Two Shifts, One Plant-Wide Standard Deviation
The Peenya unit of a Bengaluru garment exporter runs two shifts on identical machines. Last month shift A (40 operators) produced a mean of 52 pieces per operator per day with a standard deviation of 6 pieces; shift B (60 operators) produced a mean of 58 pieces with a standard deviation of 9 pieces. The production manager concludes that shift B is 'the more variable shift because 9 is bigger than 6', and reports a plant-wide output of 55 pieces and a plant-wide variability of 7.5 pieces by simply averaging the two figures. The HR head wants a single consistency ranking to decide which shift gets the process-improvement incentive, and the costing department wants one plant-wide mean and standard deviation for the monthly report. Separately, the quality team proposes to deduct 2 pieces per operator per day as unrecoverable wastage before any statistic is computed, and the export order is finally priced in dozens rather than pieces.
1. Rank the two shifts on consistency using an appropriate relative measure and comment on why the manager's use of the standard deviation alone is misleading here.
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Use the coefficient of variation, CV = (SD/mean) × 100, because the two shifts have different average output levels.
- Shift A: CV = (6/52) × 100 = 11.54%
- Shift B: CV = (9/58) × 100 = 15.52%
Shift A is the more consistent shift: its day-to-day output wobbles by about 11.5% of its own mean, against 15.5% for B. The standard deviation is an absolute measure in pieces; it tends to grow with the size of the mean, so a high-output group can show a large SD and still be tightly controlled relative to its level. Comparing 9 with 6 compares two numbers measured against different centres. The incentive judgement therefore goes to A, and the manager's ranking reverses once scaling is removed. Note the caveat: CV is unstable if a mean is near zero, which is not the case here.
2. Compute the combined mean and combined standard deviation of all 100 operators, and explain why 7.5 is not the plant-wide SD.
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Combined mean: x̄ = (40×52 + 60×58)/100 = (2080 + 3480)/100 = 55.6 pieces - not 55, because the shifts are of unequal size.
Deviations of the shift means from the grand mean: d_A = 52 − 55.6 = −3.6; d_B = 58 − 55.6 = +2.4.
Combined variance = [n_A(s_A² + d_A²) + n_B(s_B² + d_B²)]/N = [40(36 + 12.96) + 60(81 + 5.76)]/100 = [40(48.96) + 60(86.76)]/100 = (1958.4 + 5205.6)/100 = 71.64
Combined SD = √71.64 = 8.46 pieces.
Averaging the two SDs, (6 + 9)/2 = 7.5, treats the plant as if all operators shared the same centre. They do not: the shifts differ by 6 pieces on average, and that between-shift spread is itself variation that the plant-wide SD must contain. Hence the correct 8.46 is appreciably larger than 7.5, though still smaller than shift B's 9.
3. Taking shift A alone, state the new mean and SD if (i) 2 pieces are deducted from every operator as wastage, and (ii) the resulting figures are then expressed in dozens.
Read the worked answer
Apply the two standard transformations in order, tracking origin and scale separately.
(i) Deducting a constant 2 from every operator is a change of origin: x → x − 2. The mean falls by 2 to 50 pieces, but the SD is unchanged at 6 pieces, because every observation moves by the same amount and the gaps between operators are untouched. Wastage deduction does not make output more uniform.
(ii) Converting pieces to dozens is a change of scale: x → x/12. Both the mean and the SD are divided by 12: mean = 50/12 = 4.17 dozen, SD = 6/12 = 0.50 dozen.
Rule to quote: addition or subtraction of a constant shifts only the mean; multiplication or division by a constant rescales both mean and SD by that constant. The CV, being a ratio, is unaffected by either operation: 6/52 = 0.50/4.17 = 11.54%.