(a)     (i)     Using l’Hôpital’s rule, show that

\[\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^{\lambda x}}}} = 0;{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}\lambda  \in {\mathbb{R}^ + }\]

(ii)     Using mathematical induction on \(n\), prove that

\[\int_0^\infty  {{x^n}{{\text{e}}^{ - \lambda x}}{\text{d}}x = \frac{{n!}}{{{\lambda ^{n + 1}}}};{\text{ }}} n \in \mathbb{N},{\text{ }}\lambda  \in {\mathbb{R}^ + }\]

(b)     The random variable \(X\) has probability density function

\[f(x) = \left\{ \begin{array}{l}\frac{{{\lambda ^{n + 1}}{x^n}{{\rm{e}}^{ - \lambda x}}}}{{n!}}x \ge 0,n \in {\mathbb{Z}^ + },\lambda  \in {\mathbb{R}^ + }\\{\rm{otherwise}}\end{array} \right.\]

Giving your answers in terms of \(n\) and \(\lambda \), determine

(i)     \({\text{E}}(X)\);

(ii)     the mode of \(X\).

(c)     Customers arrive at a shop such that the number of arrivals in any interval of duration \(d\) hours follows a Poisson distribution with mean \(8d\). The third customer on a particular day arrives \(T\) hours after the shop opens.

(i)     Show that \({\text{P}}(T > t) = {{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right)\).

(ii)     Find an expression for the probability density function of \(T\).

(iii)     Deduce the mean and the mode of \(T\).

(a)     (i)     using l’Hopital’s rule once,

\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^{\lambda x}}}} = \mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n{x^{n - 1}}}}{{\lambda {{\text{e}}^{\lambda x}}}}\)     (A1)(A1)

 

Note: Award A1 for numerator, A1 for denominator.

 

if \(n > 1\), this still gives \(\frac{\infty }{\infty }\) so differentiate again giving

\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n(n - 1){x^{n - 2}}}}{{{\lambda ^2}{{\text{e}}^{\lambda x}}}}\)     (A1)

if \(n > 2\), this still gives \(\frac{\infty }{\infty }\) so differentiate a further \(n - 2\) times giving     M1

\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n!}}{{{\lambda ^n}{{\text{e}}^{\lambda x}}}}\)     A1

\( = 0\)     AG

(ii)     first prove the result true for \(n = 0\)

\(\int_0^\infty  {{{\text{e}}^{ - \lambda x}}{\text{d}}x =  - \frac{1}{\lambda }\left[ {{{\text{e}}^{ - \lambda x}}} \right]_0^\infty  = \frac{1}{\lambda }} \) as required     M1A1

assume the result is true for \(n = k\)     M1

\(\int_0^\infty  {{x^k}{{\text{e}}^{ - \lambda x}}{\text{d}}x = \frac{{k!}}{{{\lambda ^{k + 1}}}}} \)

consider, for \(n = k + 1\),

\(\int_0^\infty  {{x^{k + 1}}{{\text{e}}^{ - \lambda x}}{\text{d}}x =  - \frac{1}{\lambda }\left[ {{x^{k + 1}}{{\text{e}}^{ - \lambda x}}} \right]_0^\infty  + \frac{{k + 1}}{\lambda }\int_0^\infty  {{x^k}{{\text{e}}^{ - \lambda x}}{\text{d}}x} } \)     M1A1

\( = (0 + )\frac{{k + 1}}{\lambda } \times \frac{{k!}}{{{\lambda ^{k + 1}}}}\)     A1

\( = \frac{{(k + 1)!}}{{{\lambda ^{k + 2}}}}\)     A1

therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 0\), the result is proved by induction     R1

 

Note: Only award the R1 if at least 4 of the previous marks have been awarded.

 

Note: If a candidate starts at \(n = 1\), do not award the first 2 marks but follow through thereafter.

 

[13 marks]

 

(b)     (i)     \({\text{E}}(X) = \frac{{{\lambda ^{n + 1}}}}{{n!}}\int_0^\infty  {{x^{n + 1}}} {{\text{e}}^{ - \lambda x}}{\text{d}}x\)     M1

\( = \frac{{{\lambda ^{n + 1}}}}{{n!}} \times \frac{{(n + 1)!}}{{{\lambda ^{n + 2}}}}\)     A1

\( = \frac{{(n + 1)}}{\lambda }\)     A1

(ii)     the mode satisfies \(f'(x) = 0\)     M1

\(f'(x) = \frac{{{\lambda ^{n + 1}}}}{{n!}}\left( {n{x^{n - 1}}{{\text{e}}^{ - \lambda x}} - \lambda {x^n}{{\text{e}}^{ - \lambda x}}} \right)\)     A1

mode \( = \frac{n}{\lambda }\)     A1

[6 marks]

 

(c)     (i)     \({\text{P}}(T > t) = {\text{P}}(0,{\text{ }}1{\text{ or 2 arrivals in }}[0,{\text{ }}t])\)     (M1)

\( = {{\text{e}}^{ - 8t}} + {{\text{e}}^{ - 8t}} \times 8t + {{\text{e}}^{ - 8t}} \times \frac{{{{(8t)}^2}}}{2}\)     A1

\( = {{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right)\)     AG

(ii)     differentiating,

\( - f(t) =  - 8{{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right) + {{\text{e}}^{ - 8t}}(8 + 64t)\)     A1A1

 

Note: Award A1 for LHS, A1 for RHS.

 

\(f(t) = 256{t^2}{{\text{e}}^{ - 8t}}\)     A1

(iii)     with the previous notation, \(n = 2,{\text{ }}\lambda  = 8\).     (M1)

mean \( = \frac{3}{8}\)     A1

mode \( = \frac{1}{4}\)     A1

[8 marks]

 

Note: Do not follow through if they use a negative probability density function.

 

[27 marks]

[N/A]