Café wait-time model against a complaint log
A café owner has 300 timed orders from last month and models the wait between ordering and receiving a drink as normally distributed with mean 4.2 minutes and standard deviation 1.1 minutes. The owner wants to advertise that '90% of drinks are served within five minutes' and is also considering a five-minute guarantee that refunds the price of any drink that takes longer. A drink sells for an average of £3.40 and the café sells about 620 drinks a week. Meanwhile the complaints log shows that 47 of the last 400 orders took longer than five minutes - a period that includes one week when the espresso machine was failing. Staff say the model is 'too pessimistic'; the owner says the log is 'unreliable'. Both want a number they can defend to the accountant, and neither wants to publish a claim the data cannot support.
1. Using the model, find the probability that a drink takes longer than five minutes and test the proposed advertisement.
Read the worked answer
Let X be the wait, X ~ N(4.2, 1.1²).
Standardise: z = (5 − 4.2)/1.1 = 0.8/1.1 = 0.727
P(X > 5) = P(Z > 0.727) = 1 − 0.7665 = 0.2335
So about 23% of drinks take longer than five minutes, not 10%. The advertisement is false as written: the true 'within five minutes' rate is about 76.6%, not 90%.
The mistake to avoid is treating 0.2335 as the answer to the advertisement question - the claim is about the complement, and 76.6% falls well short of 90%.
2. Find the time within which 90% of drinks are served, and state what would have to change for the five-minute claim to be true.
Read the worked answer
We need t with P(X ≤ t) = 0.90. The inverse normal value is z = 1.2816.
t = μ + zσ = 4.2 + 1.2816 × 1.1 = 4.2 + 1.410 = 5.61 minutes
So the honest version of the claim is '90% of drinks within 5.6 minutes'. Rounding down to 5.5 or 5.0 would overstate performance again.
For the five-minute claim to be true with the same spread, the mean would have to fall: with σ = 1.1 and P(X ≤ 5) = 0.90, μ = 5 − 1.2816 × 1.1 = 5 − 1.410 = 3.59 minutes.
That is a 0.6-minute cut in average service time - a staffing or workflow change, not a wording change. This is the useful answer for the owner: the claim is attainable, but only with a genuine operational improvement.
3. Estimate the weekly cost of the five-minute guarantee under the model and under the complaint log, and advise the owner which figure to budget with.
Read the worked answer
Model: expected refunds = 0.2335 × 620 = 144.8 drinks per week; cost = 144.8 × £3.40 = £492 per week (about £2,130 per four-week month).
Log: 47/400 = 0.1175, so 0.1175 × 620 = 72.9 drinks; cost = 72.9 × £3.40 = £248 per week.
The two differ by a factor of about two, which is a signal rather than a nuisance: a gap that large suggests the process was not stable during the logged month (consistent with the failing machine). The log is direct evidence but contaminated; the model rests on 300 timings but assumes normality and a steady process.
Advice: re-time a clean week before publishing anything, and budget provisionally at the model figure of about £490 per week, because refunds are a cash commitment and under-budgeting is the costlier error. Do not publish the 90%-within-five claim on either data set.