Two Collection Centres, One Misleading 'Average' Member
A dairy cooperative at Berhampur collects milk from 60 member households through two collection centres. Centre A has 40 members whose daily supply has mean 38.2 litres and standard deviation 9.5 litres; Centre B has 20 members, three of whom are large suppliers, with mean 28.6 litres and standard deviation 12.0 litres. The secretary, in a hurry, averages the two means and writes that 'the typical member supplies 33.4 litres a day'. The board also wants to compare this year with last year, when the combined collection of the same 60 households had mean 32.0 litres and standard deviation 12.8 litres. Centre B's figures are markedly right-skewed: its median is 26.1 litres while its mean is 28.6 litres. A member asks whether the report should quote a mean at all for Centre B, and the treasurer wants a single mean, plus a statement on whether supply has become more or less variable than last year, for the annual report.
1. Compute the correct combined mean for all 60 households and explain precisely why the secretary's 33.4 litres is wrong.
Read the worked answer
The combined mean is a weighted mean, weighted by the number of households in each centre:
x̄ = (n_A·x̄_A + n_B·x̄_B)/N = (40 × 38.2 + 20 × 28.6)/60 = (1528 + 572)/60 = 2100/60 = 35.0 litres.
Check: 40 × 38.2 = 1528 and 20 × 28.6 = 572; the total collection is 2100 litres over 60 households.
The secretary computed (38.2 + 28.6)/2 = 33.4, which gives each centre equal weight (1/2) instead of each household equal weight. Centre B, with only 20 of the 60 households, receives double its due weight, and because B has the lower mean (28.6) the reported figure is dragged down by 1.6 litres. The general point the exam tests: a simple average of group means equals the overall mean only when the group sizes are equal. Here the weights should be 40/60 and 20/60, not 1/2 and 1/2. The error is in the weighting, not in the arithmetic of the two centre means, and it matters in practice - a 1.6-litre shortfall on 60 households understates daily collection by about 96 litres if anyone uses the figure for procurement planning.
2. Compute the combined standard deviation for all 60 households and explain why it is larger than the simple average of the two centre standard deviations (10.75 litres).
Read the worked answer
Use the combined-variance identity, which adds the within-centre spread to the between-centre spread measured by the squared deviations of the centre means from the overall mean:
σ² = [n_A(σ_A² + d_A²) + n_B(σ_B² + d_B²)]/N, where d_i = x̄_i − x̄.
d_A = 38.2 − 35.0 = 3.2, d_A² = 10.24; d_B = 28.6 − 35.0 = −6.4, d_B² = 40.96.
σ² = [40(90.25 + 10.24) + 20(144.00 + 40.96)]/60 = [40(100.49) + 20(184.96)]/60 = [4019.6 + 3699.2]/60 = 7718.8/60 = 128.65.
σ = √128.65 = 11.34 litres (a sample standard deviation with divisor 59 would be almost identical, about 11.44).
Why it exceeds 10.75: the centre means differ by 9.6 litres, so the pooled group is more heterogeneous than either centre is internally. Taking only the within-centre part gives √([40 × 90.25 + 20 × 144]/60) = √108.17 = 10.40 litres - an understatement, because it pretends every household is compared with its own centre mean rather than with the cooperative mean. Averages of standard deviations are never valid; variances combine, and only through the identity above.
3. Compare the variability of supply with last year using coefficients of variation, and advise on which measure of location should be printed for Centre B.
Read the worked answer
CV = (s/x̄) × 100, valid here because the variable is a ratio-scale quantity that is always positive and the means are far from zero.
This year: 11.34/35.0 × 100 = 32.4%. Last year: 12.8/32.0 × 100 = 40.0%.
So relative variability has fallen, from 40% to about 32%, even though the absolute standard deviation only fell from 12.8 to 11.3 litres. Both parts of the change matter: the mean rose (32.0 → 35.0) and the spread narrowed. The CV is the right comparison because it asks about variability relative to the level of supply. It must not be used to compare, say, litres with rupees, or to compare a mean that could be near zero.
For Centre B: mean 28.6 > median 26.1 is the classic signature of right skew - the three large suppliers pull the mean up. The median is the better description of what a typical member supplies, because it is not affected by the three extreme values. But do not throw the mean away: it is still the input required for the combined collection totals and the combined mean in the paragraph above. The defensible report prints both for Centre B ('median 26.1 litres, mean 28.6 litres, the difference reflecting three large suppliers'), while for the cooperative as a whole the mean of 35.0 litres is the figure the treasurer needs.