(a) Obtain the control limits for X̄-chart and R-chart and describe the significance of joint study of these charts. 20 marks

(b) Find the reliability and hazard functions of Weibull distribution with scale parameter θ and shape parameter β, and int

Question

(a) Obtain the control limits for X̄-chart and R-chart and describe the significance of joint study of these charts. 20 marks

(b) Find the reliability and hazard functions of Weibull distribution with scale parameter θ and shape parameter β, and interpret the findings. 10 marks

(c) Use simplex method to solve the following LPP :
Maximize z = 5x₁ + 2x₂
subject to
6x₁ + x₂ ≥ 6
4x₁ + 3x₂ ≥ 12
x₁ + 2x₂ ≥ 4
x₁ ≥ 0, x₂ ≥ 0
20 marks

UPSC Answer Check · Accepted Answer

(a) Let samples of size n have mean X̄ and range R. For an in-control normal process with mean μ and standard deviation σ, the sample mean satisfies X̄ ~ N(μ, σ²/n), so its 3-sigma limits are μ ± 3σ/√n. Since μ and σ are unknown, estimate μ by X̄̄ and σ by R̄/d₂, because E(R)=d₂σ. Hence for the X̄-chart: CL = X̄̄ UCL = X̄̄ + A₂R̄ LCL = X̄̄ − A₂R̄ where A₂ = 3/(d₂√n). For the R-chart, E(R)=d₂σ and SD(R)=d₃σ, so the 3-sigma limits are (d₂ ± 3d₃)σ. Replacing σ by R̄/d₂ gives: CL = R̄ UCL = D₄R̄ LCL = D₃R̄ where D₄ = 1 + 3d₃/d₂ and D₃ = max(0, 1 − 3d₃/d₂). In standard tables D₃=0 for n≤6; if D₃<0, take LCL=0. Conditions: independent observations, fixed subgroup size, and approximate normality. Significance of joint study: The X̄-chart detects shifts in the process mean, while the R-chart detects changes in process variability. They must be used together because the X̄-chart limits depend on R̄. If the R-chart is out of control, R̄ is not a reliable estimate of σ, so X̄-chart signals may be misleading. Usually the R-chart is examined first. If R is in control and X̄ is out, the location has shifted. If R is out, variability has changed and may explain apparent mean shifts. Joint patterns help diagnose assignable causes such as sudden shifts, increasing spread, or gradual drift. (b) For a Weibull distribution with scale θ and shape β, the density is f(t) = (β/θ)(t/θ)^(β−1) exp[−(t/θ)^β], t≥0, θ>0, β>0. The distribution function is F(t) = 1 − exp[−(t/θ)^β]. Hence the reliability function is R(t) = P(T>t) = 1 − F(t) = exp[−(t/θ)^β]. The hazard function is h(t) = f(t)/R(t) = (β/θ)(t/θ)^(β−1) = β t^(β−1)/θ^β, t≥0, with unit “per unit time” if t is time. Interpretation: β<1 gives a decreasing hazard, representing infant mortality. β=1 gives a constant hazard h(t)=1/θ, the exponential memoryless case. β>1 gives an increasing hazard, representing wear-out or aging. The scale θ is the characteristic life because R(θ)=exp(−1). (c) Introduce surplus variables s₁,s₂,s₃≥0: 6x₁ + x₂ − s₁ = 6 4x₁ + 3x₂ − s₂ = 12 x₁ + 2x₂ − s₃ = 4 x₁,x₂,s₁,s₂,s₃ ≥ 0. A feasible basic solution is x₁=4, x₂=0, s₁=18, s₂=4, s₃=0. Taking x₂ and s₃ as nonbasic, solve for the basic variables: x₁ = 4 − 2x₂ + s₃ s₁ = 18 − 11x₂ + 6s₃ s₂ = 4 − 5x₂ + 4s₃. Then z = 5x₁ + 2x₂ = 20 − 8x₂ + 5s₃. The coefficient of s₃ in z is +5, so s₃ enters. For fixed x₂=0, as s₃ increases: x₁ = 4 + s₃, s₁ = 18 + 6s₃, s₂ = 4 + 4s₃, all of which remain nonnegative and increase. No basic variable decreases to zero, so there is no leaving variable. Thus the objective increases without bound: z = 20 + 5s₃ → ∞ as s₃ → ∞. Therefore the LPP has no finite maximum; z is unbounded above.

(a) Obtain the control limits for X̄-chart and R-chart and describe the significance of joint study of these charts. 20…

Model answer

What "Solve" 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

Practice this exact question

More from Statistics 2024 Paper II