Outlet Consistency Award: Equal Means, Different Spread
A retail chain compares two outlets, P and Q, over five working days. The table gives daily sales in hundreds of rupees.
| Day | Outlet P | Outlet Q | | Mon | 120 | 160 | | Tue | 180 | 150 | | Wed | 150 | 170 | | Thu | 210 | 140 | | Fri | 140 | 180 |
Both outlets have the same total sales and the same mean for the week. The regional manager must decide which outlet to recommend for a consistency award, defined as the outlet whose daily sales are less dispersed around its own mean. A supervisor argues that because the totals are equal, the award should be shared. Another argues that the outlet with the higher median should win. The manager wants the coefficient of variation used as the tie-breaker because it is scale-free, and also wants to know how a sixth day with sales of 300 for P and 0 for Q would change the mean and range. Treat the five days as the complete population for dispersion calculations.
1. Compute the mean, median and range for each outlet over the five days.
Read the worked answer
Outlet P: 120, 180, 150, 210, 140. Sum = 800, so mean = 800 ÷ 5 = 160. Sorted values: 120, 140, 150, 180, 210; median = 150. Range = 210 − 120 = 90.
Outlet Q: 160, 150, 170, 140, 180. Sum = 800, so mean = 800 ÷ 5 = 160. Sorted values: 140, 150, 160, 170, 180; median = 160. Range = 180 − 140 = 40.
The means are equal, but the medians and ranges differ. Q's daily sales are concentrated closer to 160, while P has a wider spread from 120 to 210. Range alone suggests Q is more consistent, but the manager wants a dispersion measure that uses every observation.
2. Compute the population standard deviation and coefficient of variation for each outlet. Which outlet should receive the consistency award?
Read the worked answer
For P, deviations from 160 are −40, 20, −10, 50, −20. Squares: 1,600; 400; 100; 2,500; 400. Sum = 5,000. Population variance = 5,000 ÷ 5 = 1,000, so SD = √1,000 ≈ 31.62. CV = (31.62 ÷ 160) × 100 ≈ 19.76%.
For Q, deviations from 160 are 0, −10, 10, −20, 20. Squares: 0; 100; 100; 400; 400. Sum = 1,000. Population variance = 1,000 ÷ 5 = 200, so SD = √200 ≈ 14.14. CV = (14.14 ÷ 160) × 100 ≈ 8.84%.
Q has the lower SD and the lower CV, so Q should receive the consistency award. The supervisor's 'equal totals' argument is invalid: equal means can coexist with very different spreads. The higher-median argument also points to Q, but the median alone does not measure dispersion; it happens to agree here and should not replace the SD/CV comparison.
3. Add a sixth day with sales 300 for P and 0 for Q. Recompute each mean and range. Does the consistency conclusion still hold on the range measure?
Read the worked answer
New P sum = 800 + 300 = 1,100. Mean = 1,100 ÷ 6 ≈ 183.33. New P minimum = 120, maximum = 300, so range = 300 − 120 = 180.
New Q sum = 800 + 0 = 800. Mean = 800 ÷ 6 ≈ 133.33. New Q minimum = 0, maximum = 180, so range = 180 − 0 = 180.
On the range measure alone, the two outlets are now tied at 180. The means have also moved in opposite directions: P's mean rises, Q's falls. A single extreme day has changed the range drastically, which shows why range is a fragile measure of consistency. The original SD/CV comparison used all five days and found Q more consistent; if the sixth day is treated as part of the population, the dispersion should be recomputed rather than relying on the tied range.