Question

12. A control chart is to be established on a process producing refrigerators. The inspection unit is one refrigerator, and a common chart for nonconformities is to be used. As preliminary data, 16 nonconformities were counted in inspecting 30 refrigerators.

    (a) What are the three-sigma control limits?
    (b) What is the α-risk for this control chart?
    (c) What is the β-risk if the average number of defects is actually 2 (i.e., if c = 2.0)? (d) Find the average run length if the average number of defects is actually 2.

Answer 1

Answer: Given that A data, with 16 non com formities were counted in inspecting 30 refrigerators. a) To find the three-sigma control limits Inspection unit nal and EDG = 16 ē - let = 0.5333 caleulated as follows. sfama limits is Hence the three CL 20 = 0.5333 UCL = Ttko =0.5333 + 30.5333 = 0.5333+ 2.1906 = 2.7239 LLL = C-KE = 0,5333-3001533 =0.5333 - 2.1906 =1.66 b) to find the x- risk for this control chart a = pixelella) +P(x>0C4/) Here, Xis a poissiop, random Variable with parameter c. x=P(XCLCLICH.P (X>uch/a) a = P(XCLCL (0) +CI- PCX SUCLLL} =Pfx = 07 =0,5333] + [l-P{x=0,1,2]+= 0.5338}] Where, x is a poission random Variable with parametric

=0+{\-P(X=1) +P (X=2).. =;- $ 80.8833 (0:5333)' 20.5333(0.5333)27 y!t2 = |-[0.3129+0:083401 = 1-0.3963 2.0.6037 c) To find ß-risk - where c=2 Cao = its =0.0667 ::B3p{XEOCLIC}-P{xsicclc} Here ; x is a poission random yariable with parameteric' B=P{LCL excuci) 720.06673 =P3 x=121 € = 0.0669} S 20.0667.6.0667) 10667.0 667)? = [0.0624+0.6021] -0.0645 d) ARL-i 1-0.064 = 1,0689 11.07