Enclosure 1: Solved Numerical Problems (Chapter-wise) Statistical Process Control: A Step-by-Step Guide

Solved Numerical Problems (Chapter-wise)

Statistical Process Control: A Step-by-Step Guide

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🔹 Chapter 1: Introduction to Statistical Process Control

Problem 1.1

A process produces metal rods with varying lengths. Explain whether SPC should be applied for inspection or improvement.

Solution

SPC is applied for process monitoring and improvement, not merely inspection. It helps:

• Identify process variation

• Distinguish between common and special causes

• Reduce variability systematically

✅ SPC is preventive, not corrective.

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🔹 Chapter 2: Statistical Foundations for SPC

Problem 2.1

Five observations of a quality characteristic are:
12, 14, 15, 13, 16
Find the mean and range.

Solution

[
\bar{X} = \frac{12+14+15+13+16}{5} = \frac{70}{5} = 14
]

[
Range = 16 - 12 = 4
]

✅ Mean = 14, Range = 4

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Problem 2.2

If the standard deviation of a process is 2 mm, find the natural tolerance.

Solution

[
\text{Natural tolerance} = \mu \pm 3\sigma = \pm 6
]

✅ Process spread = 6σ = 12 mm

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🔹 Chapter 3: Process Variation and Rational Subgrouping

Problem 3.1

A stable process shows random variation within limits. Should the operator adjust the machine?

Solution

No.
This variation is common cause variation. Adjusting a stable process results in tampering, which increases variability.

✅ Correct action: No adjustment

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Problem 3.2

Why should samples in a rational subgroup be collected close in time?

Solution

To ensure:

• Within-subgroup variation reflects common causes

• Between-subgroup variation reveals special causes

✅ This improves sensitivity of control charts

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🔹 Chapter 4: Control Charts for Variables

Problem 4.1 (X̄–R Chart)

Five samples of size 4 were collected.

Sample	Mean (X̄)	Range (R)
1	20	4
2	22	5
3	21	3
4	23	6
5	24	4

Given:
A₂ = 0.729, D₃ = 0, D₄ = 2.282

Find control limits.

Solution

[
\bar{X̄} = \frac{20+22+21+23+24}{5} = 22
]

[
\bar{R} = \frac{4+5+3+6+4}{5} = 4.4
]

X̄-chart limits:
[
UCL = 22 + (0.729)(4.4) = 25.21
]
[
LCL = 22 - (0.729)(4.4) = 18.79
]

R-chart limits:
[
UCL = 2.282 \times 4.4 = 10.04
]
[
LCL = 0
]

✅ Process is in control if all points lie within limits

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Problem 4.2 (I–MR Chart)

Mean = 50, Average Moving Range = 4
Given: d₂ = 1.128

Solution

[
UCL = 50 + 3 \left(\frac{4}{1.128}\right) = 60.64
]
[
LCL = 50 - 3 \left(\frac{4}{1.128}\right) = 39.36
]

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🔹 Chapter 5: Control Charts for Attributes

Problem 5.1 (p-Chart)

In samples of size 200, the average fraction defective is 0.05.
Find control limits.

Solution

[
UCL = 0.05 + 3\sqrt{\frac{0.05(0.95)}{200}} = 0.097
]
[
LCL = 0.05 - 3\sqrt{\frac{0.05(0.95)}{200}} = 0.003
]

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Problem 5.2 (c-Chart)

Average number of defects per unit = 9.

Solution

[
UCL = 9 + 3\sqrt{9} = 18
]
[
LCL = 9 - 3\sqrt{9} = 0
]

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🔹 Chapter 6: Process Capability Analysis

Problem 6.1

Given:

• USL = 110

• LSL = 90

• σ = 3

Find Cp.

Solution

[
C_p = \frac{110-90}{6(3)} = \frac{20}{18} = 1.11
]

✅ Process is marginally capable

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Problem 6.2

If mean = 105, σ = 3, find Cpk.

Solution

[
C_{pk} = \min\left(\frac{110-105}{9}, \frac{105-90}{9}\right)
]
[
= \min(0.56, 1.67) = 0.56
]

❌ Process not capable due to poor centering

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🔹 Chapter 7: SPC-Based Process Improvement

Problem 7.1

Control chart shows all points within limits but process output is unsatisfactory. What action is required?

Solution

• Process is stable

• Improvement requires management action

• Reduce common cause variation

✅ Redesign or process improvement needed

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🔹 Chapter 8: Acceptance Sampling

Problem 8.1

A single sampling plan has n = 100, c = 2.
If 3 defectives are found, should the lot be accepted?

Solution

Since defectives > c
❌ Reject the lot

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Problem 8.2

Explain producer’s risk in one sentence.

Solution

Producer’s risk is the probability of rejecting a good lot.

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🔹 Chapter 9: Measurement System Analysis

Problem 9.1

Total variation = 10
Gage R&R variation = 2

Find % Gage R&R.

Solution

[
%GRR = \frac{2}{10} \times 100 = 20%
]

✅ Measurement system is conditionally acceptable

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Problem 9.2

If NDC = 3, comment on the measurement system.

Solution

NDC < 5 indicates poor discrimination.
Measurement system needs improvement.

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🔹 Chapter 10: SPC Implementation

Problem 10.1

Why should SPC not be used for employee performance appraisal?

Solution

SPC reflects process behavior, not individual performance.
Using SPC for appraisal promotes fear and data manipulation.