(a) Samples of size n = 5 units are taken from a process every hour. The x̄ and R̄ values for a particular quality characteristic…

(a) Control limits, process capability and shift detection

(i) x̄ chart: x̄̄ ± A₂R̄ = 20 ± 0.577 × 4.56 = 20 ± 2.631, so LCL = 17.369, CL = 20, UCL = 22.631. R chart: CL = R̄ = 4.56, UCL = D₄R̄ = 2.114 × 4.56 = 9.640, LCL = D₃R̄ = 0.

(ii) With both charts in control, σ̂ = R̄/d₂ = 4.56/2.326 = 1.960.

(iii) LSL = 14 and USL = 24, so the specification width is 10, while the natural tolerance is 6σ̂ = 11.76. Cp = 10/11.76 = 0.850 < 1. The mean 20 is off the target 19, nearer USL: Cpk = min(24 − 20, 20 − 14)/(3σ̂) = 4/5.881 = 0.680. Fraction outside specifications = Φ((14 − 20)/1.960) + 1 − Φ((24 − 20)/1.960) = Φ(−3.06) + 1 − Φ(2.04) = 0.0011 + 0.0207 = 0.0218. About 2.2% of output is nonconforming, mostly above 24, so the process is not capable; its spread must be reduced and its mean re-centred.

(iv) After the shift x̄ ~ N(24, σ̂²/5) with standard error 1.960/√5 = 0.877. The shift goes undetected if x̄ stays within the limits: β = P(17.369 < x̄ < 22.631 | μ = 24) = Φ((22.631 − 24)/0.877) − Φ((17.369 − 24)/0.877) = Φ(−1.56) − Φ(−7.56) = 0.0592 − 0 ≈ 0.059. So the shift is missed on the first sample with probability about 0.059 (detected with probability 0.941).

(b) Weibull distribution: hazard and reliability

A random variable X has the Weibull distribution with scale α > 0 and shape β > 0 if its density is f(x) = (β/α)(x/α)^(β−1) exp[−(x/α)^β], x > 0, and its distribution function is F(x) = 1 − exp[−(x/α)^β].

Reliability function: R(x) = 1 − F(x) = exp[−(x/α)^β]. Hazard function: h(x) = f(x)/R(x) = (β/α)(x/α)^(β−1) = βx^(β−1)/α^β. The cumulative hazard is H(x) = (x/α)^β, so R(x) = exp[−H(x)]. At x = α, R = e⁻¹ ≈ 0.368 for every β, so α is the characteristic life.

Differentiating, h′(x) = [β(β − 1)/α²](x/α)^(β−2), which has the sign of β − 1:

• 0 < β < 1: h′(x) < 0, so the failure rate decreases from ∞ at x = 0 (early-life failures).
• β = 1: h(x) = 1/α, constant; the Weibull reduces to the exponential distribution with mean α.
• β > 1: h′(x) > 0, so the failure rate increases from h(0) = 0 (wear-out); for β = 2, h(x) = 2x/α², linear in x. Thus the shape parameter alone fixes the failure-rate behaviour; α only rescales time.

(c) Protection given by the 10% sampling plan

The plan has n = 0.10N and c = 0.02n = 0.002N: for N = 5000, n = 500 and c = 10; for N = 10000, n = 1000 and c = 20. For a lot with fraction defective p the number defective in the sample is hypergeometric, and Pa(p) = P(D ≤ c) = Σ_d=0^c C(Np, d) C(N − Np, n − d)/C(N, n). Exact values for N = 5000 and N = 10000:

• p = 0.01: Pa = 0.991 and 0.999
• p = 0.02: Pa = 0.583 and 0.559
• p = 0.03: Pa = 0.102 and 0.027
• p = 0.05: Pa = 0.0003 and 3 × 10⁻⁷

Conclusions:

1. The protection is not uniform. Both OC curves give Pa ≈ 0.56–0.58 at p = c/n = 2%, but the curve for the larger lot is steeper; a lot with 3% defective is accepted 10% of the time if N = 5000 but only 2.7% if N = 10000.
2. At LTPD = 0.05 the consumer's risk Pa(0.05) is at most 0.0003, far below the usual 0.10, so the consumer is well protected for every lot size from 5000 to 10000. Lots of about 3% defective are accepted at most about 10% of the time.
3. The price is paid by the producer: lots at 2% defective are rejected about 42–44% of the time, and a fixed percentage of a large lot means heavy inspection. A better plan fixes n and c from the required AQL and LTPD points, whatever the lot size.