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 A_{2} 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

A

_{2}= 0.577

Upper control limit (UCL) = x̿ + A

_{2}* 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 + A_{2} * R

b) x̿ + A_{2} – R

c) x̿ – A_{2} * R

d) x̿ * A_{2} + R

View Answer

Explanation: The formula to calculate the lower control limit (LCL) for x chart is x̿ – A

_{2}* 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 A_{2} value depends upon the number of items in the sample.

a) True

b) False

View Answer

Explanation: The factor A

_{2}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 A

_{2}is 0.729. If the sample size is 5, then A

_{2}is 0.577. The values of A

_{2}can be obtained from the S.Q.C tables.

5. Match the following.

p) x – chart | i) x̿ + A_{2} * R |

q) R chart | ii) D_{4} * 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 + A

_{2}* R

q) R chart – D

_{4}* 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 d_{2} as 2.059.

a) 2.986

b) 29.86

c) 0.2896

d) 2.896

View Answer

Explanation: Given,

R = 1.025

d

_{2}= 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 d_{2} is 2.059.

a) 1.1506

b) 0.1506

c) 1.5056

d) 15.056

View Answer

Explanation: Given,

R = 3.1

d

_{2}= 2.059

σ = \(\frac {\overline{R}}{d_2}\) = 1.5056.

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