This set of Industrial Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Control Charts and their Applications – Set 3”.
1. Calculate the upper control limit (UCL) for x chart. Take average of averages as 42.12, average of range as 4.8 and A2 as 0.577.
a) 39.35
b) 3.935
c) 44.8
d) 448.8
View Answer
Explanation: Given,
x̿ = 42.12
R= 4.8
A2 = 0.577
Upper control limit (UCL) = x̿ + A2 * R = 42.12 + (0.577 * 4.8) = 44.8
2. Which of the following is the formula to calculate the lower control limit (LCL) for x chart?
a) x̿ + A2 * R
b) x̿ + A2 – R
c) x̿ – A2 * R
d) x̿ * A2 + R
View Answer
Explanation: The formula to calculate the lower control limit (LCL) for x chart is x̿ – A2 * R. It denotes the lower boundary of the space in the region where the points denoting the process is plotted.
3. In x – chart, the center line is drawn with the value x̿.
a) True
b) False
View Answer
Explanation: The center line in the x – chart is the line that is drawn in the middle of the region. It is located in between the UCL and LCL lines. In x – chart, the center line is drawn with the value x̿.
4. The factor A2 value depends upon the number of items in the sample.
a) True
b) False
View Answer
Explanation: The factor A2 is used in the plotting of x – chart. It depends upon the number of items in the sample. For instance, if the sample size is 4, then A2 is 0.729. If the sample size is 5, then A2 is 0.577. The values of A2 can be obtained from the S.Q.C tables.
5. Match the following.
| p) x – chart | i) x̿ + A2 * R |
| q) R chart | ii) D4 * R |
| r) P chart | iii) p + 3 * \(\sqrt {\frac {\overline{p} * (1-\overline{p})}{n}}\) |
| s) np chart | iv) np + 3 * \(\sqrt {n\overline{p} * (1-\overline{p})}\) |
a) p-i, q-ii, r-iv, s-iii
b) p-ii, q-iv, r-iii, S-i
c) p-i, q-ii, r-iii, s-iv
d) p-i, q-iv, r-iii, s-ii
View Answer
Explanation: The criterion for matching is based on the upper control limit (UCL) of charts. They are matched as follows:
p) x – chart – x̿ + A2 * R
q) R chart – D4 * R
r) P chart – P + 3 * \(\sqrt {\frac {\overline{p} * (1-\overline{p})}{n}}\)
s) np chart – nP + 3 * \(\sqrt {n\overline{p} * (1-\overline{p})}\)
6. Match the following according to the central line.
p) x chart and R chart i) c q) p chart ii) x̿, R r) C chart iii) p s) U chart iv) u
a) p-i, q-ii, r-iv, s-iii
b) p-ii, q-iv, r-iii, S-i
c) p-ii, q-iii, r-i, s-iv
d) p-iv, q-iii, r-ii, s-i
View Answer
Explanation: Criterion for matching is the central line (CL) of control charts. They are as follows:
| Chart | Central line |
| x chart and R chart | x̿, R |
| p chart | p |
| C chart | c |
| U chart | u |
7. Find the process capability. Take R as 1.025 and d2 as 2.059.
a) 2.986
b) 29.86
c) 0.2896
d) 2.896
View Answer
Explanation: Given,
R = 1.025
d2 = 2.059
σ = \(\frac {1.025}{2.059}\) = 0.498
Process capability = 6σ = 6 × 0.498 = 2.986
8. The p and np charts control the fraction defectives in the product whereas the c chart controls the number of defects in the products.
a) True
b) False
View Answer
Explanation: The above statement describes the use of charts. The p and np charts control the fraction defectives in the product whereas the c chart controls the number of defects in the products.
9. Calculate the standard deviation (σ) when the average of the ranges is 3.1 and d2 is 2.059.
a) 1.1506
b) 0.1506
c) 1.5056
d) 15.056
View Answer
Explanation: Given,
R = 3.1
d2 = 2.059
σ = \(\frac {\overline{R}}{d_2}\) = 1.5056.
Sanfoundry Global Education & Learning Series – Industrial Engineering.
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