Please provide work, math and reason for the answers: 3. A company making tires
ID: 359975 • Letter: P
Question
Please provide work, math and reason for the answers:
3. A company making tires for bikes is concerned about the exact width of its cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and mean is 23 mm. What is the process capability index for the process?
4. Consider again that the company making tires for bikes is concerned about the exact width of its cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm.
6. For 120 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for serving. Each day, she took a sample of 8 bottles. The average across all 960 bottles (120 days, 8 bottles per day) was 46 degrees Fahrenheit. The standard deviation across all bottles was 0.8 degree. When constructing an X-bar chart, what would be the center line?
7. For 120 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for serving. Each day, she took a sample of 8 bottles. The average across all 960 bottles (120 days, 8 bottles per day) was 46 degrees Fahrenheit. The standard deviation across all bottles was 0.8 degree. When constructing an X-bar chart, what would be the upper control limit?
11. For 40 days in the summer, you are working in a small, student-run company that sends out merchandise with university branding to alumni around the world. Every day, you take a sample of 50 shipments that are ready to be shipped to the alumni and inspect them for correctness. Across all days, the average percentage of incorrect shipments is 5 percent. What would be the center line for a p-chart?
40
50
0.05
2.5
None of the above
12. You are working in a small, student-run company that sends out merchandise with university branding to alumni around the world. Every day, you take a sample of 50 shipments that are ready to be shipped to the alumni and inspect them for correctness. Across all days, the average percentage of incorrect shipments is 5 percent. What would be the upper control limit for a p-chart?
0
0.05
0.03082207
0.142466
Explanation / Answer
Answer to question 3 :
Process Capability Index
= Minimum ( ( USL – m)/ 3 x Sd , ( m – LSL) / 3x sd )
Where ,
USL = Upper Specification Limit = 23.2 mm
LSL = Lower Specification Limit = 22.8 mm
M = Mean = 23 mm
Sd = Standard Deviation = 0.15 mm
Process Capability Index
= Minimum ( ( 23.2 – 23 )/ 3 x 0.15 , ( 23 – 22.8 )/ 3 x 0.15 )
= Minimum ( 0.444, 0.444)
= 0.444
Answer to Question 4 :
No question is featuring
Answer to question 6 :
Centerline would be = Xbar-bar
Where Xbar-bar = Mean of sample means
Since average across all bottles ( total 960 bottles ) was 46 degree Farenheit
Xbar_bar = 46 degree Farenheit
Answer to question 7:
Upper Control limit ( UCL ) will be defined as :
UCL = Xbar-bar + 3 x Standard deviation/ Square root ( Sample size)
= 46 + 3 x 0.8 / Square root ( 8)
= 46 + 3 x0.8/2.828
= 46 + 0.848
= 46.848
Upper Control Limit = 46.848 Farenheit
Answer to question 11 :
Center line of a p chart = pbar = Average of percentage error in decimal
= 5 /100 = 0.05
CENTERLINE FOR A P CHART WOULD BE = 0.05
Answer to question 12 :
Upper Control Limit ( UCL) of a p chart
= pbar + 3 x Square root ( pbar x ) 1 – pbar)/n)
Pbar = Center line of p chart = 0.05
N = sample size = 50 shipments
Therefore ,
UCL = 0.05 + 3 x Square root ( 0.05 x 0.95 / 50)
= 0.05 + 3 x Square root ( 0.00095)
= 0.05 + 3 x 0.030822
= 0.05 + 0.092466
= 0.142466
UPPER CONTROL LIMIT FOR P CHART = 0.142466
UPPER CONTROL LIMIT FOR P CHART = 0.142466
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