(a) (i) What are control charts by variables and control charts by attributes? 5 marks
(ii) Derive the control limits for the construction of control charts for the mean and variability based on sample standard deviation. 15 marks

(b) (i) State the

Question

(a) (i) What are control charts by variables and control charts by attributes? 5 marks
(ii) Derive the control limits for the construction of control charts for the mean and variability based on sample standard deviation. 15 marks

(b) (i) State the assumptions involved under sampling inspection plans by variables and describe the operating procedure of a single sampling plan by variables under the assumption of normal distribution for a quality characteristic. 5 marks
(ii) Establish the relationship between the fraction defective and the acceptance probability under a single sampling plan by variables when the quality characteristic follows a normal distribution with mean μ and variance σ², where σ² is unknown, and when an upper specification limit is specified. Using the relationship, obtain the formula for finding the parameters of the sampling plan. 10 marks

(c) (i) Given a system consisting of n components, define the state vector and the structure function of the system. What do they indicate? 5 marks
(ii) Defining (1) a series system, (2) a parallel system and (3) a k-out-of-n system, obtain the associated expressions for the structure functions and the reliability functions. 10 marks

UPSC Answer Check · Accepted Answer

(a)(i) 
A control chart by variables is used when a quality characteristic is measurable on a continuous scale, e.g. length, weight, tensile strength. Samples give numerical values; charts for mean, range or standard deviation are plotted. A control chart by attributes is used when items are classified as conforming/nonconforming or defects are counted; e.g. p, np, c and u charts. Variables charts are more informative but need measurement and often normality; attributes charts are simpler but need larger samples.

(a)(ii) 
Let X₁,…,Xₙ be a random sample from N(μ,σ²). Define
X̄ = (1/n) Σ Xᵢ,
S = √[(1/(n−1)) Σ (Xᵢ−X̄)²].
For normal population: X̄ ~ N(μ, σ²/n), E(X̄)=μ, SD(X̄)=σ/√n. Also S is independent of X̄, with
E(S)=c₄σ, SD(S)=σ√(1−c₄²),
where c₄ = √(2/(n−1)) Γ(n/2)/Γ((n−1)/2).

For 3-sigma limits: limits = mean of statistic ± 3 standard error.

For the mean chart: if σ known,
UCL_X̄ = μ + 3σ/√n, LCL_X̄ = μ − 3σ/√n.
If unknown, estimate μ by M (grand mean of sample means), and σ by s̄/c₄, where s̄ is average sample standard deviation. Then
UCL_X̄ = M + A₃ s̄, LCL_X̄ = M − A₃ s̄, A₃ = 3/(c₄√n).

For the S chart: if σ known, centre line c₄σ,
UCL_S = c₄σ + 3σ√(1−c₄²), LCL_S = max(0, c₄σ − 3σ√(1−c₄²)).
If unknown, estimate σ by s̄/c₄. Then centre line = s̄,
UCL_S = B₄ s̄, LCL_S = max(0, B₃ s̄),
B₄ = 1 + (3/c₄)√(1−c₄²), B₃ = 1 − (3/c₄)√(1−c₄²).
Validity requires independent normal samples and known sample size n.

(b)(i) 
Assumptions for variables sampling inspection: the quality characteristic is normally distributed; observations are independent; the lot is homogeneous; measurement error is negligible; σ may be known or unknown; one or two specification limits are prescribed.
For a single sampling plan by variables with upper specification U, choose sample size n and acceptance constant k. Draw a random sample of n items. If σ is known, compute X̄ and accept the lot if X̄ ≤ U − k σ/√n, i.e. (U−X̄)/(σ/√n) ≥ k. If σ is unknown, compute X̄ and S and accept if X̄ + kS ≤ U, i.e. (U−X̄)/S ≥ k. For a lower specification L, accept if X̄ ≥ L + k σ/√n (σ known) or X̄ − kS ≥ L (σ unknown). Otherwise reject.

(b)(ii) 
Let X~N(μ,σ²), upper limit U, σ² unknown. The fraction defective is
p = P(X>U) = 1 − Φ((U−μ)/σ).
Put z_p = Φ⁻¹(1−p) = (U−μ)/σ.

The plan accepts if X̄ + kS ≤ U, i.e. (U−X̄)/S ≥ k. Define
T = √n (U−X̄)/S.
Since X̄ ~ N(μ, σ²/n) and (n−1)S²/σ² ~ χ² with n−1 df independently, T follows a noncentral t-distribution with n−1 degrees of freedom and noncentrality parameter
δ = √n z_p.
Thus
P_a(p) = P(T ≥ k√n) = 1 − F(k√n; √n z_p),
where F is the noncentral t cdf. This is the required relation.

To determine n and k, specify producer’s risk α at acceptable fraction p₁ and consumer’s risk β at lot tolerance fraction p₂. Solve
F(k√n; √n z_p₁) = α,
F(k√n; √n z_p₂) = 1 − β,
with z_pᵢ = Φ⁻¹(1−pᵢ), i=1,2. The simultaneous solution gives n and k.

(c)(i) 
For n components, the state vector is x=(x₁,x₂,…,xₙ), where xᵢ=1 if component i works and xᵢ=0 if it fails. The structure function φ(x)=1 if the system works, and φ(x)=0 if it fails. It indicates the system state as a Boolean function of component states. The reliability is R=P(φ(X)=1).

(c)(ii) 
1. Series system: it works only if all components work.
φ(x) = ∏ (i=1 to n) xᵢ = min(x₁,…,xₙ).
If components work independently with reliabilities pᵢ,
R_s = ∏ (i=1 to n) pᵢ. If identical, R_s = pⁿ.
2. Parallel system: it works if at least one component works.
φ(x) = 1 − ∏ (i=1 to n) (1−xᵢ) = max(x₁,…,xₙ).
Reliability: R_p = 1 − ∏ (i=1 to n) (1−pᵢ). If identical, R_p = 1 − (1−p)ⁿ.
3. k-out-of-n system: it works if at least k of n components work.
φ(x) = 1 if Σ (i=1 to n) xᵢ ≥ k, else 0.
For independent components,
R = Σ (j=k to n) Σ over all subsets A of 1,…,n with |A|=j [∏ (i∈A) pᵢ ∏ (i∉A) (1−pᵢ)].
If identical,
R = Σ (j=k to n) C(n,j) pʲ (1−p)ⁿ⁻ʲ.
Here k=1 gives parallel, k=n gives series.

(a) (i) What are control charts by variables and control charts by attributes? 5 marks (ii) Derive the control limits for the…

Model answer

What "Derive" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Statistics 2025 Paper II