This set of Industrial Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Control Charts and their Applications – Set 2”.
1. Which of the following is not an example of variable charts?
a) x chart
b) R chart
c) σ chart
d) C – chart
View Answer
Explanation: Control charts are of two types.
i. Variables or Measurements chart – x chart, R chart, σ chart
ii. Attribute charts – p chart, np chart, C chart, U chart.
2. Attribute charts require comparatively ________ sample size.
a) Very small
b) Small
c) Very big
d) Big
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Explanation: Attributes are plotted based on the data collected through go and no go gauges. This data is less effective compared to measured values. Hence, they require comparatively bigger sample size.
3. An attribute chart involves the measurement of dimensions of a product which forms the base for its acceptance or rejection.
a) True
b) False
View Answer
Explanation: Attribute charts are plotted using attribute data. As the attribute data is collected using go or no go gauges. Hence, an attribute chart only differentiates between a defective item and non-defective item without going into the measurement of its dimensions.
4. Variable charts contain more information compared to attribute charts.
a) True
b) False
View Answer
Explanation: Variable charts are plotted using variable data. They are more detailed and contain more information than attribute charts.
5. Match the following according to their use.
p) x chart and R chart i) Number of defects in a sample of constant size
q) p chart ii) Process control
r) C chart iii) Analyses fraction defectives
s) U chart iv) Average number of defects in a sample of variable size
a) p-iv, q-ii, r-i, s-iii
b) p-iv, q-i, r-iii, s-ii
c) p-ii, q-iii, r-i, s-iv
d) p-iv, q-iii, r-ii, s-i
View Answer
Explanation: Each control chart is defined according to its specific use. They are as follows:
Chart | Use |
x chart and R chart | Process control |
p chart | Analyses fraction defectives |
C chart | Number of defects in a sample of constant size |
U chart | Average number of defects in a sample of variable size |
6. Match the following.
p) x chart i) Monitors the centering of the process to control its accuracy
q) R chart ii) Monitors the dispersion or precision of the process
r) σ chart iii) Shows the variation of the process
a) p-iv, q-ii, r-i
b) p-i, q-ii, r-iii
c) p-ii, q-iii, r-i
d) p-iv, q-iii, r-ii
View Answer
Explanation: All are variable control charts.
Chart | Use |
x chart | Monitors the centering of the process to control its accuracy |
R chart | Monitors the dispersion or precision of the process |
σ chart | Shows the variation of the process |
7. In case of variables charts, for a process to be in control bothx chart and R chart should be in control.
a) True
b) False
View Answer
Explanation: In case of variables charts, for a process to be in control both x chart and R chart should be in control because x chart gives information about the accuracy of the process and R chart about the precision of the process.
8. What is the formula to calculate the sub-group average of a sub-group containing the items x1, x2, x3, x4 and x5?
a) x = (x1 * x2 * x3 * x4 * x5) / 5
b) x = (x1 * x2 + x3 + x4 + x5) / 5
c) x = (x1 + x2 + x3 + x4 + x5) / 5
d) x = (x1 – x2 – x3 – x4 – x5) / 5
View Answer
Explanation: The formula to calculate the sub-group average of a sub-group containing the items x1, x2, x3, x4 and x5 is x = (x1 + x2 + x3 + x4 + x5) / 5.
9. Match the following.
Consider a sub-group containing the items x1, x2, x3, x4 and x5. N is the size of the sub-group.
p) Sub-group average i) R = Maximum value – Minimum value q) Sub-group range ii) x= (x1 + x2 + x3 + x4 + x5) / 5 r) Average of averages iii) x̿ = ∑x / N s) Average of range iv) R = ∑R / N
a) p-iv, q-ii, r-i, s-iii
b) p-iv, q-i, r-iii, s-ii
c) p-ii, q-i, r-iii, s-iv
d) p-iv, q-iii, r-ii, s-i
View Answer
Explanation: The criterion for matching is formula.
Sub-group average – x = (x1 + x2 + x3 + x4 + x5) / 5
Sub-group range – R = (Maximum value – Minimum value)
Average of averages – x̿ = ∑x / N
Average of range – R = ∑R / N
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