Setting the Fill Mean Under a 5% Underweight Rule
A roastery fills nominal 250 g bags of coffee on an automated line. The filled mass is modelled as normally distributed with standard deviation 4.0 g, and the mean μ is adjustable. Bags under 245 g are classed as underweight and must be refilled at a cost of £0.60 each. Trading standards require that no more than 5% of bags be underweight. The line manager, keen not to give away coffee, proposes μ = 250 g, pointing out that 'the machine is set to the label weight, so on average the customer gets exactly what is paid for'. A sample of 12 consecutive bags is then taken from the line for a check. Separately, the production director asks whether rebuilding the filler to a standard deviation of 2.0 g would allow the mean to be lowered, since the coffee in the fill costs £4.20 per kilogram.
1. With μ = 250 g, find the proportion of underweight bags and rule on the manager's proposal. Then find the mean required to meet the 5% rule.
Read the worked answer
With μ = 250, σ = 4: z = (245 − 250)/4 = −1.25. P(Z < −1.25) = 0.1056, so 10.6% of bags are underweight - more than twice the permitted 5%. The proposal fails the rule.
The manager's reasoning has a specific flaw: setting the mean equal to the label guarantees the average is correct but says nothing about the lower tail. Since the distribution is continuous and symmetric, about half the bags fall below the mean, so a mean of 250 forces a large underweight tail. The requirement is one-sided.
Required μ: we need P(X < 245) = 0.05, so (245 − μ)/4 = −1.6449 (the 5% point of the standard normal).
245 − μ = −6.5796 → μ = 251.6 g (1 d.p.).
The rule forces deliberate overfill of about 1.6 g per bag; that overfill is the cost of the standard deviation, not slack in the process.
2. With μ set correctly, find the probability that at least 2 of the next 12 bags are underweight, and state the assumptions behind your model.
Read the worked answer
Each bag is underweight with probability p = 0.05. Modelling the number X of underweight bags in n = 12 as binomial(n = 12, p = 0.05):
P(X ≥ 2) = 1 − P(0) − P(1)
P(0) = 0.95¹² = 0.5404 P(1) = 12 × 0.05 × 0.95¹¹ = 0.6 × 0.5688 = 0.3413
P(X ≥ 2) = 1 − 0.5404 − 0.3413 = 0.118 (3 s.f.).
Assumptions: (i) each bag's mass is independently drawn from the same normal distribution - constant μ, constant σ; (ii) the bag is 'underweight' with the same probability p for every bag; (iii) bags are independent of one another.
Assumption (iii) is the one to challenge in practice. The 12 bags are consecutive, so they share the same hopper load, the same vibration setting and the same operator. If the filler drifts, underweight bags arrive in runs, and the count is over-dispersed relative to the binomial. In that case 0.118 is not a reliable figure - the true probability of seeing at least two in a row-correlated sample is higher than the binomial value, so 0.118 should be treated as a best case, not a working figure.
3. If the standard deviation were reduced to 2.0 g, find the new required mean and estimate the annual saving on a production of 1 000 000 bags.
Read the worked answer
With σ = 2.0 and the same 5% rule: (245 − μ)/2.0 = −1.6449 → μ = 245 + 3.2898 = 248.3 g.
Reduction in mean fill = 251.58 − 248.29 = 3.29 g per bag.
Cost of that coffee = 3.29 g × (£4.20/1000 g) = 3.29 × 0.0042 = £0.01382 per bag.
Over 1 000 000 bags: 0.01382 × 1 000 000 ≈ £13 800 per year (about £13 820 unrounded).
The underweight cost does not change the comparison, because both settings hold the underweight rate at 5%: 0.05 × £0.60 = £0.03 per bag either way. The saving comes entirely from giving away less coffee while still satisfying the rule.
Judgement for the report: the £13.8k figure is a gross saving and must be set against the capital cost of the rebuild and any requalification trials. It is also sensitive to the normal assumption - if the fill distribution is skewed rather than normal, the 5% point moves and the quoted mean is wrong. Present it as an upper bound on the benefit, contingent on verifying normality with a larger sample than 12.