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Submitted by Clay1203 on Sun, 2012-05-06 01:31
due on Thu, 2012-05-10 01:30
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# Evaluate 5.2 Problem 11

Jim Outfitters, Inc. makes custom fancy shirts for cowboys. The shirts could be flawed in various ways, including flaws in the weave or color of the fabric, loose buttons or decorations, wrong dimensions, and uneven stitches. Jim randomly examined 10 shirts, with the following results which are in the attached table.

a. Assuming that 10 observations are adequate for these purposes, determine the three sigma control limits for defects per shirt.

b. Suppose that the next shirt has 13 flaws. What can you say about the process now?

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Submitted by shahimermaid on Sun, 2012-05-06 02:41
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file1.2_problem_11_data_set.xlsx preview (78 words)

# xxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
 xxxxx Defects xxxx bar xxxxx xxxxxx xxxxx Defects xxxx bar (xi-x xxxxxx x x 2 x x 8 1.3600000000000003 1.85 x x xx xx x 0 xxxxx 44.09 x x x x x 7 0.36000000000000032 xxxx 4 12 x 36 x xx xxxx 28.73 5 xx 1 x -1.6399999999999997 xxxx 6 10 4 xx x 10 3.3600000000000003 11.29 x 2 xx 16 x x xxxxxxxxxxxxxxxxxxx 21.53 8 4 xx 4 x 4 xxxxxxxxxxxxxxxxxxx 6.97 9 6 x x 9 6 -0.63999999999999968 xxxx xx x 0 x xx x xxxxxxxxxxxxxxxxxxxx xxxx 60 xxx xx xx xxxx 40.45 xx xxxxxx xxxxx 60/10 x 6 mean= 73/11= xxxx sd= xx xx xxxxxx 3.56 sd= sq rt 158.55/10=3.98 3 -sigma limits xxx means xx xxxxxx xx xxx 3 xxxxxx limits xxx xxxxx of sample of xxx UCL x mean x xxxxxxxxxxx 6 x 3(3.56/3.16) x xxxx x defects xxx = xxxx x 3(SD/Sqrtn) xxxx xxxxxxxxxxxxxx 6.64 + 3.6 x 10 xxxxxxx central line = mean 6 central xxxx = mean xxxxx 7 defects LCL=mean -3(SD/Sqrtn) xxx =3 xxxxxxx xxxxxxxx -3(SD/Sqrtn) 6.64-3.6= x xxxxxxx