Question

15 points. The specifications for A, B, and C are 2+0.1,2.5+0.1, and 5+0.3, respectively. Assume the lengths of the parts A, B, and C are all independent and normally distributed, and the tolerances are based on 3 times the standard deviation of the individual lengths. Gap (a) What is the expected tolerance of the gap? Show calculations. (b) If the customer requires a tolerance of t 0.2 on the gap, what proportion of the assemblies will be outside specification? Show calculations.

Answer 1

Solution

Back-up Theory

If X, Y and Z are independent, V(X + Y + Z) = V(X) + V(Y) + V(Z) ……………………….. (1)

Now to work out the solution,

Part (a)

If A, B, C and G represent the lengths of segments A, B, C and Gap. Then, C = A + B + G. So, vide (1), V(C) = V(A + B + G) = V(A) + V(B) + V(G) Or, V(G) = V(C) - V(A) - V(B) …………… (2)

Given that the tolerances are based on 3 times the standard deviation (SD) of the legths and tolerances for A, B and C are 0.1, 0.1 and 0.3 respectively,

SD(A) = 0.1/3, SD(B) = 0.1/3 and SD(C) = 0.3/3 = 0.1 …………………………………………… (3)

(2) and (3) => V(G) = 0.1² – 2(0.1/3)²

= 0.01 – 0.02/9

= 0.07/9

So, SD(G) = (1/3)√0.07 …………………………………………… (4)

and hence tolerance of G is:

3 x SD(G)

= √0.07

= 0.2645.

Thus, the expected tolerance for the gap = ± 0.26454   Answer

Part (b)

Vide (4), SD(G) = 0.0882

Given customer tolerance of G is ± 0.2, the customer tolerance = (0.2/0.0882) = 2.2676 times SD(G).

So, the expected proportion outside specification = P(| Z | > mean + 2.2676 SD), where Z ~ N(0, 1)

= 2 x 0.0117 [Using Excel Function: Statistical NORMSDIST]

= 0.0234 Answer

DONE