# IB DP Maths Topic 7.2 Mean of linear combinations of n random variables HL Paper 3

## Question

Anna has a fair cubical die with the numbers 1, 2, 3, 4, 5, 6 respectively on the six faces. When she tosses it, the score is defined as the number on the uppermost face. One day, she decides to toss the die repeatedly until all the possible scores have occurred at least once.

(a)     Having thrown the die once, she lets $${X_2}$$ denote the number of additional throws required to obtain a different number from the one obtained on the first throw. State the distribution of $${X_2}$$ and hence find $${\text{E}}({X_2})$$ .

(b)     She then lets $${X_3}$$ denote the number of additional throws required to obtain a different number from the two numbers already obtained. State the distribution of $${X_3}$$ and hence find $${\text{E}}({X_3})$$ .

(c)     By continuing the process, show that the expected number of tosses needed to obtain all six possible scores is 14.7.

## Markscheme

(a)     $${X_2}$$ is a geometric random variable     A1

with $$p = \frac{5}{6}.$$     A1

Therefore $${\text{E}}({X_2}) = \frac{6}{5}.$$     A1

[3 marks]

(b)     $${X_3}$$ is a geometric random variable with $$p = \frac{4}{6}.$$     A1

Therefore $${\text{E}}({X_3}) = \frac{6}{4}.$$     A1

[2 marks]

(c)     $${\text{E}}({X_4}) = \frac{6}{3},{\text{ E}}({X_5}) = \frac{6}{2},{\text{ E}}({X_6}) = \frac{6}{1}$$     A1A1A1

$${\text{E}}({X_1}) = 1\,\,\,\,\,{\text{(or }}{X_1} = 1)$$     A1

Expected number of tosses $$\sum\limits_{n = 1}^6 {{\text{E}}({X_n})}$$     M1

$$= 14.7$$     AG

[5 marks]

Total [10 marks]

## Examiners report

Many candidates were unable even to start this question although those who did often made substantial progress.

## 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)     $${\text{E}}({Y^2})$$ .

[6]
a.

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 = 3S – 4R.

Calculate P(V > 5).

[6]
b.

## 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]

a.

E(V) = E(3S – 4R)

= 3E(S) – 4E(R)     M1

= 24 – 20 = 4     A1

Var(3S – 4R) = 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]

b.

## 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 $$P(V > 5) = 1 – P(V \leqslant 5)$$.

a.

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 $$P(V > 5) = 1 – P(V \leqslant 5)$$.

b.

## 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.

## 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 $${\text{Var}}(X) = \lambda$$.

[6]
a.

$$Y$$ 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)     $${\text{Var}}(S)$$.

[3]
b.

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 $$S$$.

[3]
c.

Could either $$S$$ or $$T$$ model a Poisson distribution? Justify your answer.

[1]
d.

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

[3]
e.

Find

(i)     $${G_{X + Y}}(1)$$;

(ii)     $${G_{X + Y}}( – 1)$$.

[2]
f.

Hence find the probability that $$X + Y$$ is an even number.

[3]
g.

## 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]

a.

(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]

b.

(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]

c.

no, mean does not equal the variance     R1

[1 mark]

d.

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

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

[3 marks]

e.

(i)     $${G_{X + Y}}(1) = 1$$     A1

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

[2 marks]

f.

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

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

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]

g.

## 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.

a.

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

b.

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

c.

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

d.

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 $${\text{E}}(X + Y) = 2\lambda$$.

e.

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 $${G_{X + Y}}( – 1)$$.

f.

[N/A]

g.

## 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.

[2]
a.

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.

[6]
b.

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.

[5]
c.

## 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]

a.

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]

b.

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]

c.

## 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).

a.

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).

b.

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).

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.

## 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]

[N/A]

## 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 $$a$$.

[3]
a.

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

[3]
b.

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.

[6]
c.

## Markscheme

$$z = 0.841 \ldots$$    (A1)

$$a = \mu + z\sigma$$    (M1)

$$= 26.2$$    A1

[3 marks]

a.

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 A1 for the variance.

$${\text{P}}(T > 120) = 0.186$$    A1

[3 marks]

b.

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]

c.

## Examiners report

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

a.

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

b.

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 $${\text{Var}}(Y) – 2(or{\text{ }}4){\text{Var}}(X)$$.

c.

## Question

Alun answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is always $$\frac{6}{7}$$, independently of all other questions. He will stop for coffee immediately following a second incorrect answer. Let $$X$$ be the number of questions Alun answers before he stops for coffee.

Nic answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is initially $$\frac{6}{7}$$. After his first incorrect answer, Nic loses confidence in his own ability and from this point onwards, the probability that he answers any question correctly is now only $$\frac{4}{7}$$.

Both before and after his first incorrect answer, the result of each question is independent of the result of any other question. Nic will also stop for coffee immediately following a second incorrect answer. Let $$Y$$ be the number of questions Nic answers before he stops for coffee.

(i)     State the distribution of $$X$$, including its parameters.

(ii)     Calculate $${\text{E}}(X)$$.

(iii)     Calculate $${\text{P}}(X = 5)$$.

[6]
a.

(i)     Calculate $${\text{E}}(Y)$$.

(ii)     Calculate $${\text{P}}(Y = 5)$$.

[9]
b.

## Markscheme

(i)     $${\text{NB}}\left( {2,{\text{ }}\frac{1}{7}} \right)$$     A1A1A1

Note: The final A1 mark can be awarded for knowing that $$p = \frac{1}{7}$$ independent of the other two marks.

(ii)     $${\text{E}}(X) = \frac{r}{p} = 14$$     A1

(iii)     $$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){\left( {\frac{6}{7}} \right)^3}{\left( {\frac{1}{7}} \right)^2} = 0.0514$$     (M1)A1

Note: Accept any number that rounds to this 3sf number.

[6 marks]

a.

(i)     $$Y = {Y_1} + {Y_2}$$ (number up to1st + number up to 2nd)     (M1)

$${Y_1} \sim Geo\left( {\frac{1}{7}} \right),{\text{ }}{Y_2} \sim Geo\left( {\frac{3}{7}} \right)$$    (A1)

Notes: The above (A1) is independent of the (M1).

Could have $${\text{NB }}(1,{\text{ }}p)$$, instead of $$Geo(p)$$.

$${\text{E}}(Y) = \frac{1}{{\left( {\frac{1}{7}} \right)}} + \frac{1}{{\left( {\frac{3}{7}} \right)}} = 7 + \frac{7}{3} = 9\frac{1}{3}{\text{ (9.33)}}$$    M1A1

(ii)     $$Y = {Y_1} + {Y_2} = 5$$ happens when     (M1)

$${Y_1} = 1,{\text{ }}{Y_2} = 4$$ or $${Y_1} = 2,{\text{ }}{Y_2} = 3$$ or $${Y_1} = 3,{\text{ }}{Y_2} = 2$$ or $${Y_1} = 4,{\text{ }}{Y_2} = 1$$     (A1)

so probability is $$\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{3}{7}$$     (M1)(A1)

$$= 0.0928{\text{ }}\left( {\frac{{1560}}{{16807}}} \right)$$    A1

Note: Accept any answer that rounds to 0.093.

[9 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

Alun answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is always $$\frac{6}{7}$$, independently of all other questions. He will stop for coffee immediately following a second incorrect answer. Let $$X$$ be the number of questions Alun answers before he stops for coffee.

Nic answers mathematics questions and checks his answer after doing each one.

The probability that he answers any question correctly is initially $$\frac{6}{7}$$. After his first incorrect answer, Nic loses confidence in his own ability and from this point onwards, the probability that he answers any question correctly is now only $$\frac{4}{7}$$.

Both before and after his first incorrect answer, the result of each question is independent of the result of any other question. Nic will also stop for coffee immediately following a second incorrect answer. Let $$Y$$ be the number of questions Nic answers before he stops for coffee.

(i)     State the distribution of $$X$$, including its parameters.

(ii)     Calculate $${\text{E}}(X)$$.

(iii)     Calculate $${\text{P}}(X = 5)$$.

[6]
a.

(i)     Calculate $${\text{E}}(Y)$$.

(ii)     Calculate $${\text{P}}(Y = 5)$$.

[9]
b.

## Markscheme

(i)     $${\text{NB}}\left( {2,{\text{ }}\frac{1}{7}} \right)$$     A1A1A1

Note: The final A1 mark can be awarded for knowing that $$p = \frac{1}{7}$$ independent of the other two marks.

(ii)     $${\text{E}}(X) = \frac{r}{p} = 14$$     A1

(iii)     $$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){\left( {\frac{6}{7}} \right)^3}{\left( {\frac{1}{7}} \right)^2} = 0.0514$$     (M1)A1

Note: Accept any number that rounds to this 3sf number.

[6 marks]

a.

(i)     $$Y = {Y_1} + {Y_2}$$ (number up to1st + number up to 2nd)     (M1)

$${Y_1} \sim Geo\left( {\frac{1}{7}} \right),{\text{ }}{Y_2} \sim Geo\left( {\frac{3}{7}} \right)$$    (A1)

Notes: The above (A1) is independent of the (M1).

Could have $${\text{NB }}(1,{\text{ }}p)$$, instead of $$Geo(p)$$.

$${\text{E}}(Y) = \frac{1}{{\left( {\frac{1}{7}} \right)}} + \frac{1}{{\left( {\frac{3}{7}} \right)}} = 7 + \frac{7}{3} = 9\frac{1}{3}{\text{ (9.33)}}$$    M1A1

(ii)     $$Y = {Y_1} + {Y_2} = 5$$ happens when     (M1)

$${Y_1} = 1,{\text{ }}{Y_2} = 4$$ or $${Y_1} = 2,{\text{ }}{Y_2} = 3$$ or $${Y_1} = 3,{\text{ }}{Y_2} = 2$$ or $${Y_1} = 4,{\text{ }}{Y_2} = 1$$     (A1)

so probability is $$\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{3}{7}$$     (M1)(A1)

$$= 0.0928{\text{ }}\left( {\frac{{1560}}{{16807}}} \right)$$    A1

Note: Accept any answer that rounds to 0.093.

[9 marks]

b.

[N/A]

a.

[N/A]

b.

## 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.

[1]
a.

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

[6]
b.

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.

[4]
c.

## 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]

a.

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

consider the random variable X − 2Y     (M1)

E(X − 2Y) =  − 0.6     (A1)

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

= 0.13     (A1)

X − 2Y ∼ N(−0.6, 0.13)

P(X − 2Y > 0)     (M1)

= 0.0480     A1

[6 marks]

b.

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

let W = X1 + X2 + Y1 + Y2 + Y3 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]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.