Direct binomial probability calculation

5

1. At a particular checkout in a supermarket, the probability that the barcode reader fails to read the barcode first time on any item is 0.07 , and is independent from item to item.
   1. Calculate the probability that, from a shopping trolley containing 17 items, the reader fails to read the barcode first time on exactly 2 of the items.
   2. Determine the probability that, from a shopping trolley containing 50 items, the reader fails to read the barcode first time on at most 5 of the items.
2. At another checkout in the supermarket, the probability that a faulty barcode reader fails to read the barcode first time on any item is 0.55 , and is independent from item to item. Determine the probability that, from a shopping trolley containing 50 items, this reader fails to read the barcode first time on at least 30 of the items.
3. At a third checkout in the supermarket, a record is kept of \(X\), the number of times per 50 items that the barcode reader fails to read a barcode first time. An analysis of the records gives a mean of 10 and a standard deviation of 6.8.
   1. Estimate \(p\), the probability that the barcode reader fails to read a barcode first time.
   2. Using your estimate of \(p\) and assuming that \(X\) can be modelled by a binomial distribution, estimate the standard deviation of \(X\).
   3. Hence comment on the assumption that \(X\) can be modelled by a binomial distribution.

3 The lengths of snakes on a tropical island were measured and found to be normally distributed with a mean of 160 cm and a standard deviation of 6 cm . Find the probability that a randomly selected snake has a length of less than 170 cm .

It is known that on average 85% of seeds of a particular variety of tomato will germinate. Ramesh selects 15 of these seeds at random and sows them.

1. 1. Find the probability that exactly 12 germinate. [3]
   2. Find the probability that fewer than 12 germinate. [2]

The following year Ramesh finds that he still has many seeds left. Because the seeds are now one year old, he suspects that the germination rate will be lower. He conducts a trial by randomly selecting \(n\) of these seeds and sowing them. He then carries out a hypothesis test at the 1% significance level to investigate whether he is correct.

1. Write down suitable null and alternative hypotheses for the test. Give a reason for your choice of alternative hypothesis. [4]
2. In a trial with \(n = 20\), Ramesh finds that 13 seeds germinate. Carry out the test. [4]
3. Suppose instead that Ramesh conducts the trial with \(n = 50\), and finds that 33 seeds germinate. Given that the critical value for the test in this case is 35, complete the test. [3]
4. If \(n\) is small, there is no point in carrying out the test at the 1% significance level, as the null hypothesis cannot be rejected however many seeds germinate. Find the least value of \(n\) for which the null hypothesis can be rejected, quoting appropriate probabilities to justify your answer. [3]

A survey has found that, of the 2400 households in Growmore, 1680 eat home-grown fruit and vegetables.

1. Using the binomial distribution, find the probability that, out of a random sample of 25 households in Growmore, exactly 22 eat home-grown fruit and vegetables. [2 marks]
2. Give a reason why you would not expect your calculation in part (a) to be valid for the 25 households in Gifford Terrace, a residential road in Growmore. [1 mark]

In a particular country, 8% of the population has blue eyes. A random sample of 20 people is selected from this population. Find the probability that exactly two of these people have blue eyes. [2]

The random variable \(X\) has the binomial distribution B(16, 0·3). Showing your calculation, find \(P(X = 7)\). [2]