## Question

The random variable *Y* is such that \({\text{E}}(2Y + 3) = 6{\text{ and Var}}(2 – 3Y) = 11\).

Calculate

(i) E(*Y*) ;

(ii) \({\text{Var}}(Y)\) ;

(iii) \({\text{E}}({Y^2})\) .

Independent random variables *R* and *S* are such that

\[R \sim {\text{N}}(5,{\text{ 1}}){\text{ and }}S \sim {\text{N(8, 2).}}\]

The random variable *V* is defined by *V* = 3*S* – 4*R*.

Calculate P(*V* > 5).

**Answer/Explanation**

## Markscheme

(i) \({\text{E}}(2Y + 3) = 6\)

\(2{\text{E}}(Y) + 3 = 6\) *M1*

\({\text{E}}(Y) = \frac{3}{2}\) *A1*

* *

(ii) \({\text{Var}}(2 – 3Y) = 11\)

\({\text{Var}}( – 3Y) = 11\) *(M1)*

\(9{\text{Var}}(Y) = 11\)

\({\text{Var}}(Y) = \frac{{11}}{9}\) *A1*

* *

(iii) \({\text{E}}({Y^2}) = {\text{Var}}(Y) + {\left[ {{\text{E}}(Y)} \right]^2}\) *M1*

\( = \frac{{11}}{9} + \frac{9}{4}\)

\( = \frac{{125}}{{36}}\) *A1 N0*

*[6 marks]*

E(*V*) = E(3*S – *4*R*)

= 3E(*S*) – 4E(*R*) *M1*

= 24 – 20 = 4 *A1*

Var(3*S – *4*R*) = 9Var(*S*) + 16Var(*R*) , since *R *and *S *are independent random variables *M1*

=18 + 16 = 34 *A1*

\(V \sim {\text{N}}(4,{\text{ 34}})\)

\({\text{P}}(V > 5) = 0.432\) *A2 N0*

*[6 marks]*

## Examiners report

*E*(*Y*) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. *V* was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 – P(V \leqslant 5)\).

*E*(*Y*) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. *V* was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 – P(V \leqslant 5)\).

## Question

(a) A random variable, *X* , has probability density function defined by

\[f(x) = \left\{ {\begin{array}{*{20}{l}}

{100,}&{{\text{for }} – 0.005 \leqslant x < 0.005} \\

{0,}&{{\text{otherwise}}{\text{.}}}

\end{array}} \right.\]

Determine *E*(*X*) and Var(*X*) .

(b) When a real number is rounded to two decimal places, an error is made.

Show that this error can be modelled by the random variable *X* .

(c) A list contains 20 real numbers, each of which has been given to two decimal places. The numbers are then added together.

(i) Write down bounds for the resulting error in this sum.

(ii) Using the central limit theorem, estimate to two decimal places the probability that the absolute value of the error exceeds 0.01.

(iii) State clearly any assumptions you have made in your calculation.

**Answer/Explanation**

## Markscheme

(a) *f*(*x*)is even (symmetrical about the origin) *(M1)*

\({\text{E}}(X) = 0\) *A1*

\({\text{Var}}(X) = {\text{E}}({X^2}) = \int_{ – 0.005}^{0.005} {100{x^2}{\text{d}}x} \) *(M1)(A1)*

\( = 8.33 \times {10^{ – 6}}\left( {{\text{accept }}0.83 \times {{10}^{ – 5}}{\text{ or }}\frac{1}{{120\,000}}} \right)\) *A1*

*[5 marks]*

* *

(b) rounding errors to 2 decimal places are uniformly distributed *R1*

and lie within the interval \( – 0.005 \leqslant x < 0.005.\) *R1*

this defines *X* *AG*

*[2 marks]*

* *

(c) (i) using the symbol *y* to denote the error in the sum of 20 real numbers each rounded to 2 decimal places

\( – 0.1 \leqslant y( = 20 \times x) < 0.1\) *A1*

* *

(ii) \(Y \approx {\text{N}}(20 \times 0,{\text{ }}20 \times 8.3 \times {10^{ – 6}}) = {\text{N}}(0,{\text{ }}0.00016)\) *(M1)(A1)*

\({\text{P}}\left( {\left| Y \right| > 0.01} \right) = 2\left( {1 – {\text{P}}(Y < 0.01)} \right)\) *(M1)(A1)*

\( = 2\left( {1 – {\text{P}}\left( {Z < \frac{{0.01}}{{0.0129}}} \right)} \right)\)

\( = 0.44\) to 2 decimal places *A1**N4*

* *

(iii) it is assumed that the errors in rounding the 20 numbers are independent *R1*

and, by the central limit theorem, the sum of the errors can be modelled approximately by a normal distribution *R1*

*[8 marks]*

*Total [15 marks]*

## Examiners report

This was the only question on the paper with a conceptually ‘hard’ final part. Part(a) was generally well done, either by integration or by use of the standard formulae for a uniform distribution. Many candidates were not able to provide convincing reasoning in parts (b) and (c)(iii). Part(c)(ii), the application of the Central Limit Theorem was only very rarely tackled competently.

## Question

If \(X\) is a random variable that follows a Poisson distribution with mean \(\lambda > 0\) then the probability generating function of \(X\) is \(G(t) = {e^{\lambda (t – 1)}}\).

(i) Prove that \({\text{E}}(X) = \lambda \).

(ii) Prove that \({\text{Var}}(X) = \lambda \).

\(Y\) is a random variable, independent of \(X\), that also follows a Poisson distribution with mean \(\lambda \).

If \(S = 2X – Y\) find

(i) \({\text{E}}(S)\);

(ii) \({\text{Var}}(S)\).

Let \(T = \frac{Y}{2} + \frac{Y}{2}\).

(i) Show that \(T\) is an unbiased estimator for \(\lambda \).

(ii) Show that \(T\) is a more efficient unbiased estimator of \(\lambda \) than \(S\).

Could either \(S\) or \(T\) model a Poisson distribution? Justify your answer.

By consideration of the probability generating function, \({G_{X + Y}}(t)\), of \(X + Y\), prove that \(X + Y\) follows a Poisson distribution with mean \(2\lambda \).

Find

(i) \({G_{X + Y}}(1)\);

(ii) \({G_{X + Y}}( – 1)\).

Hence find the probability that \(X + Y\) is an even number.

**Answer/Explanation**

## Markscheme

(i) \(G'(t) = \lambda {e^{\lambda (t – 1)}}\) *A1*

\({\text{E}}(X) = G'(1)\) *M1*

\( = \lambda \) *AG*

(ii) \(G”(t) = {\lambda ^2}{e^{\lambda (t – 1)}}\) *M1*

\( \Rightarrow G”(1) = {\lambda ^2}\) *(A1)*

\({\text{Var}}(X) = G”(1) + G'(1) – {\left( {G'(1)} \right)^2}\) *(M1)*

\( = {\lambda ^2} + \lambda – {\lambda ^2}\) *A1*

\( = \lambda \) *AG*

*[6 marks]*

(i) \({\text{E}}(S) = 2\lambda – \lambda = \lambda \) *A1*

(ii) \({\text{Var}}(S) = 4\lambda + \lambda = 5\lambda \) *(A1)A1*

**Note: **First ** A1 **can be awarded for either \(4\lambda \) or \(\lambda \).

**[3 marks]**

(i) \({\text{E}}(T) = \frac{\lambda }{2} + \frac{\lambda }{2} = \lambda \;\;\;\)(so *\(T\) *is an unbiased estimator) *A1*

(ii) \({\text{Var}}(T) = \frac{1}{4}\lambda + \frac{1}{4}\lambda = \frac{1}{2}\lambda \) *A1*

this is less than \({\text{Var}}(S)\)*, *therefore \(T\) is the more efficient estimator *R1AG*

**Note: **Follow through their variances from (b)(ii) and (c)(ii).

**[3 marks]**

no, mean does not equal the variance *R1*

*[1 mark]*

\({G_{X + Y}}(t) = {e^{\lambda (t – 1)}} \times {e^{\lambda (t – 1)}} = {e^{2\lambda (t – 1)}}\) *M1A1*

which is the probability generating function for a Poisson with a mean of \(2\lambda \) *R1AG*

*[3 marks]*

(i) \({G_{X + Y}}(1) = 1\) *A1*

(ii) \({G_{X + Y}}( – 1) = {e^{ – 4\lambda }}\) *A1*

*[2 marks]*

\({G_{X + Y}}(1) = p(0) + p(1) + p(2) + p(3) \ldots \)

\({G_{X + Y}}( – 1) = p(0) – p(1) + p(2) – p(3) \ldots \)

so \({\text{2P(even)}} = {G_{X + Y}}(1) + {G_{X + Y}}( – 1)\) *(M1)(A1)*

\({\text{P(even)}} = \frac{1}{2}(1 + {e^{ – 4\lambda }})\) *A1*

*[3 marks]*

*Total [21 marks]*

## Examiners report

Solutions to the different parts of this question proved to be extremely variable in quality with some parts well answered by the majority of the candidates and other parts accessible to only a few candidates. Part (a) was well answered in general although the presentation was sometimes poor with some candidates doing the differentiation of \(G(t)\) and the substitution of \(t = 1\) simultaneously.

Part (b) was well answered in general, the most common error being to state that \({\text{Var}}(2X – Y) = {\text{Var}}(2X) – {\text{Var}}(Y)\).

Parts (c) and (d) were well answered by the majority of candidates.

Parts (c) and (d) were well answered by the majority of candidates.

Solutions to (e), however, were extremely disappointing with few candidates giving correct solutions. A common incorrect solution was the following:

\(\;\;\;{G_{X + Y}}(t) = {G_X}(t){G_Y}(t)\)

Differentiating,

\(\;\;\;{G’_{X + Y}}(t) = {G’_X}(t){G_Y}(t) + {G_X}(t){G’_Y}(t)\)

\(\;\;\;{\text{E}}(X + Y) = {G’_{X + Y}}(1) = {\text{E}}(X) \times 1 + {\text{E}}(Y) \times 1 = 2\lambda \)

This is correct mathematics but it does not show that \(X + Y\) is Poisson and it was given no credit. Even the majority of candidates who showed that \({G_{X + Y}}(t) = {{\text{e}}^{2\lambda (t – 1)}}\) failed to state that this result proved that \(X + Y\) is Poisson and they usually differentiated this function to show that \({\text{E}}(X + Y) = 2\lambda \).

In (f), most candidates stated that \({G_{X + Y}}(1) = 1\) even if they were unable to determine \({G_{X + Y}}(t)\) but many candidates were unable to evaluate \({G_{X + Y}}( – 1)\). Very few correct solutions were seen to (g) even if the candidates correctly evaluated \({G_{X + Y}}(1)\) and \({G_{X + Y}}( – 1)\).

[N/A]

## Question

Engine oil is sold in cans of two capacities, large and small. The amount, in millilitres, in each can, is normally distributed according to Large \( \sim {\text{N}}(5000,{\text{ }}40)\) and Small \( \sim {\text{N}}(1000,{\text{ }}25)\).

A large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.

A large can and a small can are selected at random. Find the probability that the large can contains at least \(30\) milliliters more than five times the amount contained in the small can.

A large can and five small cans are selected at random. Find the probability that the large can contains at least \(30\) milliliters less than the total amount contained in the small cans.

**Answer/Explanation**

## Markscheme

\({\text{P}}(L \ge 4995) = 0.785\) *(M1)A1*

**Note:** Accept any answer that rounds correctly to \(0.79\).

Award ** M1A0** for \(0.78\).

**Note:** Award ** M1A0** for any answer that rounds to \(0.55\) obtained by taking \({\text{SD}} = 40\).

**[2 marks]**

we are given that \(L \sim {\text{N}}(5000,{\text{ }}40)\) and \(S \sim {\text{N}}(1000,{\text{ }}25)\)

consider \(X = L – 5S\) (ignore \( \pm 30\)) *(M1)*

\({\text{E}}(X) = 0\) (\( \pm 30\) consistent with line above) *A1*

\({\text{Var}}(X) = {\text{Var}}(L) + 25{\text{Var}}(S) = 40 + 625 = 665\) *(M1)A1*

require \({\text{P}}(X \ge 30)\;\;\;({\text{or P}}(X \ge 0){\text{ if }} – 30{\text{ above}})\) *(M1)*

obtain \(0.122\) *A1*

**Note:** Accept any answer that rounds correctly to \(2\) significant figures.

**[6 marks]**

consider \(Y = L – ({S_1} + {S_2} + {S_3} + {S_4} + {S_5})\) (ignore \( \pm 30\)) *(M1)*

\({\text{E}}(Y) = 0\) (\( \pm 30\) consistent with line above) *A1*

\({\text{Var}}(Y) = 40 + 5 \times 25 = 165\) *A1*

require \({\text{P}}(Y \le – 30){\text{ (or P}}(Y \le 0){\text{ if }} + 30{\text{ above)}}\) *(M1)*

obtain \(0.00976\) *A1*

**Note:** Accept any answer that rounds correctly to \(2\) significant figures.

**Note:** Condone the notation \(Y = L – 5S\) if the variance is correct.

**[5 marks]**

**Total [13 marks]**

## Examiners report

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\)*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\)*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

Most candidates solved (a) correctly. In (b) and (c), however, many candidates made the usual error of confusing \(\sum\limits_{i = 1}^n {{X_i}} \) and \(nX\)*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

## Question

Engine oil is sold in cans of two capacities, large and small. The amount, in millilitres, in each can, is normally distributed according to Large \( \sim {\text{N}}(5000,{\text{ }}40)\) and Small \( \sim {\text{N}}(1000,{\text{ }}25)\).

A large can is selected at random. Find the probability that the can contains at least \(4995\) millilitres of oil.

A large can and a small can are selected at random. Find the probability that the large can contains at least \(30\) milliliters more than five times the amount contained in the small can.

A large can and five small cans are selected at random. Find the probability that the large can contains at least \(30\) milliliters less than the total amount contained in the small cans.

**Answer/Explanation**

## Markscheme

\({\text{P}}(L \ge 4995) = 0.785\) *(M1)A1*

**Note:** Accept any answer that rounds correctly to \(0.79\).

Award ** M1A0** for \(0.78\).

**Note:** Award ** M1A0** for any answer that rounds to \(0.55\) obtained by taking \({\text{SD}} = 40\).

**[2 marks]**

we are given that \(L \sim {\text{N}}(5000,{\text{ }}40)\) and \(S \sim {\text{N}}(1000,{\text{ }}25)\)

consider \(X = L – 5S\) (ignore \( \pm 30\)) *(M1)*

\({\text{E}}(X) = 0\) (\( \pm 30\) consistent with line above) *A1*

\({\text{Var}}(X) = {\text{Var}}(L) + 25{\text{Var}}(S) = 40 + 625 = 665\) *(M1)A1*

require \({\text{P}}(X \ge 30)\;\;\;({\text{or P}}(X \ge 0){\text{ if }} – 30{\text{ above}})\) *(M1)*

obtain \(0.122\) *A1*

**Note:** Accept any answer that rounds correctly to \(2\) significant figures.

**[6 marks]**

consider \(Y = L – ({S_1} + {S_2} + {S_3} + {S_4} + {S_5})\) (ignore \( \pm 30\)) *(M1)*

\({\text{E}}(Y) = 0\) (\( \pm 30\) consistent with line above) *A1*

\({\text{Var}}(Y) = 40 + 5 \times 25 = 165\) *A1*

require \({\text{P}}(Y \le – 30){\text{ (or P}}(Y \le 0){\text{ if }} + 30{\text{ above)}}\) *(M1)*

obtain \(0.00976\) *A1*

**Note:** Accept any answer that rounds correctly to \(2\) significant figures.

**Note:** Condone the notation \(Y = L – 5S\) if the variance is correct.

**[5 marks]**

**Total [13 marks]**

## Examiners report

*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

*. *Indeed some candidates even use the second expression to mean the first. This error leads to an incorrect variance and of course an incorrect answer. Some candidates had difficulty in converting the verbal statements into the correct probability statements, particularly in (c).

## Question

Two students are selected at random from a large school with equal numbers of boys and girls. The boys’ heights are normally distributed with mean \(178\) cm and standard deviation \(5.2\) cm, and the girls’ heights are normally distributed with mean \(169\) cm and standard deviation \(5.4\) cm.

Calculate the probability that the taller of the two students selected is a boy.

**Answer/Explanation**

## Markscheme

let \(X\) denote boys’ height and \(Y\) denote girls’ height

if \(BB,{\text{ P(taller is boy)}} = 1\) *(A1)*

if \(GG,{\text{ P(taller is boy)}} = 0\) *(A1)*

if \(BG\) or \(GB\):

consider \(X – Y\) *(M1)*

\(E(X – Y) = 178 – 169 = 9\) *A1*

\({\text{Var}}(X – Y) = {5.2^2} + {5.4^2}\;\;\;( = 56.2)\) *(M1)A1*

\({\text{P}}(X – Y > 0) = 0.885\) *A1*

answer is \(\frac{1}{4} \times 1 + \frac{1}{2} \times 0.885 = 0.693\) *(M1)A1*

*[9 marks]*

## Examiners report

## Question

Adam does the crossword in the local newspaper every day. The time taken by Adam, \(X\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(22,{\text{ }}{5^2})\).

Beatrice also does the crossword in the local newspaper every day. The time taken by Beatrice, \(Y\) minutes, to complete the crossword is modelled by the normal distribution \({\text{N}}(40,{\text{ }}{6^2})\).

Given that, on a randomly chosen day, the probability that he completes the crossword in less than \(a\) minutes is equal to 0.8, find the value of \(a\).

Find the probability that the total time taken for him to complete five randomly chosen crosswords exceeds 120 minutes.

Find the probability that, on a randomly chosen day, the time taken by Beatrice to complete the crossword is more than twice the time taken by Adam to complete the crossword. Assume that these two times are independent.

**Answer/Explanation**

## Markscheme

\(z = 0.841 \ldots \) *(A1)*

\(a = \mu + z\sigma \) *(M1)*

\( = 26.2\) *A1*

*[3 marks]*

let \(T\) denote the total time taken to complete 5 crosswords.

\(T\) is \({\text{N}}(110,{\text{ }}125)\) *(A1)(A1)*

**Note: A1 **for the mean and

**for the variance.**

*A1*\({\text{P}}(T > 120) = 0.186\) *A1*

*[3 marks]*

consider the random variable \(U = Y – 2X\) *(M1)*

\({\text{E}}(U) = – 4\) *A1*

\({\text{Var}}(U) = {\text{Var}}(Y) + 4{\text{Var}}(X)\) *(M1)*

\( = 136\) *A1*

\({\text{P}}(Y > 2X) = {\text{P}}(U > 0)\) *(M1)*

\( = 0.366\) *A1*

*[6 marks]*

## Examiners report

Part (a) was very well answered with only a very few weak candidates using 0.8 instead of 0.841…

Part (b) was well answered with only a few candidates calculating the variance incorrectly.

Part (c) was again well answered. The most common errors, not often seen, were writing the variance of \(Y – 2X\) as either \({\text{Var}}(Y) + 2{\text{Var}}(X)\) or \({\text{Var}}(Y) – 2(or{\text{ }}4){\text{Var}}(X)\).

## Question

The weights, *X* kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.

The weights, *Y* kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.

Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.

Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.

Two randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.

**Answer/Explanation**

## Markscheme

**Note:** In question 1, accept answers that round correctly to 2 significant figures.

P(4.75 < *X* < 4.85) = 0.197 **A1**

**[1 mark]**

**Note:** In question 1, accept answers that round correctly to 2 significant figures.

consider the random variable *X* − 2*Y* **(M1)**

E(*X* − 2*Y*) = − 0.6 **(A1)**

Var(*X* − 2*Y*) = Var(*X*) + 4Var(*Y*) **(M1)**

= 0.13 **(A1)**

*X* − 2*Y* ∼ N(−0.6, 0.13)

P(*X* − 2*Y* > 0) **(M1)**

= 0.0480 ** A1**

**[6 marks]**

**Note:** In question 1, accept answers that round correctly to 2 significant figures.

let *W* = *X*_{1} + *X*_{2} + *Y*_{1} + *Y*_{2} + *Y*_{3} be the total weight

E(*W*) = 17.7 **(A1)**

Var(*W*) = 2Var(*X*) + 3Var(*Y*) = 0.1475 ** (M1)(A1)**

W ∼ N(17.7, 0.1475)

P(*W* > 18) = 0.217 **A1**

**[4 marks]**

## Examiners report

[N/A]

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