Unformatted text preview: ts to use an
chart to control the average length of the bolts
manufactured. He has recently collected the six samples given below. Calculate the average range.
A. .036
B. .18
C. .030
D. .05
0.03 AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 4
Topic: Xbar Control Charts 11730 Chapter 01  An Introduction to Business Statistics 90. A foreman wants to use an
chart to control the average length of the bolts
manufactured. He has recently collected the six samples given below. Determine the LCL and the UCL for the
A. (1.975 2.035)
B. (1.915 2.095)
C. (1.983 2.027)
D. (1.991 2.019) chart. 1.983 to 2.027. All six sample means fall within the control limits. 2.005 ± (.483)(.03) = 1.983 to 2.027
All six sample means fall within the control limits. AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 4
Topic: Xbar Control Charts 11731 Chapter 01  An Introduction to Business Statistics 91. A foreman wants to use an
chart to control the average length of the bolts
manufactured. He has recently collected the six samples given below. Determine the LCL and the UCL for the R chart.
A. (0 .0685)
B. (0 .076)
C. (0 .03)
D. (0 .0601)
0 and .0685. All six sample ranges fall within the control limits. UCLR= (2.004) (.03) = .0685
LCLR = none
All six sample ranges fall within the control limits. AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 4
Topic: Range Control Charts 11732 Chapter 01  An Introduction to Business Statistics 92. A fastener company produces bolts with a nominal (target) length of 2.00 inches. The
specifications are 2.00 ± .006 inches.
Determine the upper specification limit and the lower specification limit for this process.
A. (1.982 2.018)
B. (1.94 2.06)
C. (1.994 2.006)
D. (1.99 2.01)
1.994 and 2.006
USL = 2 + .006 = 2.006
LSL = 2  .006 = 1.994 AACSB: Analytic
Bloom's: Application
Difficulty: Easy
Learning Objective: 6
Topic: Process capability 11733 Chapter 01  An Introduction to Business Statistics 93. A fastener company produces bolts with a nominal (target) length of 2.00 inches. The
specifications are 2.00 ± .006 inches.
If the process mean is equal to the target value and the process standard deviation is .0016
determine the value of Cpk.
A. 1.250
B. 3.750
C. 2.00
D. 1.998
1.25
USL = 2 + .006 = 2.006
LSL = 2  .006 = 1.994 AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 6
Topic: Process capability 11734 Chapter 01  An Introduction to Business Statistics 94. A fastener company produces bolts with a nominal (target) length of 2.00 inches. The
specifications are 2.00 ± .006 inches.
If the process mean is 2.001 and the process standard deviation is .0016, determine the value
of Cpk.
A. 1.042
B. 1.25
C. 2.001
D. 1.458
1.042
USL = 2 + .006 = 2.006
LSL = 2  .006 = 1.994 AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 6
Topic: Process capability 11735 Chapter 01  An Introduction to Business Statistics 95. A fastener company produces bolts with a nominal (target) length of 2.00 inches. The
specifications are 2.00 ± .006 inches.
If the process mean is 2.001 and the process standard deviation is .0016, determine the
estimated standard deviations of leeway for this process.
A. .0016
B. .0048
C. 1.375
D. .125
0.125
USL = 2 + .006 = 2.006
LSL = 2  .006 = 1.994 Estimated number of standard deviations of process leeway = 3.125  3 = 0.125 AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 6
Topic: Process capability 11736 Chapter 01  An Introduction to Business Statistics 96. A fastener company produces a certain type of bolt for the automobile industry with a
nominal (target) length of 2.00 inches. The specifications for the length of the bolt are 2.00
± .006 inches. An automobile manufacturing company will only purchase from this company
if the sigma level of capability of the process is at least 4. If the process mean is equal to
2.001, determine the maximum process standard deviation necessary for the fastener
manufacturing company in order to qualify as a supplier for the automobile manufacturing
company.
A. .0125
B. .005
C. .00125
D. .00025
.00125 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 6
Topic: Process capability 11737 Chapter 01  An Introduction to Business Statistics 97. Suppose that and R charts are based on subgroups of size four are being used to monitor the tire diameter size of a new radial tire manufactured by a tire company. The
and R charts are found to be in statistical control with
inches. A histogram
of the tire diameter measurements indicates that distribution of these measurements is
approximately normally distributed.
If the tire diameter specifications are 50 inches ± 1 inch, is the process capable of meeting the
specifications?
A. Yes
B. No
No, Natural tolerance limits are not within the specification limits. LSL = 50  1 = 49
USL = 50 + 1 = 51
No, 51.42 > 51. Natural tolerance limits are not within the specification limits. AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 6
Topic: Process capabil...
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 Winter '14
 Frequency, Frequency distribution, Histogram, AACSB, Statistical charts and diagrams

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