Two companies: wages and combined SD

Let n₁ = 100, x̄₁ = 460, σ₁ = 50 for Company A, and n₂ = 80, x̄₂ = 490, σ₂ = 40 for Company B.

i) Which company pays the larger amount?

Total weekly wages = number of workers × mean wage.

Company A: 100 × 460 = Rs 46,000
Company B:  80 × 490 = Rs 39,200

Company A pays the larger total weekly wage bill (Rs 46,000). Company B pays the higher average wage per worker (Rs 490).

ii) Which shows greater variability?

Compare using the coefficient of variation, CV = σ / x̄ × 100.

CV(A) = 50/460 × 100 = 10.87 %
CV(B) = 40/490 × 100 =  8.16 %

Company A has greater variability (higher CV). Its SD is also higher (50 against 40).

iii) Combined mean and SD

Combined mean = (n1 x̄1 + n2 x̄2) / (n1 + n2)
              = (46,000 + 39,200) / 180
              = 473.33

Deviations of each mean from the combined mean:

d1 = 460 − 473.33 = −13.33,  d1² = 177.78
d2 = 490 − 473.33 =  16.67,  d2² = 277.78

σ² = [ n1(σ1² + d1²) + n2(σ2² + d2²) ] / (n1 + n2)
   = [ 100(2500 + 177.78) + 80(1600 + 277.78) ] / 180
   = (267,777.8 + 150,222.2) / 180
   = 418,000 / 180
   = 2322.2
σ  = √2322.2 = 48.19

Answer: (i) Company A has the larger wage bill (Rs 46,000 against Rs 39,200). (ii) Company A is more variable (CV 10.87 % against 8.16 %). (iii) Combined mean = Rs 473.33 and combined SD = Rs 48.19.