(a) What do you understand by control chart for fraction defective? Explain its construction. Give the theoretical distribution…

Part (a) A control chart for fraction defective, or p-chart, is an attribute control chart used when each unit is inspected and classified as defective or acceptable, and the interest is in the proportion defective rather than a measurement. It is appropriate when sample sizes are constant or known; if they vary, the limits must be adjusted for each sample. For sample i, pᵢ = dᵢ/nᵢ. The chart shows how pᵢ varies over time around a centre line. Construction: collect m samples, compute pᵢ, estimate p̄ = Σdᵢ/Σnᵢ (or Σdᵢ/(mn) if n constant), compute 3-sigma limits p̄ ± 3√[p̄(1-p̄)/nᵢ], plot points with CL and limits. The theoretical distribution is binomial: dᵢ ~ Binomial(nᵢ,p), because each item is a Bernoulli trial with probability p of being defective; hence E(pᵢ)=p and Var(pᵢ)=p(1-p)/nᵢ. For large nᵢ, the binomial is approximated by a normal distribution, so the 3-sigma limits capture about 99.7% of common-cause variation. A point beyond limits, or a run/trend, means the probability of defectives has changed due to an assignable cause; if none, the process is stable and limits can be used for future monitoring.

Part (b) Here n=50 for all 24 days, so a p-chart is suitable. Total defectives = 88; p̄=88/1200=0.0733. Standard error = √[0.0733(0.9267)/50]=0.0369. Initial limits: CL=0.0733; UCL=0.0733+3(0.0369)=0.1839; LCL=0.0733-0.1106<0, so 0. The 24 plotted points pᵢ=dᵢ/50 are: 0.12, 0.04, 0.10, 0.02, 0.04, 0.04, 0.06, 0.10, 0.06, 0.08, 0.24, 0.08, 0.08, 0.02, 0.06, 0.10, 0.08, 0.02, 0.08, 0.06, 0.10, 0.08, 0.04, 0.06. All lie between 0 and 0.1839 except day 11, p=12/50=0.24, which exceeds UCL. This single point indicates an assignable cause on day 11, such as a machine fault or material batch, and the process was not in control then. After investigating and removing day 11, total defectives=76 over 23 samples; p̄=76/1150=0.0661. New SE=√[0.0661(0.9339)/50]=0.0351. Revised limits: CL=0.0661; UCL=0.0661+3(0.0351)=0.1715; LCL=0. The remaining 23 points are within these limits and show no obvious run or trend, so the process may be regarded as in statistical control. For subsequent use, adopt CL=0.0661, UCL=0.1715, LCL=0; if future sample sizes differ, use p̄ ± 3√[p̄(1-p̄)/nᵢ].

Part (c) The given failure probabilities are cumulative by week, not weekly. Therefore weekly failure probabilities fᵢ = pᵢ - pᵢ₋₁ are 0.10, 0.15, 0.25, 0.20, 0.30. For 1000 bulbs, expected weekly failures N fᵢ are 100, 150, 250, 200, 300; cumulative expected failures N p_T by end of weeks 1-5 are 100, 250, 500, 700, 1000. Under a policy of group replacement every T weeks plus individual replacement of failures in between, group cost per cycle is 1000×$1=$1000. Individual replacement cost is $3 multiplied by expected failures during the T-week interval. Average weekly cost = (group cost + individual cost)/T.

T | Cumulative failures | Total cost | Average cost/week 1 | 100 | 1000+300=1300 | 1300 2 | 250 | 1000+750=1750 | 875 3 | 500 | 1000+1500=2500 | 833.33 4 | 700 | 1000+2100=3100 | 775 5 | 1000 | 1000+3000=4000 | 800

The minimum average cost is $775 per week at T=4. Thus the optimum replacement policy is to replace all 1000 bulbs simultaneously at the end of every 4 weeks and replace failed bulbs individually in the intervening weeks. This is the economic balance: if group replacement is done too often (T=1 or 2), the fixed group cost is spread over too few weeks and average cost is high; if it is delayed to week 5, many bulbs fail individually before group replacement, raising the $3 individual cost. Week 4 minimizes the total expected cost per week.