5.02b Expectation and variance: discrete random variables

Members of a library may borrow up to 6 books. Past experience has shown that the number of books borrowed, \(X\), follows the distribution shown in the table.

\(x\)	0	1	2	3	4	5	6
P(X = x)	0	0.19	0.26	0.20	0.13	0.07	0.15

1. Find the probability that a member borrows more than 3 books. [1 mark]
2. Assume that the numbers of books borrowed by two particular members are independent. Find the probability that one of these members borrows more than 3 books and the other borrows fewer than 3 books. [3 marks]
3. Show that the mean of \(X\) is 3.08, and calculate the variance of \(X\). [4 marks]
4. One of the library staff notices that the values of the mean and the variance of \(X\) are similar and suggests that a Poisson distribution could be used to model \(X\). Without further calculations, give two reasons why a Poisson distribution would not be suitable to model \(X\). [2 marks]
5. The library introduces a fee of 10 pence for each book borrowed. Assuming that the probabilities do not change, calculate:
   1. the mean amount that will be paid by a member;
   2. the standard deviation of the amount that will be paid by a member.
   [3 marks]

When a park is redeveloped, it is claimed that 70\% of the local population approve of the new design. Assuming this to be true, find the probability that, in a group of 10 residents selected at random,

1. 6 or more approve, [3 marks]
2. exactly 7 approve. [3 marks]

A conservation group, however, carries out a survey of 20 people, and finds that only 9 approve.

1. Use this information to carry out a hypothesis test on the original claim, working at the 5\% significance level. State your conclusion clearly. [5 marks]

If the conservationists are right, and only 45\% approve of the new park,

1. use a suitable approximation to the binomial distribution to estimate the probability that in a larger survey, of 500 people, less than half will approve. [7 marks]

In an orchard, all the trees are either apple or pear trees. There are four times as many apple trees as pear trees. Find the probability that, in a random sample of 10 trees, there are

1. equal numbers of apple and pear trees, [3 marks]
2. more than 7 apple trees. [3 marks]

In a sample of 60 trees in the orchard,

1. find the expected number of pear trees. [1 mark]
2. Calculate the standard deviation of the number of pear trees and compare this result with the standard deviation of the number of apple trees. [2 marks]
3. Find the probability that exactly 35 in the sample of 60 trees are pear trees. [4 marks]
4. Find an approximate value for the probability that more than 15 of the 60 trees are pear trees. [5 marks]

1. Briefly describe the main features of a binomial distribution. [2 marks]

I conduct an experiment by randomly selecting 10 cards, without replacement, from a normal pack of 52.

1. Explain why the distribution of \(X\), the number of hearts obtained, is not \(\text{B}(10, \frac{1}{4})\). [2 marks]

After making the appropriate adjustment to the experiment, which should be stated, so that the distribution is \(\text{B}(10, \frac{1}{4})\), find

1. the probability of getting no hearts, [3 marks]
2. the probability of getting 4 or more hearts. [2 marks]
3. If the modified experiment is repeated 50 times, find the total number of hearts that you would you expect to have drawn. [2 marks]

A shop receives weekly deliveries of 120 eggs from a local farm. The proportion of eggs received from the farm that are broken is 0.008

1. Explain why it is reasonable to use the binomial distribution to model the number of eggs that are broken in each delivery. [3 marks]
2. Use the binomial distribution to calculate the probability that at most one egg in a delivery will be broken. [4 marks]
3. State the conditions under which the binomial distribution can be approximated by the Poisson distribution. [1 mark]
4. Using the Poisson approximation to the binomial, find the probability that at most one egg in a delivery will be broken. Comment on your answer. [5 marks]

An advert for Tatty's Crisps claims that 1 in 10 bags contain a free scratchcard game. Tatty's Crisps can be bought in a Family Pack containing 10 bags. Find the probability that the bags in one of these Family Packs contain

1. no scratchcards, [2]
2. more than 2 scratchcards. [2]

Tatty's Crisps can also be bought wholesale in boxes containing 50 bags. A pub Landlord notices that her customers only found 2 scratchcards in the crisps from one of these boxes.

1. Stating your hypotheses clearly, test at the 10\% level of significance whether or not this gives evidence of there being fewer free scratchcards than is claimed by the advert. [4]

A bag contains 40 beads of the same shape and size. The ratio of red to green to blue beads is \(1 : 3 : 4\) and there are no beads of any other colour. In an experiment, a bead is picked at random, its colour noted and the bead replaced in the bag. This is done ten times.

1. Suggest a suitable distribution for modelling the number of times a blue bead is picked out and give the value of any parameters needed. [2]
2. Explain why this distribution would not be suitable if the beads were not replaced in the bag. [1]
3. Find the probability that of the ten beads picked out
   1. five are blue,
   2. at least one is red. [6]

The experiment is repeated, but this time a bead is picked out and replaced \(n\) times.

1. Find in the form \(a^n < b\), where \(a\) and \(b\) are exact fractions, the condition which \(n\) must satisfy in order to have at least a 99\% chance of picking out at least one red bead. [3]

A researcher collects data on the height of boys aged between nine and nine and-a-half years and their diet. The data on the height, \(V\) cm, of the 80 boys who had always eaten a vegetarian diet is summarised by $$\Sigma V = 10\,367, \quad \Sigma V^2 = 1\,350\,314.$$

1. Calculate unbiased estimates of the mean and variance of \(V\). [5]

The researcher calculates unbiased estimates of the mean and variance of the height of boys whose diet has included meat from a sample of size 280, giving values of 130.5 cm and 96.24 cm\(^2\) respectively.

1. Stating your hypotheses clearly, test at the 1% level whether or not there is a significant difference in the heights of boys of this age according to whether or not they have a vegetarian diet. [8]

The discrete random variable \(X\) has the probability distribution given below.

1. Find the mean of \(X\) in terms of \(k\). [2 marks]
2. Find the bias in using \((2\overline{X} - 5)\) as an estimator of \(k\). [3 marks]

Fifty observations of \(X\) were made giving a sample mean of 8.34 correct to 3 significant figures.

1. Calculate an unbiased estimate of \(k\). [2 marks]

A physicist believes that the number of particles emitted by a radioactive source with a long half-life can be modelled by a Poisson distribution. She records the number of particles emitted in 80 successive 5-minute periods and her results are shown in the table below.

1. Comment on the suitability of a Poisson distribution for this situation. [3 marks]
2. Show that an unbiased estimate of the mean number of particles emitted in a 5-minute period is 1.2 and find an unbiased estimate of the variance. [5 marks]
3. Explain how your answers to part (b) support the fitting of a Poisson distribution. [1 mark]
4. Stating your hypotheses clearly and using a 5\% level of significance, test whether or not these data can be modelled by a Poisson distribution. [11 marks]

A continuous uniform distribution on the interval \([0, k]\) has mean \(\frac{k}{2}\) and variance \(\frac{k^2}{12}\). A random sample of three independent variables \(X_1\), \(X_2\) and \(X_3\) is taken from this distribution.

1. Show that \(\frac{2}{3}X_1 + \frac{1}{2}X_2 + \frac{5}{6}X_3\) is an unbiased estimator for \(k\). [3]

An unbiased estimator for \(k\) is given by \(\hat{k} = aX_1 + bX_2\) where \(a\) and \(b\) are constants.

1. Show that Var(\(\hat{k}\)) = \((a^2 - 2a + 2) \frac{k^2}{6}\) [6]
2. Hence determine the value of \(a\) and the value of \(b\) for which \(\hat{k}\) has minimum variance, and calculate this minimum variance. [6]

Faults occur in a roll of material at a rate of \(\lambda\) per m\(^2\). To estimate \(\lambda\), three pieces of material of sizes 3 m\(^2\), 7 m\(^2\) and 10 m\(^2\) are selected and the number of faults \(X_1\), \(X_2\) and \(X_3\) respectively are recorded. The estimator \(\hat{\lambda}\), where $$\hat{\lambda} = k(X_1 + X_2 + X_3)$$ is an unbiased estimator of \(\lambda\).

1. Write down the distributions of \(X_1\), \(X_2\) and \(X_3\) and find the value of \(k\). [4]
2. Find Var(\(\hat{\lambda}\)). [3]

A random sample of \(n\) pieces of this material, each of size 4 m\(^2\), was taken. The number of faults on each piece, \(Y\), was recorded.

1. Show that \(\frac{1}{4}\bar{Y}\) is an unbiased estimator of \(\lambda\). [2]
2. Find Var(\(\frac{1}{4}\bar{Y}\)). [3]
3. Find the minimum value of \(n\) for which \(\frac{1}{4}\bar{Y}\) becomes a better estimator of \(\lambda\) than \(\hat{\lambda}\). [2]

A technician is trying to estimate the area \(\mu^2\) of a metal square. The independent random variables \(X_1\) and \(X_2\) are each distributed \(\text{N}(\mu, \sigma^2)\) and represent two measurements of the sides of the square. Two estimators of the area, \(A_1\) and \(A_2\), are proposed where $$A_1 = X_1X_2 \text{ and } A_2 = \left(\frac{X_1 + X_2}{2}\right)^2.$$ [You may assume that if \(X_1\) and \(X_2\) are independent random variables then $$\text{E}(X_1X_2) = \text{E}(X_1)\text{E}(X_2)$$]

1. Find \(\text{E}(A_1)\) and show that \(\text{E}(A_2) = \mu^2 + \frac{\sigma^2}{2}\). [4]
2. Find the bias of each of these estimators. [2]

The technician is told that \(\text{Var}(A_1) = \sigma^4 + 2\mu^2\sigma^2\) and \(\text{Var}(A_2) = \frac{1}{2}\sigma^4 + 2\mu^2\sigma^2\). The technician decided to use \(A_1\) as the estimator for \(\mu^2\).

1. Suggest a possible reason for this decision. [1]

A statistician suggests taking a random sample of \(n\) measurements of sides of the square and finding the mean \(\overline{X}\). He knows that \(\text{E}(\overline{X}^2) = \mu^2 + \frac{\sigma^2}{n}\) and $$\text{Var}(\overline{X}^2) = \frac{2\sigma^4}{n^2} + \frac{4\sigma^2\mu^2}{n}.$$

1. Explain whether or not \(\overline{X}^2\) is a consistent estimator of \(\mu^2\). [3]

A random sample of three independent variables \(X_1\), \(X_2\) and \(X_3\) is taken from a distribution with mean \(\mu\) and variance \(\sigma^2\).

1. Show that \(\frac{2}{5}X_1 - \frac{1}{5}X_2 + \frac{4}{5}X_3\) is an unbiased estimator for \(\mu\). [3]

An unbiased estimator for \(\mu\) is given by \(\hat{\mu} = aX_1 + bX_2\) where \(a\) and \(b\) are constants.

1. Show that Var(\(\hat{\mu}\)) = \((2a^2 - 2a + 1)\sigma^2\). [6]
2. Hence determine the value of \(a\) and the value of \(b\) for which \(\hat{\mu}\) has minimum variance. [5]

When a tree seed is planted the probability of it germinating is \(p\). A random sample of size \(n\) is taken and the number of tree seeds, \(X\), which germinate is recorded.

1. 1. Show that \(\hat{p}_1 = \frac{X}{n}\) is an unbiased estimator of \(p\).
   2. Find the variance of \(\hat{p}_1\). [4]
   A second sample of size \(m\) is taken and the number of tree seeds, \(Y\), which germinate is recorded. Given that \(\hat{p}_2 = \frac{Y}{m}\) and that \(\hat{p}_3 = a(3\hat{p}_1 + 2\hat{p}_2)\) is an unbiased estimator of \(p\),
2. show that
   1. \(a = \frac{1}{5}\),
   2. \(\text{Var}(\hat{p}_3) = \frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right)\). [6]
3. Find the range of values of \(\frac{n}{m}\) for which $$\text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_1) \text{ and } \text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_2)$$ [3]
4. Given that \(n = 20\) and \(m = 60\), explain which of \(\hat{p}_1\), \(\hat{p}_2\) or \(\hat{p}_3\) is the best estimator. [3]

The discrete random variable \(X\) has the following probability distribution.

1. Find P\((X \leq 6)\) [1 mark]
2. Let \(Y = 3X + 2\) Show that Var\((Y) = 32.49\) [5 marks]
3. The continuous random variable \(T\) is independent of \(Y\). Given that Var\((T) = 5\), find Var\((T + Y)\) [1 mark]

The probability distribution of a discrete random variable \(W\) is given in the table.

\(w\)	0	1	2	3
\(\mathrm{P}(W = w)\)	0.19	0.18	\(x\)	\(y\)

Given that \(\mathrm{E}(W) = 1.61\), find the value of \(\mathrm{Var}(3W + 2)\). [7]

The number of televisions of a particular model sold per week at a retail store can be modelled by a random variable \(X\) with the probability function shown in the table.

\(x\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X = x)\)	\(0.05\)	\(0.2\)	\(0.5\)	\(0.2\)	\(0.05\)

1. 1. Explain why \(\text{E}(X) = 2\). [1]
   2. Find \(\text{Var}(X)\). [3]
2. The profit, measured in pounds made in a week, on the sales of this model of television is given by \(Y\), where \(Y = 250X - 80\). Find

The remote controls for the televisions are quality tested by the manufacturer to see how long they last before they fail.

1. Explain why it would be inappropriate to test all the remote controls in this way. [1]
2. State an advantage of using random sampling in this context. [1]

A website awards a random number of loyalty points each time a shopper buys from it. The shopper gets a whole number of points between \(0\) and \(10\) (inclusive). Each possibility is equally likely, each time the shopper buys from the website. Awards of points are independent of each other.

1. Let \(X\) be the number of points gained after shopping once. Find
2. Let \(Y\) be the number of points gained after shopping twice. Find
3. Find the probability of the most likely number of points gained after shopping twice. Justify your answer. [4]

A fair coin has \(+1\) written on the heads side and \(-1\) on the tails side. The coin is tossed \(100\) times. The sum of the numbers showing on the \(100\) tosses is the random variable \(Y\). Show that the variance of \(Y\) is \(100\). [4]

The random variable \(X\) has the binomial distribution B(12, 0·3). The independent random variable \(Y\) has the Poisson distribution Po(4). Find

1. \(E(XY)\), [2]
2. Var\((XY)\). [6]

A game at a school fete is played with a fair coin and a random number generator which generates random integers between 1 and 52 inclusive. It costs 50 pence to play the game. First, the player tosses the coin. If it lands on tails, the player loses. If it lands on heads, the player is allowed to generate a random number. If the number is 1, the player wins £5. If the number is between 2 and 13 inclusive, the player wins £1. If the number is greater than 13, the player loses.

1. Find the probability distribution of the player's profit. [5]
2. Find the mean and standard deviation of the player's profit. [4]
3. Given that 200 people play the game, calculate
   1. the expected number of players who win some money,
   2. the expected profit for the fete. [2]

The random variable \(X\) has mean14 and standard deviation 5. The independent random variable \(Y\) has mean 12 and standard deviation 3. The random variable \(W\) is given by \(W = XY\). Find the value of

1. E(W), [1]
2. Var(W). [6]

The random variable \(X\) has probability density function $$f(x) = 1 + \frac{3\lambda x}{2} \quad \text{for } -\frac{1}{2} \leqslant x \leqslant \frac{1}{2},$$ $$f(x) = 0 \quad \text{otherwise,}$$ where \(\lambda\) is an unknown parameter such that \(-1 \leqslant \lambda \leqslant 1\).

1. 1. Find E\((X)\) in terms of \(\lambda\).
   2. Show that \(\text{Var}(X) = \frac{16 - 3\lambda^2}{192}\). [6]
2. Show that P\((X > 0) = \frac{8 + 3\lambda}{16}\). [2]

In order to estimate \(\lambda\), \(n\) independent observations of \(X\) are made. The number of positive observations obtained is denoted by \(Y\) and the sample mean is denoted by \(\overline{X}\).

1. 1. Identify the distribution of \(Y\).
   2. Show that \(T_1\) is an unbiased estimator for \(\lambda\), where $$T_1 = \frac{16Y}{3n} - \frac{8}{3}.$$ [4]
2. 1. Show that \(\text{Var}(T_1) = \frac{64 - 9\lambda^2}{9n}\).
   2. Given that \(T_2\) is also an unbiased estimator for \(\lambda\), where $$T_2 = 8\overline{X},$$ find an expression for Var\((T_2)\) in terms of \(\lambda\) and \(n\).
   3. Hence, giving a reason, determine which is the better estimator, \(T_1\) or \(T_2\). [6]

The discrete random variable \(X\) has the following probability distribution, where \(\theta\) is an unknown parameter belonging to the interval \(\left(0, \frac{1}{3}\right)\).

1. Obtain an expression for \(E(X)\) in terms of \(\theta\) and show that $$\text{Var}(X) = 4\theta(3 - \theta).$$ [4] In order to estimate the value of \(\theta\), a random sample of \(n\) observations on \(X\) was obtained and \(\bar{X}\) denotes the sample mean.
2. 1. Show that $$V = \frac{\bar{X} - 3}{2}$$ is an unbiased estimator for \(\theta\).
   2. Find an expression for the variance of \(V\). [4]
3. Let \(Y\) denote the number of observations in the random sample that are equal to 1. Show that $$W = \frac{Y}{n}$$ is an unbiased estimator for \(\theta\) and find an expression for \(\text{Var}(W)\). [5]
4. Determine which of \(V\) and \(W\) is the better estimator, explaining your method clearly. [4]