A small company makes 50 glass ornaments each working day. Some of the ornaments turn out to be faulty.
State, in the context of the question, two assumptions needed for the number of faulty ornaments made in a day to be well modelled by a binomial distribution.[2]
Assume now that the number of faulty ornaments produced each day has the distribution \(\mathrm{B}(50,0.04)\).
Show that the numerical values of the mean and variance of this distribution differ by \(0.08\).[1]
Find the probability that no more than 2 faulty ornaments are produced on a randomly chosen working day.[1]
Find the probability that no more than 2 faulty ornaments are produced on at least 3 days in a randomly chosen 5-day working week. State the distribution you use.[3]
Find the probability that no more than 10 faulty items are produced in a randomly chosen 5-day working week. State the distribution you use.[2]
The company also makes pens which are sold in randomly packed boxes of one hundred pens. The probability of a pen being not faulty is \(p\), where \(0<p<1\).
For quality control purposes, a random sample of pens from each box is tested. Mr Lu and Mrs Ming carry out the tests but they use different methods.
Mr Lu tests a random sample of 6 pens from a box. If there are no faulty pens or only 1 faulty pen the box is accepted.
Mrs Ming tests a random sample of 3 pens from a box.
If there are no faulty pens in her sample the box is accepted.
If there are 2 or 3 faulty pens in her sample the box is rejected.
If there is 1 faulty pen in her sample she takes a second random sample of 3 pens. She accepts the box if there are no faulty pens in this second sample.
Show algebraically that Mrs Ming accepts a greater proportion of boxes than Mr Lu does.[6]