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4). A manufacturer has been sampling output from a process to help measure the number of defective items the process is producing. The quality manager has taken 20 samples of 20 items each, and has found a total of 10 defective items. Given this data, what would be the lower and upper control limits for a p-chart of this process if the limits are based on 2 standard deviations?
Answer: ________________________

5). A manufacturer of cylinder valves for GM has been sampling output from their process. So far, they have collected 30 samples of 10 valves each, and measured the length of each valve in the sample. The mean length of all 300 valves so far is 45.2 mm, and the mean range of measurements within each sample has been .52 mm. What would be the 3-sigma upper control limit for an X-bar control chart based on this sample data?
Answer: ________________________

6). A manufacturer of piston rings for Toyota has been sampling from their process. They have collected 15 samples of 20 rings each, and have been measuring the outside diameter of the ring. The overall mean diameter of the rings samples is 105.1 mm, and the mean range of measurements within the samples is .28 mm. What would the correct upper limit for an R-chart?

please see the attached file, please provide me with detailed and clear answer, if you can please do not use excel.

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Answers to questions on p-chart , X-bar control chart , R-chart. It calculates 3-sigma upper control limit an X-bar control chart , upper limit for an R-chart and lower and upper control limits for a p-chart

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4). A manufacturer has been sampling output from a process to help measure the number of defective items the process is producing. The quality manager has taken 20 samples of 20 items each, and has found a total of 10 defective items. Given this data, what would be the lower and upper control limits for a p-chart of this process if the limits are based on 2 standard deviations?
<br>Answer:
<br>Upper Control Limit=p + z *?Ð = 0.04062;
<br>Lower Control Limit=p - z *?Ð= 0.00938
<br>
<br>
<br>No of samples= 20
<br>No of items in each sample= 20
<br>Total no of items=n= 400 =20*20
<br>No of defectives= 10
<br>Therefore Proportion of ...

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