Upper Lontrol Limit (UCL)-X+za Lower Control Limit (LCL)-X-zax ox Standard devia
ID: 368129 • Letter: U
Question
Upper Lontrol Limit (UCL)-X+za Lower Control Limit (LCL)-X-zax ox Standard deviation of distribution of sample means Estimate of the population standard deviation n-Sample size z Standard Normal deviate 3 Average of the sample means - Zine wh ere x = UCL or LCL (take absolute value if using LCL) and s given. Z:5 1) Specifications for a part for a DVD player state that the part should weigh between 24 and 25 ounces. The process that produces the parts has a mean of 24.5 ounces and a . standard deviation of .2 ounces. The distribution of output is normal. a. What percentage of parts will not meet the weight specs? 6-151. b. Within what values will 95.44 percent of the sample means of this process fall, if samples of n-to are taken and the process is in control? 2.) An automatic filling machine is used to fill 1-liter bottles of cola. The machine's output is approximately normal with a mean of 1.0 liter and a standard deviation of.01 liter. Output is monitored using means of samples with 25 observations in each sample. a. Determine the upper and lower control limits that will include roughly 97 percent of the sample means when the process is in control. Given these sample means: 1.005, 1.001, .998, 1.002, .995, and .999, is the process in control? b.Explanation / Answer
The corresponding z values in the case
Z= 25-24.5 /0.2 and 24-24.5/0.2
z = 2.5 to -2.5, which corresponds to 0.62% on each side.
Total percentage = 0.62x2=1.24
Hence 1.24% of samples will be out of range.
Control limits = Xbar +- z(0.2/2)
Z value for 95.44 confidence limits = 1.97
Control limits = 24.5 +- 0.1x1.97
= 24.697 and 24.303
Part 2
Control limits = Xbar +-z x s.d./ ( sample size)1/2
= 1 +- 2.17x 0.01/5
= 1.00434 and 0.9956 litres
Part b
Since 33% of the given samples (2 out of 6 )is out of limits against the norm of 97%, the process is out of control.
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