MECH$\mathrm{.}5710$ Quality Engineering

Homework Assignment week #$4$

Elements of $6$ Sigma

$1.$ For the following processes, complete the given table:

Specification Process $\%$ above $\%$ below

Nominal Tol Mean StdD$.\%$OK USL LSL Cp Cpk

$10\mathrm{.}00+0\mathrm{.}0410\mathrm{.}000\mathrm{.}0295\mathrm{.}45=(1002*2\mathrm{.}275)2\mathrm{.}275\%2\mathrm{.}275\%0\mathrm{.}670\mathrm{.}67$

$10\mathrm{.}00+0\mathrm{.}049\mathrm{.}990\mathrm{.}02$

$10\mathrm{.}00+0\mathrm{.}0410\mathrm{.}010\mathrm{.}02$

$10\mathrm{.}00+0\mathrm{.}0510\mathrm{.}000\mathrm{.}02$

$10\mathrm{.}00+0\mathrm{.}0610\mathrm{.}010\mathrm{.}02$

C$\_$p$=($USL$-$LSL$)/6\backslash$sigma

C$\_$pk$=$min$[($USL$-\backslash$mu $)/3\backslash$sigma $,(\backslash$mu $-$LSL$)/3\backslash$sigma $]$

$$Z$(\backslash$pm SL$)=3\backslash$times C$\_$pk $(\backslash$pm SL$)$

Row a$.$

C$\_$p$=(+0\mathrm{.}04-(-0\mathrm{.}04))/(6\backslash$times $0\mathrm{.}02)=0\mathrm{.}67$

C$\_$pk$=$min$[(10\mathrm{.}04-10\mathrm{.}00)/(3\backslash$times $0\mathrm{.}02),(10\mathrm{.}00-9\mathrm{.}96)/(3\backslash$times $0\mathrm{.}02)]=$min$[0\mathrm{.}67,0\mathrm{.}67]=0\mathrm{.}67$

$$Z$=3\backslash$times C$\_$pk$=2$

$\%$above USL,f$(2)=0\mathrm{.}00275$; $\%$below LSL$,$f$(2)=0\mathrm{.}00275$

$\%$OK$=1-0\mathrm{.}00275-0\mathrm{.}00275=0\mathrm{.}9545$