Page 456 - Six Sigma Demystified
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436        Six SigMa  DemystifieD


                          15.	 d.
                          16.	 a.
                          17.	 c.
                          18.	 d.
                          19.	 c.
                          20.	 a.	Choices	b	and	c	are	incorrect.	A	negative	interaction	implies	that	as	one
                             requirement	decreases,	the	other	increases.
                          21.	 b.	Cook	staff	training	has	a	higher	importance	than	number	of	cook	staff.	Choice
                             c	is	incorrect	because	number	of	cook	staff	has	a	moderate	relationship	with
                             taste,	which	has	a	customer	importance	of	4.
                          22.	 d.
                          23.	 b.
                          24.	 d.	Step	2b1	is	not	on	the	critical	path,	so	any	improvement	will	not	improve
                             capacity,	nor	will	it	improve	the	total	delivery	time.
                          25.	 b.	Step	1a	will	remain	on	the	critical	path.	The	effect	of	moving	step	2b0	to	run
                             concurrent	with	step	2a	is	to	move	it	off	the	critical	path,	so	the	cycle	time	for
                             the	critical	path	is	reduced	by	1	day.
                          26.	 b.
                          27.	 c.
                          28.	 a.
                          29.	 c.
                          30.	 c.
                          31.	 d.
                          32.	 b.	The	average	expected	time	for	step	1	is	[2	+	(4	×	5)	+	9]/6	=	5.17,	and	the
                             average	expected	time	for	step	2	is	[4	+	(4	×	8)	+	13]/6	=	8.17.	The	sum	of	the
                             two	steps	then	is	13.34	days.
                          33.	 c.	The	expected	standard	deviation	of	the	cycle	time	for	step	1	is	(9	–	2)/6	=
                             1.17,	and	the	expected	standard	deviation	of	the	cycle	time	for	step	2	is
                             (13	–	4)/6	=	1.5.	The	sum	of	the	variances	of	the	two	steps	then	is	(1.17	×	2)	+
                             (1.5	×	2)	=	3.6.	The	standard	deviation	of	the	two	steps	is	the	square	root	of
                             3.6	=	1.9	days.
                          34.	 d.
                          35.	 a.	Walter	Shewhart’s	main	reason	for	inventing	the	control	chart	technique	was
                             to	detect	assignable	causes	of	variation	in	the	process.
                          36.	 b.
                          37.	 a.
                          38.	 b.
                          39.	 d.
                          40.	 b.
                          41.	 c.
                          42.	 d.
                          43.	 c.
                          44.	 d.
                          45.	 a.
                          46.	 c.
                          47.	 d.
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