## Answers

solution 15. No- x 22 تل 24 25 XI R 1 1 9 6 6 6-2 8 9 0 2 9 4 3 3 3.8 9 3 9 9 2.

0 O 2.8 NO 4 1 2 1 0 2 5 -3 0 4 0 -4 -1.6 G 2 0 2 O -0.6 9 7-3 0 -1.4 3 Mo -2 -2 3 -2 -3 --3 -2 3 5 9 2 0 -1 -3 -1 10 0 2 -1 0-4 3 2 2 IL -3 -2 - 1 -1 -1 S 12-16 2 -4 -1 - 3.8 18 0 0 113 09 -3 -9 8 Oo -3.4 0.4 114-3 -5 5 S 10 - F15 - - -2 -1-2 -1 21+ 22 + x3 + x4 + X5 5 R = Mox ( 21, 22, 23, 24, 25) - Min (2, X2, 73, 74.125) B = EB 97 = 6.47 15 15

Now 2.UCLE 9) - UCL = Upper contool limit - (DIR) LCL = Lower ". = (D2 xŔ) for sample size 15, DL = 1.65 D2 = 0.35 UCL = 1.65 X 6.47 = 10.67 LCL = 0.35 X 6.47 = 2.26 b) for the given problem, UCL = 1067 LCL = 226 points ane from the table we com see that mamy not lying between the limits. Honce, it is out of control. We need to modify our data so that it should lie between the UCL and LOL. 6) standard deviation (6) = R dz for n=5 Honce, Given (USL - LSL) d) Pooces Capability (Cp) = 2.326 6.47 2.326 - 2.78 USL = 0.3220 LSL -0.3200 USL = Upper Specification limit LSL Lower 0.3220 -0.3200 6 X 278 0.00011 Cp

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