Assume that the Length (L) is the most important characteristic to the quality o
ID: 3171110 • Letter: A
Question
Assume that the Length (L) is the most important characteristic to the quality of product X. The following Table shows the measurements of the length of 55 pieces of this product. Use these results to perform the tasks below: A. Estimate the mean and the variance of the distribution of the corresponding quality characteristics. B. Use the Minitab, SPSS, or Excel to construct the Histogram. The Excel sheet of these measurements is available on the Blackboard (file name:Measurements_Homewrok1) C. Assume that the specification limits of the length are mu_0 plusminus allowances = 9.5 plusminus 1 and N_L ~ (mu_1, sigma_0^2) = (9.5mm, 0.490 mm^2). Is the process centered? your comments D. Estimate the percentage of scrap and rework at the current situation N_L ~ (mu_1, sigma_0^2). Assume that the Length (L) is the most important characteristic to the quality of product X. The length is normally distributed variable N_L ~ (mu_0, sigma_0^2)=(20 mm, 4 mm^2). The specification limits are 20 mm plusminus 3 mm. Assuming that if an item exceeds the upper specification limit it can be reworked, and if it is below the lower specification limit it must be scrapped. A. The process is running now at N_L ~ (mu_0, sigma_0^2) = (22 mm, 4 mm^2), What percent of scrap and rework is the process now producing? Show the drawings. B. The process is running now at N_L ~ (mu_0, sigma_0^2) = (23 mm, 4 mm^2). What percent of scrap and rework is the process now producing? Show the drawings. C. (Optional) The process is running now at N_L ~ (mu_0, sigma_0^2) = (20 mm, 6 mm^2). What percent of scrap and rework is the process now producing? Show the drawings.Explanation / Answer
Given, the Lengths of the product X
Slno Length
1 9.5
2 10.98
3 10.2
4 9
5 9.5
6 9.5
7 9.6
8 10.31
9 9.38
10 10.1
11 10.54
12 9
13 11.71
14 9.89
15 10.04
16 10.02
17 9.75
18 10.03
19 11.34
20 10.2
21 10.03
22 10.03
23 10
24 11.04
25 10.5
26 10.2
27 10
28 9.84
29 10.22
30 9.74
31 9.8
32 10
33 10.2
34 10
35 10
36 9.9
37 10.19
38 8.52
39 8.07
40 11.82
41 10.02
42 10
43 10.2
44 10
45 10
46 8.76
47 9.93
48 10.76
49 11.23
50 10.42
51 9.04
52 10.09
53 9.43
54 9
55 10
A.a) mean=sum of lentghs/count of lengths
mean=549.57/55
mean=9.992
Variance=(Xi-Xavg)^2/n-1
computing for the data in excel using stdev.s function
we get variance=0.490732
A.b) The histogram for the data distribution is given as
Bin Frequency
8.07 1
8.61 1
9.14 5
9.68 5
10.21 30
10.75 5
11.28 4
More 3
A.c) specification limits are given as 8.5 to 10.5
mean(X)=9.992
Standard deviation=0.7005
we need to calculate
Cpl, Cpu and Cpk to assess whether the process is under control
Cpl=(mean-LSL)/3*SD
Cpl=9.992-8.5/3*0.7005
Cpl=0.7099
Cpu=10.5-9.992/3*0.7005
=0.2417
Cpk is minimum of Cpl and Cpu
hence, Cpk=0.2417
as Cpk is less than 2 we can conclude the process is not centered
A.d) Any value below the LSL or above USL will be rejected in inspection and can be called as scrap
LSL=8.5
USL=10.5
from the given data we have,
8.07
10.54
10.76
10.98
11.04
11.23
11.34
11.71
11.82
a total of 9 reections
Scrap percentage =9/55
Scrap percentage =16.36%
hence the scrap percentage is 16.36%