5.04a Linear combinations: E(aX+bY), Var(aX+bY)

Ben is a fencing contractor who is often required to repair a garden fence by replacing a broken post between fence panels, as illustrated. \includegraphics{figure_4} The tasks involved are as follows. \(U\): detach the two fence panels from the broken post \(V\): remove the broken post \(W\): insert a new post \(X\): attach the two fence panels to the new post The mean and the standard deviation of the time, in minutes, for each of these tasks are shown in the table.

Task	Mean	Standard deviation
\(U\)	15	5
\(V\)	40	15
\(W\)	75	20
\(X\)	20	10

The random variables \(U\), \(V\), \(W\) and \(X\) are pairwise independent, except for \(V\) and \(W\) for which \(\rho_{VW} = 0.25\).

1. Determine values for the mean and the variance of:
   1. \(R = U + X\);
   2. \(F = V + W\);
   3. \(T = R + F\);
   4. \(D = W - V\).
   [8 marks]
2. Assuming that each of \(R\), \(F\), \(T\) and \(D\) is approximately normally distributed, determine the probability that:
   1. the total time taken by Ben to repair a garden fence is less than 3 hours;
   2. the time taken by Ben to insert a new post is at least 1 hour more than the time taken by him to remove the broken post.
   [5 marks]

Geoffrey is a university lecturer. He has to prepare five questions for an examination. He knows by experience that it takes about 3 hours to prepare a question, and he models the time (in minutes) taken to prepare one by the Normally distributed random variable \(X\) with mean 180 and standard deviation 12, independently for all questions.

1. One morning, Geoffrey has a gap of 2 hours 50 minutes (170 minutes) between other activities. Find the probability that he can prepare a question in this time. [3]
2. One weekend, Geoffrey can devote 14 hours to preparing the complete examination paper. Find the probability that he can prepare all five questions in this time. [3]

A colleague, Helen, has to check the questions.

1. She models the time (in minutes) to check a question by the Normally distributed random variable \(Y\) with mean 50 and standard deviation 6, independently for all questions and independently of \(X\). Find the probability that the total time for Geoffrey to prepare a question and Helen to check it exceeds 4 hours. [3]
2. When working under pressure of deadlines, Helen models the time to check a question in a different way. She uses the Normally distributed random variable \(\frac{1}{2}X\), where \(X\) is as above. Find the length of time, as given by this model, which Helen needs to ensure that, with probability 0.9, she has time to check a question. [4]

Ian, an educational researcher, suggests that a better model for the time taken to prepare a question would be a constant \(k\) representing "thinking time" plus a random variable \(T\) representing the time required to write the question itself, independently for all questions.

1. Taking \(k\) as 45 and \(T\) as Normally distributed with mean 120 and standard deviation 10 (all units are minutes), find the probability according to Ian's model that a question can be prepared in less than 2 hours 30 minutes. [2]

Juliet, an administrator, proposes that the examination should be reduced in time and shorter questions should be used.

1. Juliet suggests that Ian's model should be used for the time taken to prepare such shorter questions but with \(k = 30\) and \(T\) replaced by \(\frac{2}{3}T\). Find the probability as given by this model that a question can be prepared in less than \(1\frac{1}{4}\) hours. [3]

An electronics company purchases two types of resistor from a manufacturer. The resistances of the resistors (in ohms) are known to be Normally distributed. Type A have a mean of 100 ohms and standard deviation of 1.9 ohms. Type B have a mean of 50 ohms and standard deviation of 1.3 ohms.

1. Find the probability that the resistance of a randomly chosen resistor of type A is less than 103 ohms. [3]
2. Three resistors of type A are chosen at random. Find the probability that their total resistance is more than 306 ohms. [3]
3. One resistor of type A and one resistor of type B are chosen at random. Find the probability that their total resistance is more than 147 ohms. [3]
4. Find the probability that the total resistance of two randomly chosen type B resistors is within 3 ohms of one randomly chosen type A resistor. [5]
5. The manufacturer now offers type C resistors which are specified as having a mean resistance of 300 ohms. The resistances of a random sample of 100 resistors from the first batch supplied have sample mean 302.3 ohms and sample standard deviation 3.7 ohms. Find a 95\% confidence interval for the true mean resistance of the resistors in the batch. Hence explain whether the batch appears to be as specified. [4]

The random variable \(X\) has an \(F\)-distribution with 8 and 12 degrees of freedom. Find P\(\left(\frac{1}{5.67} < X < 2.85\right)\). [4]

The random variable \(X\) has a \(\chi^2\)-distribution with 9 degrees of freedom.

1. Find P(2.088 < \(X\) < 19.023). [3]

The random variable \(Y\) follows an \(F\)-distribution with 12 and 5 degrees of freedom.

1. [(b)] Find the upper and lower 5\% critical values for \(Y\). [3]

(Total 6 marks)

A bag contains marbles of which an unknown proportion \(p\) is red. A random sample of \(n\) marbles is drawn, with replacement, from the bag. The number \(X\) of red marbles drawn is noted. A second random sample of \(m\) marbles is drawn, with replacement. The number \(Y\) of red marbles drawn is noted. Given that \(p_1 = \frac{aX}{n} + \frac{bY}{m}\) is an unbiased estimator of \(p_1\),

1. show that \(a + b = 1\). [4]

Given that \(p_2 = \frac{(X + Y)}{n + m}\)

1. [(b)] show that \(p_2\) is an unbiased estimator for \(p\). [3]
2. Show that the variance of \(p_1\) is p(1 - \(p\))\(\left(\frac{a^2}{n} + \frac{b^2}{m}\right)\). [3]
3. Find the variance of \(p_2\). [3]
4. Given that \(a = 0.4\), \(m = 10\) and \(n = 20\), explain which estimator \(p_1\) or \(p_2\) you should use. [4]

(Total 17 marks)

The value of orders, in £, made to a firm over the internet has distribution N(\(\mu, \sigma^2\)). A random sample of \(n\) orders is taken and \(\bar{X}\) denotes the sample mean.

1. Write down the mean and variance of \(\bar{X}\) in terms of \(\mu\) and \(\sigma^2\). [2]

A second sample of \(m\) orders is taken and \(\bar{Y}\) denotes the mean of this sample. An estimator of the population mean is given by $$U = \frac{n\bar{X} + m\bar{Y}}{n + m}$$

1. [(b)] Show that \(U\) is an unbiased estimator for \(\mu\). [3]
2. Show that the variance of \(U\) is \(\frac{\sigma^2}{n + m}\). [4]
3. State which of \(\bar{X}\) or \(U\) is a better estimator for \(\mu\). Give a reason for your answer. [2]

Find the value of the constant \(a\) such that $$\text{P}(a < F_{8,10} < 3.07) = 0.94$$ [2]

The random variable \(X\) has an \(F\) distribution with 10 and 12 degrees of freedom. Find \(a\) and \(b\) such that \(\text{P}(a < X < b) = 0.90\). [3]

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]

The random variable \(X\) has mean 17 and variance 64. The independent random variable \(Y\) has mean 10 and variance 16. Find the value of

1. E\((4Y - 2X + 1)\), [2]
2. Var\((4Y - 5X + 3)\), [2]
3. E\((X^2 Y)\). [3]

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 \(D\) has the distribution Geo\((p)\). It is given that Var\((D) = \frac{40}{9}\). Determine

1. Var\((3D + 5)\). [1]
2. E\((3D + 5)\). [6]
3. \(\text{P}(D > \text{E}(D))\). [3]

The random variable \(X\) is equally likely to take any of the \(n\) integer values from \(m + 1\) to \(m + n\) inclusive. It is given that E\((3X) = 30\) and Var\((3X) = 36\). Determine the value of \(m\) and the value of \(n\). [7]