May I have help with 1-5 T My Tech Raiderlink x 0 outlook Web App x Take Home Ex
ID: 3231037 • Letter: M
Question
May I have help with 1-5
T My Tech Raiderlink x 0 outlook Web App x Take Home Exam 9-20 x Y Take home exam g.pdf x C Chegg Study l Guided S x New Tab Secure https:// ttu black board.com /bbcswebdav/pid-2595092-dt-content-rid-1400328 courses/201757-MATH-2345-DO1/lake%20home%20exam%209.pdf l (Q1-q3) Product filling weights are normally distributed with a mean of 300 grams and a standard deviation of 10 grams. A sample of size 6 is taken from the process. 1. Which is the standard deviation of the sample? A 2.45 B) 30 C) 4.08 D) 10 2. Which are the control limits for the X chart of the process? A) 287.76 312.24 B) 290 310 D) 276.23 324.78 C) 295 305 3. For the sample below, the process is 287 294 334 323 298 312 A) In control B) Out of control ll. (q4-q5). A process sampled 10 times with a sample of size 6 resulted in 22.6 and R -1.3 4. The upper and lower control limits for chart are: A 19.8 24.7 B) 21.97 23.22 C) 20.23 21.65 D) 24. 3-27.8 5. The upper and lower control limits for R chart are: A) 0.56-3.4 B) 0-2.004 C) 0.23 1.56 D) 0-2.6 Ill. (q6-q10) For a process in control, 10 samples of 200 packages each of tennis strings were tested for breaking strength. A total of 160 packages of the 2000 tested failed to conform to th manufacturer's specifications. 6. The process proportion defective when the process is in control is: A) 16 B) 100 C) 0.08 D) 0.02 7. The upper and lower control limits for the p chart are: A) 0.02 0.14 B) 0-2 C) 2.4-4.8 D) 0.16 1.64 8. What conclusion should be made about the process if a new sample of 200 packages is taken and Type here to search 7:34 PM 5/2/2017Explanation / Answer
1)
Std. deviation = 10
2)
Control limits
Upper control limit = x bar + A2 * r bar
= 300 + 0.483 * 10
= 304.83 = 305
Lower control limit = x bar - A2 * r bar
= 300 - 0.483 * 10
= 295.17 = 295
Answer is option C)
4)
Upper control limit = x bar + A2 * r bar
= 22.6 + 0.483 * 1.3
= 23.22
Lower control limit = x bar - A2 * r bar
= 22.6 - 0.483 * 1.3
= 21.97
Answer is option B)
5)
Upper control limit = D4 * R bar
= 2.004 * 1.3
= 2.6052
Lower Control Limit = D3 * R bar
= 0
Answer is option D)