The time to replace vehicle wiper blades at a service center was monitored using
ID: 326563 • Letter: T
Question
The time to replace vehicle wiper blades at a service center was monitored using a mean and a range chart. Six samples of n = 20 observations have been obtained and the sample means and ranges computed: Sample MeanRange Sample Mean 3.13 3.06 3.09 Range 3.06 3.15 3.11 .42 .50 41 46 .45 Factors for three-sigma control limits for and R charts FACTORS FOR R CHARTS Lower Control Limit, Number of Upper Control Limit, Factor for observations in Chart, Subgroup, D3 1.88 1.02 0.73 0.58 0.48 0.42 0.37 0.34 0.31 0.29 0.27 0.25 0.24 0.22 0.21 0.20 0.19 0.19 0.18 0.08 0.14 0.18 0.22 0.26 0.28 0.31 0.33 0.35 0.36 0.38 0.39 0.40 0.41 3.27 2.57 2.28 2.11 2.00 1.92 1.86 1.82 1.78 1.74 1.72 1.69 1.67 10 12 14 15 16 17 18 19 20 1.64 1.62 1.61 1.60 1.59Explanation / Answer
Answer: a
For we need to calculate the Average of means and Average of Ranges with below formula
S
Mean Data
Range Data
1
3.06
0.42
2
3.15
0.50
3
3.11
0.41
4
3.13
0.46
5
3.06
0.46
6
3.09
0.45
Average
3.10
0.45
So we got the data as
N = 20
Average of Means = 3.10
Average of range = 0.45
A2 = 0.18 is the value from the 3 sigma factor table with n=20.
UCL and LCL for mean can be calculated with the below formula as
UCL (mean) = 3.10 + 0.18 * 0.45 = 3.1810
LCL (mean) = 3.10 - 0.18 * 0.45 = 3.0190
The value of D3 and 4 from the 3 sigma table with n=20 are as below
UCL and LCL for range can be calculated with the below formula as
UCL (range) = 1.59* 0.45 =0.7155
UCL (range) = 0.41* 0.45 =0.1845
Answers are as below
S
Mean Data
Range Data
1
3.06
0.42
2
3.15
0.50
3
3.11
0.41
4
3.13
0.46
5
3.06
0.46
6
3.09
0.45
Average
3.10
0.45
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