(a) A company gives on-the-job training to its salesmen, which is followed by a test. The company is considering whether there is…

(a) Using Karl Pearson's product-moment correlation coefficient, let X = test score and Y = sales ('000). For n = 10: ΣX = 520, ΣY = 555, ΣX² = 28076, ΣY² = 31395, ΣXY = 29339. Means: X̄ = 52, Ȳ = 55.5. r = (nΣXY − ΣXΣY) / √[(nΣX² − (ΣX)²)(nΣY² − (ΣY)²)]. nΣXY = 293390, ΣXΣY = 288600, so numerator = 4790. nΣX² − (ΣX)² = 280760 − 270400 = 10360. nΣY² − (ΣY)² = 313950 − 308025 = 5925. Denominator = √(10360 × 5925) = √61,383,000 = 7834.73. r = 4790/7834.73 = 0.6114, i.e. r = 479/√613830 ≈ 0.611.

This is a moderate positive relationship; r² = 0.374, so the test score explains about 37.4% of the variation in sales. For n = 10, the 5% two-tail critical r ≈ 0.632; here r = 0.611, so it is not statistically significant at 5%. Also, a salesman with test score 33 had sales 58, showing poor test performance does not necessarily mean poor sales. Hence the company should not terminate any salesman solely because of poor test performance. The test may be used for training, counselling, probation, and re-testing, along with actual sales records.

(b) This is a binomial distribution problem. Let X = number of correct guesses. Here n = 9, p = 1/4, q = 3/4. P(X = 6) = C(9,6)(1/4)⁶(3/4)³. C(9,6) = C(9,3) = 84. P = 84 × (1/4)⁶ × (3/4)³ = 84 × 27/4⁹ = 2268/262144 = 567/65536 ≈ 0.00865. So the probability is 567/65536 ≈ 0.865%, assuming independent questions and constant p = 1/4.

(c)(i) Let x = number of boxes of type A and y = number of boxes of type B. The LPP is: Minimize C = 300x + 200y subject to 3x + y ≥ 10 (chemical X) 2x + 2y ≥ 12, i.e. x + y ≥ 6 (chemical Y) x + 2y ≥ 7 (chemical Z) x ≥ 0, y ≥ 0; x, y integers.

Graphical method: the boundary lines are: 3x + y = 10: intercepts (10/3, 0), (0, 10) x + y = 6: intercepts (6, 0), (0, 6) x + 2y = 7: intercepts (7, 0), (0, 3.5).

The feasible region lies above all three lines. Key intersections: 3x + y = 10 and x + y = 6 give 2x = 4, so x = 2, y = 4. Check Z: 2 + 2(4) = 10 ≥ 7. Feasible. C = 300(2) + 200(4) = ₹1400. 3x + y = 10 and x + 2y = 7 give x = 13/5, y = 11/5, but x + y = 24/5 = 4.8 < 6, so not feasible. x + y = 6 and x + 2y = 7 give y = 1, x = 5. Feasible. C = 300(5) + 200(1) = ₹1700. Other boundary points: (0,10) gives C = ₹2000; (7,0) gives C = ₹2100.

The minimum occurs at (2, 4). Check: X = 3(2) + 1(4) = 10; Y = 2(2) + 2(4) = 12; Z = 1(2) + 2(4) = 10 ≥ 7. Therefore, the department should buy 2 boxes of A and 4 boxes of B, with minimum cost ₹1400.

(c)(ii) Sensitivity analysis studies how changes in costs or requirements affect the optimum. Here X and Y constraints are binding, while Z has surplus 10 − 7 = 3 units and zero shadow price. From dual/complementary slackness, with Z slack, shadow prices are ₹50 per extra unit of X and ₹75 per extra unit of Y. Thus small increases in X or Y raise minimum cost, but up to 3 extra units of Z cost nothing. The optimum (2, 4) lies where the X and Y constraints meet, so it stays optimal while the cost ratio cA/cB lies between the slopes of those constraints, 1 and 3; with cB = ₹200, the cost of box A can range from ₹200 to ₹600. If cost coefficients or requirements change beyond such ranges, the optimal mix changes.