Question

; = 1190 i=1 X; = 7815 and i=1 and R are to be set up for an important quality characteristic. The sample size isn = 5, and X and r are computed for each of 35 preliminary samples. The summary data are: Control charts for Round your answers to 2 decimal places (e.g. 98.76). (a) Find trial control limits for and R charts. X Control Limits: UCL = CL = LCL = R Control Limits: UCL = CL = LCL = (b) Assuming that the process is in control, estimate the process mean and standard deviation. Statistical Tables and Charts

Answer 1

- we have n=5 k=35 - and x = 7815 & re- 11503-0 give me Soy. @ we find Step 6. + Ŕ first = + the P R i and K 1 de thenfone from ee x = 75 x 7815 and R = 1 x 1190 35 y = 223.2857 & R = 34 - 2 Now Step given sy. for 50 charts control limit for Ž cuarts is : we know the formula UCLE = ř + 3 d2 In Here CL Y = d2=2.326 the factor wleich are obtain througe control cuart table LCLX = X - 3 de Jn Hence pro UCLEE 223.2857 + 34 34 2.326 55 = 223.2857 + 102 5.2010 - 223:2857 + 19-6116 UCLY = 242.8973

Si millarey LCL r = 223.2857 - 19.6116 LILY = 203.6740 Heure contoor cuan for e are. UCLX = [242.90 ery = 223.28 LOLE = 203.67] Step ③ for Conhol cuan CLR = Ř = 34 UCLR = R + 3 d3 R is givensya (from ea" 6 d2 facrer 9 d3=0.864 t + d₂ = 2.326 = 34+ 3x0.864x34 2.326 = 34+ 880128 2.326 è 34+ 37.888 2 201204 = 41.8882201204 Simulany LELRE Ř - 3d3R = 34 - 88.128 2.306 - 34 - 37.888 2201204 = -3.8882201204 LELR=0 (being negabres-

Heures connor ment for Ŕ is UUR = 11.89 CIR = [39] Rele o ; bez lower couper is being negative limit of R. and we know that If LeL is being wejective we can take it ' zero' s (ie. Rause can never be negative) Son 6 0 ELJ = { Elfi] 71 71- = 2 x 223.2857 → ie = 6.3795914286 E[m] = 6.78 1 = 6.38 cared we cam calulate. Staielerd from que denicchim formula. ů= d2 Here d2=2.326 f Rz 34 = 34 2.326 - 14.61786 88736 3 = |4.6 2. Hemel tere Resep