N2020 P2 Q9

N2020 P2 Q9

Junior College 2
12 marks
Free

A factory produces ballpoint pens. On average \(6\%\) of the pens are faulty. The pens are packed in boxes of \(100\) for sale to retail outlets. It should be assumed that the number of faulty pens in a box of \(100\) pens follows a binomial distribution.

For quality control purposes a random sample of \(10\) pens from each box is tested. If \(2\) or fewer faulty pens are found in the sample of \(10\), the box is accepted for sale. Otherwise the box is rejected.

  1. Explain what is meant by a random sample in this context.[1]
  2. Find the probability that a randomly chosen box of 100 pens is accepted for sale.[1]
  3. One morning 75 boxes are tested in this way. Find the probability that more than 5% of these boxes are rejected.[4]

An alternative testing procedure is trialled in which a random sample of 5 pens is initially taken from a box and tested.

  • If there are no faulty pens in this sample of 5 the box is accepted.
  • If there are 3 or more faulty pens in this sample of 5 the box is rejected.
  • If there are 1 or 2 faulty pens in this sample a second random sample of 5 pens is taken from the box. When the second sample has been tested, the box is accepted if the total number of faulty pens found in the combined sample of 10 is 2 or fewer and rejected otherwise.
  1. Find the probability that a randomly chosen box of 100 pens is accepted for sale when the alternative testing procedure is used.[5]
  2. Explain why the factory manager might prefer to use the alternative testing procedure.[1]

Default solution

(i) Every set of \(10\) pens from a box of \(100\) has an equal chance of selection.

(ii) With \(X\sim\mathrm{Bin}(10,0.06)\), \(P(\text{accepted})=P(X\le2)=0.981\).

(iii) The rejection probability is \(q=1-0.9811621635=0.0188378365\). For \(Y\sim\mathrm{Bin}(75,q)\), more than \(5\%\) means at least \(4\) rejections, so \(P(Y\ge4)=0.0535\).

(iv) Let \(X_1,X_2\sim\mathrm{Bin}(5,0.06)\) for the two batches. Acceptance has probability \(P(X_1=0)+P(X_1=1)P(X_2\le1)+P(X_1=2)P(X_2=0)=0.983\).

(v) The alternative procedure often needs only \(5\) tests, saving testing time and cost while maintaining a similar acceptance probability.

Answer:(ii) \(0.981\); (iii) \(0.0535\); (iv) \(0.983\)

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