0.005

min inner diameter = 2.999 - a0.006

3.000 +a0.005 - 2.999 +a0.006 =0.003

a = 0.18

min outer dia = 3 -b0.005

max inner dia = 2.999 +b0.006

3 - b0.005 -2.999 -b0.006 = -0.010

b = 1

percentage defects = 1.18

(might be i misinterpreted)

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