Question

Problem 2. An automatic filling machine is used to fill 1-liter bottles of cola. The machine's output is approxi- mately normal with a mean of 1.0 liter and a standard deviation of .01 liter. Output is monitored using means of samples of 25 observations. a. Determine upper and lower control limits that will include roughly 97 percent of the sample means when the process is in control. b. Given these sample means: 1.005, 1.001,.998, 1.002, 995, and.999, is the process in control? For Part b, draw the mean chart using excel (preferred) or by hand.

Answer 1

Answer: - According to given data

Mean = 1.0 liter , Standard deviation = 0.01 liter , sample mean = 25

a) For the 97 percent of sample mean z value = 2.17

Critical value = 1- ( (1-0.97)/2 )= 1- 0.015 = 0.985 from standard normal distribution probability table we get z as 2.17

Upper limit = mean + z * (sigma/√n) = 1.0 + 2.17*(0.01/√25) = 1.00434

Lower limit = mean - z * (sigma/√n) = 1.0 - 2.17*(0.01/√25) = 0.9957

b) From the above obtained control limits we can conclude process is out of control because two points lie outside the control limits.

A В C E 4 5 samples means UCL CL LCL 6 1.005 1.00434 1.0 0.9957 7 1.001 1.00434 1.0 0.9957 8 0.998 1.00434 0.9957 1.0 9 1.002 1.00434 1.0 0.9957 10 0.995 1.00434 1.0 0.9957 11 0.999 1.00434 1.0 0.9957 12 13 Chart Title 14 15 1.006 16 1.004 17 1.002 18 1 19 20 0.998 21 0.996 22 0.994 23 0.992 24 0.99 25 2 3 1 5 6 26 - LCL samples means UCL CL 27 28 LL LO 4 LO