Underfill inspection and a normal approximation
A pharmaceutical filling line produces vials of sterile solution. Historically 2.0% of vials are underfilled. Vials are packed in trays of 60, and the quality rule is: reject a tray if it contains more than 3 underfilled vials. Assume underfills occur independently and the rate remains constant within a production run. The line supervisor claims that 'under this rule, fewer than 5% of trays will be rejected'. A new pump is being tested; during the trial the underfill rate rises to 6.0%, and a larger tray of 150 vials is used. Management wants an approximate probability that such a tray has 12 or more underfilled vials before deciding whether to stop the trial. The approximate calculation should use a normal approximation with continuity correction, and its validity should be checked against the usual rule of thumb. Do not use a calculator's exact binomial function for the approximate part.
1. Calculate the probability that a tray of 60 is rejected under the 2.0% underfill rate, and assess the supervisor's claim.
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Let X be the number of underfilled vials in a tray of 60. Under the stated assumptions, X ~ Bin(60, 0.02). Rejection is X ≥ 4, so P(X ≥ 4) = 1 − [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]. P(X = 0) = 0.98^60 ≈ 0.2976. Using the recurrence P(k + 1)/P(k) = (60 − k)/(k + 1), we get P(X = 1) ≈ 0.3644, P(X = 2) ≈ 0.2194, and P(X = 3) ≈ 0.0866. Their sum is about 0.9680, so P(reject) ≈ 0.0320 = 3.20%. This is below 5%, so the supervisor's claim is supported under the independence and constant-rate assumptions.
2. Use a normal approximation with continuity correction to estimate P(X ≥ 12) when n = 150 and p = 0.06, and check the approximation's validity.
Read the worked answer
Let Y ~ Bin(150, 0.06). The mean is np = 9.00 and the variance is np(1 − p) = 150(0.06)(0.94) = 8.46, so the standard deviation is about 2.909. With continuity correction, P(Y ≥ 12) ≈ P(Z > (11.5 − 9)/2.909) = P(Z > 0.859) ≈ 0.195. For validity, np = 9 and n(1 − p) = 141; both are at least 5, although np = 9 is close to the stricter threshold of 10, so the approximation is rough. The estimated probability is about 19.5%, not negligible; if management's stop rule is 15%, the trial would trigger a stop. An exact binomial calculation would be preferable for a final decision.