Let \(X\) and \(Y\) be independent random variables with \(X \sim {P_o}{\text{ (3)}}\) and \(Y \sim {P_o}{\text{ (2)}}\).

Let \(S = 2X + 3Y\).

(a)     Find the mean and variance of \(S\).

(b)     Hence state with a reason whether or not \(S\) follows a Poisson distribution.

Let \(T = X + Y\).

(c)     Find \({\text{P}}(T = 3)\).

(d)     Show that \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \).

(e)     Hence show that \(T\) follows a Poisson distribution with mean 5.

Find the probability of a Type II error if the leaves are in fact from a plant of species B.

(a)     \({\text{E}}(S) = 2{\text{E}}(X) + 3{\text{E}}(Y) = 6 + 6 = 12\)     A1

\({\text{Var}}(S) = 4{\text{Var}}(X) + 9{\text{Var}}(Y) = 12 + 18 = 30\)     A1

[2 marks]

(b)     \(S\) does not have a Poisson distribution     A1

because \({\text{Var}}(S) \ne {\text{E}}(S)\)     R1

 

Note:     Follow through their \({\text{E}}(S)\) and \({\text{Var}}(S)\) if different.

 

[2 marks]

(c)     EITHER

\({\text{P}}(T = 3) = {\text{P}}\left( {(X,{\text{ }}Y) = (3,{\text{ }}0)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (2,{\text{ }}1)} \right) + \)

\( + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}2)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}3)} \right)\)     (M1)

\( = {\text{P}}(X = 3){\text{P}}(Y = 0) + {\text{P}}(X = 2){\text{P}}(Y = 1) + \)

\( + {\text{P}}(X = 1){\text{P}}(Y = 2) + {\text{P}}(X = 0){\text{P}}(Y = 3)\)     (M1)

\( = \frac{{125{e^{ - 5}}}}{6}{\text{ }}( = 0.140)\)     A2

 

Note:     Accept answers which round to 0.14.

 

OR

\(T\) is \({{\text{P}}_o}(2 + 3) = {{\text{P}}_o}(5)\)     (M1)(A1)

\({\text{P}}(T = 3) = \frac{{125{e^{ - 5}}}}{6}{\text{ }}( = 0.140)\)     A2

 

Note:     Accept answers which round to 0.14.

 

[4 marks]

(d)     \({\text{P}}(T = t) = {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}t)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}t - 1)} \right) +  \ldots {\text{P}}\left( {(X,{\text{ }}Y) = (t,{\text{ }}0)} \right)\)     (M1)

\( = {\text{P}}(X = 0){\text{P}}(Y = t) + {\text{P}}(X = 1){\text{P}}(Y = t - 1) +  \ldots  + {\text{P}}(X = t){\text{P}}(Y = 0)\)     A1

\( = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \)     AG

[2 marks]

(e)     \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \)

\( = \sum\limits_{r = 0}^t {\frac{{{e^{ - 3}}{3^r}}}{{r!}} \times \frac{{{e^{ - 2}}{2^{t - r}}}}{{(t - r)!}}} \)     M1A1

\( = \frac{{{e^{ - 5}}}}{{t!}}\sum\limits_{r = 0}^t {\frac{{t!}}{{r!(t - r)!}} \times {3^r}{2^{t - r}}} \)     M1

\( = \frac{{{e^{ - 5}}}}{{t!}}{(3 + 2)^t}\)     A1

\(\left( { = \frac{{{e^{ - 5}}{5^t}}}{{t!}}} \right)\)

hence \(T\) follows a Poisson distribution with mean 5     AG

[4 marks]

type II error probability \( = {\text{P}}(\bar X > 4.70654 \ldots |\bar X{\text{ is }}N\left( {4.6,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\)     (M1)

\( = 0.361\)     A1

Parts (a) and (b) were well answered by most candidates. The most common error in (a) was to calculate \(E(2X + 3Y)\) correctly as 12 and then state that, because the sum is Poisson, the variance is also 12. Many of these candidates then stated in (b) that the sum is Poisson because the mean and variance are equal, without apparently realising the circularity of their argument. Although (c) was intended as a possible hint for solving (d) and (e), many candidates simply noted that \(X + Y\) is \({{\text{P}}_o}{\text{(5)}}\) which led immediately to the correct answer. Some candidates tended to merge (d) and (e), often unsuccessfully, while very few candidates completed (e) correctly where the need to insert \(t!\) in the numerator and denominator was not usually spotted.

[N/A]