(a) A train is scheduled to arrive at a station at a random time between 09:00 AM and 09:30 AM. The actual arrival time is…

(a)(i) Let X be the arrival time measured in minutes after 09:00 AM. Since the arrival time is equally likely anywhere from 0 to 30 minutes, X has a continuous uniform distribution on [0, 30]. The probability density function is: f(x) = 1/30, for 0 ≤ x ≤ 30 f(x) = 0, otherwise. This is valid because f(x) ≥ 0 and ∫₀³⁰ (1/30) dx = 1.

(a)(ii) Using the PDF, P(X < 13) = ∫₀¹³ (1/30) dx = [x/30]₀¹³ = 13/30 = 0.4333 approximately. Final answer: P(arrival before 09:13 AM) = 13/30 ≈ 0.4333, i.e. 43.33%.

(a)(iii) For a continuous uniform distribution on [a, b], E[X] = (a + b)/2 and Var(X) = (b − a)²/12. Here a = 0, b = 30. E[X] = (0 + 30)/2 = 15 minutes after 09:00 AM. So expected arrival time = 09:15 AM. Var(X) = (30 − 0)²/12 = 900/12 = 75 minutes². Standard deviation = √75 = 5√3 minutes ≈ 8.660 minutes. Final answer: E[X] = 09:15 AM, Var(X) = 75 minutes², SD = 5√3 minutes.

(a)(iv) Given that the train has not arrived by 09:11 AM, we condition on X > 11. The conditional distribution of X is uniform on [11, 30]. Its conditional density is: f(x | X > 11) = 1/(30 − 11) = 1/19, for 11 ≤ x ≤ 30. Therefore, E[X | X > 11] = ∫₁₁³⁰ x (1/19) dx = (1/19)[x²/2]₁₁³⁰ = (30² − 11²)/(2 × 19) = (900 − 121)/38 = 779/38 = 20.5 minutes after 09:00 AM. 20.5 minutes after 09:00 AM is 09:20:30 AM. Final answer: expected arrival time = 09:20:30 AM.

(b) We use the finite population sample size formula for estimating a population mean. Given: N = 5000, σ² = 121 grams², so σ = 11 grams, margin of error E = 2.5 grams, confidence level = 99%, so z = 2.58 from the normal table. For finite population: n = N z² σ² / [(N − 1)E² + z² σ²]. First compute: z² σ² = (2.58)² × 121 = 6.6564 × 121 = 805.4244. E² = (2.5)² = 6.25. So, n = 5000 × 805.4244 / [(4999 × 6.25) + 805.4244] = 4,027,122 / [31,243.75 + 805.4244] = 4,027,122 / 32,049.1744 = 125.65 approximately. Since sample size must be rounded up, n = 126 containers.

If the population is assumed infinite, the finite population correction is omitted: n₀ = z² σ² / E² = 805.4244 / 6.25 = 128.8679 approximately. Rounding up gives n₀ = 129 containers.

Thus, assuming the population is infinite increases the required sample size from 126 to 129, i.e. by 3 containers, which is about a 2.38% increase. The finite population correction reduces the sample size because sampling without replacement from a finite population provides more information per observation.

(c)(i) The data are counts of defects per sheet, and the inspection unit is constant. Therefore, the appropriate control chart is a c-chart. Total defects = 5 + 6 + 4 + 4 + 6 + 7 + 0 + 6 + 5 + 3 + 1 + 4 + 5 + 3 + 6 + 4 + 3 + 1 + 3 + 4 = 80. Number of sheets = 20. Center line: CL = c̄ = total defects / number of units = 80/20 = 4 defects per sheet. Upper control limit: UCL = c̄ + 3√c̄ = 4 + 3√4 = 4 + 3 × 2 = 10. Lower control limit: LCL = c̄ − 3√c̄ = 4 − 6 = −2. Since the number of defects cannot be negative, LCL is taken as 0. Final answer: CL = 4, UCL = 10, LCL = 0.

(c)(ii) Plot sheet number 1 to 20 on the x-axis and defect count on the y-axis. Draw horizontal lines at CL = 4, UCL = 10, and LCL = 0. The observed defect counts are: 5, 6, 4, 4, 6, 7, 0, 6, 5, 3, 1, 4, 5, 3, 6, 4, 3, 1, 3, 4. All these points lie between LCL = 0 and UCL = 10. No point falls outside the control limits. There is no clear run or systematic pattern beyond common-cause variation. Therefore, the process is in statistical control. However, the average defect level of 4 defects per sheet may still be high from a quality-improvement perspective. The c-chart indicates stability, not necessarily acceptable quality. Management should investigate whether the center line itself can be reduced through process improvement.

(d) Let X be the number of breakdowns per month. From the table: P(X = 0) = 4/20, P(X = 1) = 8/20, P(X = 2) = 6/20, P(X = 3) = 2/20. Expected breakdowns per month: E[X] = (0 × 4 + 1 × 8 + 2 × 6 + 3 × 2)/20 = (0 + 8 + 12 + 6)/20 = 26/20 = 1.3 breakdowns per month. Without preventive maintenance contract, expected monthly loss: = 1.3 × ₹4500 = ₹5850.

With preventive maintenance contract: Expected breakdowns = 1 per month. Breakdown loss = 1 × ₹4500 = ₹4500. Contract cost = ₹3300 per month. Total monthly cost with contract = ₹4500 + ₹3300 = ₹7800.

Comparison: ₹7800 − ₹5850 = ₹1950 extra per month if the contract is taken. Alternatively, savings from reduced breakdowns = (1.3 − 1) × ₹4500 = 0.3 × ₹4500 = ₹1350, which is less than the contract cost of ₹3300. Final answer: The company should not go for the preventive maintenance contract on purely cost grounds, because it increases monthly cost by ₹1950. It would be advisable only if non-cost benefits such as reliability, goodwill, or reduced disruption exceed ₹1950 per month.

(e) E-governance refers to the use of information and communication technology to deliver government services, information, and transactions efficiently and transparently. Its advantages are significant, but they are not automatic; they depend on inclusion, security, and institutional capacity.

For Government, e-governance improves efficiency by automating routine processes, reducing paperwork, and lowering administrative costs. It strengthens transparency and accountability through online records, audit trails, and real-time monitoring. It also enables data-driven policy formulation and better targeting of welfare schemes. Examples include Direct Benefit Transfer, which reduced leakage and ghost beneficiaries; GSTN, which unified indirect tax administration; GeM, which made public procurement more transparent; and e-Office, which reduced file movement delays. However, poor digital infrastructure and weak data governance can limit these benefits.

For the Public, e-governance provides convenience, speed, and wider access. Citizens can avail services 24×7 without repeated visits to offices. It reduces travel cost, waiting time, and dependence on intermediaries. It also empowers citizens through access to information and faster grievance redressal. Examples include UMANG for multiple government services, DigiLocker for digital documents, CoWIN for vaccination registration, IRCTC for rail ticketing, and online income-tax filing. Yet the digital divide can exclude rural, poor, elderly, and less literate citizens. Language barriers and low digital literacy may create new forms of exclusion.

For Business, e-governance improves ease of doing business. Online registration, single-window clearance, digital payments, and electronic compliance reduce transaction costs and delays. Examples include MCA21 for company registration, Udyam for MSME registration, GST portal for tax compliance, e-way bill for goods movement, and GeM for selling to government. These reduce corruption and improve predictability. However, cybersecurity risks, data privacy concerns, frequent portal changes, and interoperability problems can raise compliance burdens for small firms.

Critically, the advantages of e-governance are real but conditional. It can widen inequality if offline channels are withdrawn too early. It can threaten privacy through excessive data collection and surveillance. It can suffer from algorithmic opacity, exclusion errors, and cyberattacks. Therefore, e-governance should be complemented by digital literacy, robust data protection, cybersecurity, multilingual interfaces, grievance mechanisms, and alternative offline service channels. When inclusive and secure, e-governance transforms governance; when exclusive, it merely digitizes existing inequalities.