Helpline staffing standard under Poisson arrivals and a binomial retraining rule
A city bank runs a chat helpline. Queries arrive at random and independently at an average of 2.5 per five-minute window in peak hours, and a Poisson model is considered appropriate. One agent can comfortably handle five queries in a window, and the bank's service standard states that the probability of six or more queries in a window must not exceed 5%; otherwise a second agent is rostered. Separately, each query handled by a new recruit has, independently, probability 0.06 of being escalated to a supervisor, and a recruit who produces two or more escalations within a shift of 20 queries is sent for retraining. For the next quarter, average traffic is expected to be 20% higher than the 2.5 figure. Management wants to know whether the single-agent roster still meets its own standard, how often recruits will be sent for retraining, and what the traffic growth does to both answers. Use e^(−2.5) = 0.0821 and e^(−3.0) = 0.0498.
1. Test the service standard at the current arrival rate of 2.5 queries per window. State the probability and whether a second agent is required.
Read the worked answer
Let X ~ Poisson(λ = 2.5). Terms P(x) = e^(−2.5)·2.5^x / x!:
| x | P(x) |
|---|---|
| 0 | 0.0821 |
| 1 | 0.2053 |
| 2 | 0.2566 |
| 3 | 0.2138 |
| 4 | 0.1336 |
| 5 | 0.0668 |
Sum P(0) to P(5) = 0.9582, so P(X ≥ 6) = 1 − 0.9582 = 0.0418 ≈ 4.2%.
4.2% ≤ 5%, so the single-agent roster just meets the standard. The margin is only 0.8 percentage points - roughly one window in 24 breaches the five-query comfort level. That is thin enough that any small rise in traffic, or any bunching of arrivals, breaks the standard.
2. For a recruit handling 20 queries with escalation probability 0.06, find the probability of being sent for retraining (two or more escalations), and the mean and variance of the number of escalations. Comment on whether the rule is well calibrated.
Read the worked answer
Let Y ~ B(n = 20, p = 0.06).
P(0) = 0.94^20 = 0.2901. P(1) = C(20,1)(0.06)(0.94^19) = 20(0.06)(0.3086) = 0.3703. P(Y ≥ 2) = 1 − 0.2901 − 0.3703 = 0.3396 ≈ 34%.
Mean = np = 20(0.06) = 1.2; Variance = np(1 − p) = 20(0.06)(0.94) = 1.128; SD ≈ 1.06.
Calibration. A recruit whose true escalation rate is exactly the acceptable 6% is flagged about one time in three. The trigger (2) sits barely above the mean (1.2) - less than one standard deviation away - so the rule mostly measures luck, not skill. If retraining is costly, raise the trigger to 3: P(Y ≥ 3) = 0.3396 − P(2), where P(2) = C(20,2)(0.06²)(0.94^18) = 190(0.0036)(0.3283) = 0.2246, giving P(Y ≥ 3) ≈ 0.115, about 11.5%. At that threshold a genuinely weak recruit (say p = 0.15, mean 3) is still caught, while good recruits are rarely penalised.
3. Traffic rises 20%. Re-test the service standard at the new arrival rate and state what the bank should do, including any assumption that the growth exposes.
Read the worked answer
New λ = 2.5 × 1.2 = 3.0. With e^(−3) = 0.0498:
P(0) = 0.0498, P(1) = 0.1494, P(2) = 0.2241, P(3) = 0.2241, P(4) = 0.1681, P(5) = 0.1008.
Sum = 0.9163, so P(X ≥ 6) = 0.0837 ≈ 8.4%, above the 5% ceiling.
Decision. The single-agent roster fails its own standard once traffic rises 20%. Two options: roster a second agent at peak, or formally re-open the standard (e.g. allow a seven-query comfort level, where P(X ≥ 8) = 1 − 0.9881 = 0.0119, comfortably under 5%). Quietly keeping one agent and hoping is not defensible.
Assumption exposed. The Poisson model assumes a constant rate and independent arrivals. Real peak traffic is bursty - a marketing push or a system outage clusters queries - which produces over-dispersion (variance greater than the mean). If that happens, the true tail probability exceeds the 8.4% calculated here, so the Poisson figure is a lower bound on the risk, not a safe estimate.