3 An electrical retailer gives customers extended guarantees on washing machines. Under this guarantee all repairs in the first 3 years are free. The retailer records the numbers of free repairs made to 80 machines.

1. Show that the sample mean is 0.4375 .
2. The sample standard deviation \(s\) is 0.6907 . Explain why this supports a suggestion that a Poisson distribution may be a suitable model for the distribution of the number of free repairs required by a randomly chosen washing machine. The random variable \(X\) denotes the number of free repairs required by a randomly chosen washing machine. For the remainder of this question you should assume that \(X\) may be modelled by a Poisson distribution with mean 0.4375.
3. Find \(\mathrm { P } ( X = 1 )\). Comment on your answer in relation to the data in the table.
4. The manager decides to monitor 8 washing machines sold on one day. Find the probability that there are at least 12 free repairs in total on these 8 machines. You may assume that the 8 machines form an independent random sample.
5. A launderette with 8 washing machines has needed 12 free repairs. Why does your answer to part (iv) suggest that the Poisson model with mean 0.4375 is unlikely to be a suitable model for free repairs on the machines in the launderette? Give a reason why the model may not be appropriate for the launderette. The retailer also sells tumble driers with the same guarantee. The number of free repairs on a tumble drier in three years can be modelled by a Poisson distribution with mean 0.15 . A customer buys a tumble drier and a washing machine.
6. Assuming that free repairs are required independently, find the probability that
   (A) the two appliances need a total of 3 free repairs between them,
   (B) each appliance needs exactly one free repair.