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</div><h2>SL Paper 2</h2><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The random variable \(X\) has cumulative distribution function\[F(x) = \left\{ {\begin{array}{*{20}{c}}<br> 0&{x < 0} \\ <br> {{{\left( {\frac{x}{a}} \right)}^3}}&{0 \leqslant x \leqslant a} \\ <br> 1&{x > a} <br>\end{array}} \right.\]where \(a\) is an unknown parameter. You are given that the mean and variance of \(X\) </span><span style="font-family: times new roman,times; font-size: medium;">are \(\frac{{3a}}{4}\)</span><span style="font-family: times new roman,times; font-size: medium;"> and \(\frac{{3{a^2}}}{{80}}\) </span><span style="font-family: times new roman,times; font-size: medium;">respectively. To estimate the value of \(a\) , a random sample of \(n\) </span><span style="font-family: times new roman,times; font-size: medium;">independent observations, \({X_1},{X_2}, \ldots {X_n}\) is taken from the distribution of \(X\) .</span></p>
<p> </p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Find an expression for \(c\) in terms of \(n\) such that \(U = c\sum\limits_{i = 1}^n {{X_i}} \) is an unbiased </span><span style="font-family: times new roman,times; font-size: medium;">estimator for \(a\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) Determine an expression for \({\text{Var}}(U)\) in this case.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \({\rm{P}}(Y \le y) = {\left( {\frac{y}{a}} \right)^{3n}},0 \le y \le a\) and deduce an expression for the </span><span style="font-family: times new roman,times; font-size: medium;">probability density function of \(Y\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \({\rm{E}}(Y)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Show that \({\rm{Var}}(Y) = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) Find an expression for \(d\) in terms of \(n\) such that \(V = dY\) is an unbiased </span><span style="font-family: times new roman,times; font-size: medium;">estimator for \(a\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (v) Determine an expression for \({\rm{Var}}(V)\) in this case.</span></p>
<div class="marks">[16]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that \(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{3n + 2}}{5}\) and hence state, with a reason, which of \(U\) or \(V\) is </span><span style="font-family: times new roman,times; font-size: medium;">the more efficient estimator for \(a\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({\rm{E}}(U) = c \times n \times \frac{{3a}}{4} = a \Rightarrow c = \frac{4}{{3n}}\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(Var(U) = \frac{{16}}{{9{n^2}}} \times n \times \frac{{3{a^2}}}{{80}} = \frac{{{a^2}}}{{15n}}\) <em><strong>M1A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [4 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({\rm{P}}(Y \le y) = {\rm{P}}({\text{all }}Xs \le y)\) <em><strong>M1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ {\rm{P}} \right.{\left. {(X \le y)} \right]^n}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(A1)</span></strong></em></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> \( = {\left( {{{\left( {\frac{y}{a}} \right)}^3}} \right)^n}\) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Only one of the two <em><strong>A1</strong></em> marks above may be implied.</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;"> </span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = {\left( {\frac{y}{a}} \right)^{3n}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG</span></strong></em></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(g(y) = \frac{{\rm{d}}}{{{\rm{d}}y}}{\left( {\frac{y}{a}} \right)^{3n}} = \frac{{3n{y^{3n - 1}}}}{{{a^{3n}}}},(0 < y < a)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p>(<span style="font-family: times new roman,times; font-size: medium;">\(g(y) = 0\) otherwise)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{E}}(Y) = \int_0^a {\frac{{3n{y^{3n}}}}{{{a^{3n}}}}} {\rm{d}}y\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ {\frac{{3n{y^{3n + 1}}}}{{(3n + 1){a^{3n}}}}} \right]_0^a\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{3na}}{{3n + 1}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) \({\rm{Var}}(Y) = \int_0^a {\frac{{3n{y^{3n + 1}}}}{{{a^{3n}}}}} {\rm{d}}y - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ {\frac{{3n{y^{3n + 2}}}}{{(3n + 2){a^{3n}}}}} \right]_0^a - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{3n{a^2}}}{{3n + 2}} - \frac{{9{n^2}{a^2}}}{{{{(3n + 1)}^2}}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{3n{a^2}(9{n^2} + 6n + 1) - 9{n^2}{a^2}(3n + 2)}}{{(3n + 2)(3n + 1)}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">AG</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) \({\rm{E}}(V) = d \times \frac{{3na}}{{3n + 1}} = a \Rightarrow d = \frac{{3n + 1}}{{3n}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(v) \(Var(V) = {\left( {\frac{{3n + 1}}{{3n}}} \right)^2} \times \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{{a^2}}}{{3n(3n + 2)}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p> </p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[16 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{\frac{{{a^2}}}{{15n}}}}{{\frac{{{a^2}}}{{3n(3n + 2)}}}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{3n + 2}}{5}\) <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(V\) is the more efficient estimator because \(3n + 2 > 5\) (for \(n > 1\) ) <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [2 marks]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Gillian is throwing a ball at a target. The number of throws she makes before hitting the target follows a geometric distribution, \(X \sim {\text{Geo}}(p)\). When she uses a cricket ball the number of throws she makes follows a geometric distribution with \(p = \frac{1}{4}\). When she uses a tennis ball the number of throws she makes follows a geometric distribution with \(p = \frac{3}{4}\)<span class="s1">. There is a box containing a large number of balls, \(80\%\)</span> of which are cricket balls and the remainder are tennis balls. The random variable \(A\) is the number of throws needed to hit the target when a single ball is chosen at random from this box and used for all throws.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \({\text{E}}(A)\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \({\text{P}}(A = r) = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \({\text{P}}(A \leqslant 5|A > 3)\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({\text{E}}(X{\text{ }}tennis) = \frac{1}{{\frac{3}{4}}} = \frac{4}{3}\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\({\text{E}}(X{\text{ }}cricket) = \frac{1}{{\frac{1}{4}}} = 4\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\({\text{E}}(A) = \frac{4}{3} \times \frac{1}{5} + 4 \times \frac{4}{5} = \frac{{52}}{{15}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({\text{P}}(X = r|cricket) = \frac{1}{4} \cdot {\left( {\frac{3}{4}} \right)^{r - 1}}\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\({\text{P}}(X = r|tennis) = \frac{3}{4} \cdot {\left( {\frac{1}{4}} \right)^{r - 1}}\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\({\text{P}}(A = r) = {\text{P}}(X = r|cricket) \times {\text{P}}(cricket) + {\text{P}}(X = r|tennis) \times {\text{P}}(tennis)\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">\( = \frac{1}{4} \times {\left( {\frac{3}{4}} \right)^{r - 1}} \times \frac{4}{5} + \frac{3}{4} \times {\left( {\frac{1}{4}} \right)^{r - 1}} \times \frac{1}{5}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({\text{P}}(A \leqslant 5)|(A < 3) = \frac{{{\text{P}}(A = 4{\text{ or }}5)}}{{1 - {\text{P}}(A = 1{\text{ or 2 or }}3)}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1A1</em></strong></p>
<p class="p1">\({\text{P}}(A = 1) = \frac{7}{{20}}\)</p>
<p class="p1">\({\text{P}}(A = 2) = \frac{15}{{80}}\)</p>
<p class="p1">\({\text{P}}(A = 3) = \frac{39}{{320}}\)</p>
<p class="p1">\({\text{P}}(A = 4) = \frac{111}{{1280}}\)</p>
<p class="p1">\({\text{P}}(A = 5) = \frac{327}{{5120}}\)</p>
<p class="p1">\( \Rightarrow {\text{P}}(A > 3)|(A \leqslant 5) = \frac{{\frac{{771}}{{5120}}}}{{\frac{{1744}}{{5120}}}}\) <span class="Apple-converted-space"> </span><strong><em>A1A2</em></strong></p>
<p class="p1"><strong>Note: </strong>Award <em><strong>A1</strong></em> for correct working for numerator and Award <em><strong>A2</strong></em> for correct working for denominator</p>
<p class="p2"> </p>
<p class="p1">\( = \frac{{771}}{{1744}}\;\;\;( = 0.442)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part c) caused more problems. Although a number of wholly correct solutions were seen, many candidates were unable to work meaningfully with the conditional probability.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>The discrete random variable \(X\) has the following probability distribution.</p>
<p style="text-align: center;">\({\text{P}}(X = x) = \frac{{kx}}{{{3^x}}}\), where \(x \in {\mathbb{Z}^ + }\) and \(k\) is a constant.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down the first three terms of the infinite series for \(G(t)\), the probability generating function for \(X\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the radius of convergence of this infinite series.</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By considering \(\left( {1 - \frac{t}{3}} \right)G(t)\), show that</p>
<p>\[G(t) = \frac{{3kt}}{{{{(3 - t)}^2}}}.\]</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence show that \(k = \frac{4}{3}\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By differentiating both sides of this equation, determine the values of \(G’(1)\) and \(G’’(1)\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find \({\text{Var}}(X)\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.iii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} + \ldots \) <strong><em>M1A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{(n + 1)k{t^{n + 1}}}}{{{3^{n + 1}}}} \times \frac{{{3^n}}}{{nk{t^n}}}\) <strong><em>M1A1</em></strong></p>
<p>\( \to \frac{t}{3}{\text{ as }}n \to \infty \) <strong><em>A1</em></strong></p>
<p>for convergence, \(\left| {\frac{t}{3}} \right| < 1\) so radius of convergence \( = 3\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} + \ldots \)</p>
<p>\(\frac{t}{3}G(t) = \frac{{k{t^2}}}{{{3^2}}} + \frac{{2k{t^3}}}{{{3^3}}} + \ldots \)</p>
<p>\(\left( {1 - \frac{t}{3}} \right)G(t) = \frac{{kt}}{3} + \frac{{k{t^2}}}{{{3^2}}} + \frac{{k{t^3}}}{{{3^3}}} + \ldots \) <strong><em>M1A1</em></strong></p>
<p>\( = \frac{{\frac{{kt}}{3}}}{{\left( {1 - \frac{t}{3}} \right)}}\) <strong><em>A1</em></strong></p>
<p>\(G(t) = \frac{{\frac{{kt}}{3}}}{{{{\left( {1 - \frac{t}{3}} \right)}^2}}} = \frac{{3kt}}{{{{(3 - t)}^2}}}\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(G(1) = 1\) <strong><em>M1</em></strong></p>
<p>so \(k = \frac{4}{3}\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\ln G(t) = \ln 4t - \ln {(3 - t)^2}\) <strong><em>M1</em></strong></p>
<p>\(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{G'(1)}}{{G(1)}} = \frac{1}{t} + \frac{2}{{3 - t}}\) <strong><em>M1A1</em></strong></p>
<p>putting \(t = 1\)</p>
<p>\(G’(1) = 2\) <strong><em>A1</em></strong></p>
<p>\(\frac{{G''(t)G(t) - {{[G'(t)]}^2}}}{{{{[G(t)]}^2}}} = - \frac{1}{{{t^2}}} + \frac{2}{{{{(3 - t)}^2}}}\) <strong><em>M1A1</em></strong></p>
<p>putting \(t = 1\)</p>
<p>\(G’’(1) - 4 = - 1 + \frac{1}{2}\)</p>
<p>\(G’’(1) = \frac{7}{2}\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({\text{Var}}(X) = G''(1) + G'(1) - {[G'(1)]^2} = \frac{3}{2}\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.iv.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. </span><span style="font-family: times new roman,times; font-size: medium;">However the weights vary with a standard deviation of \(0.54\) kg. A random sample of </span><span style="font-family: times new roman,times; font-size: medium;">\(24\)</span><span style="font-family: times new roman,times; font-size: medium;"> bags is taken to check that the mean weight is \(28\) kg.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} {\rm{ .}}\]</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Use your answer to (a) to find an approximate expression for the cumulative distributive </span><span style="font-family: times new roman,times; font-size: medium;">function of \({\rm{N}}(0,1)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>Hence</strong> find an approximate value for \({\rm{P}}( - 0.5 \le Z \le 0.5)\) , where </span><span style="font-family: times new roman,times; font-size: medium;">\(Z \sim {\rm{N}}(0,1)\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">State and justify an appropriate test procedure giving the null and alternate </span><span style="font-family: times new roman,times; font-size: medium;">hypotheses.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">What is the critical region for the sample mean if the probability of a Type I error </span><span style="font-family: times new roman,times; font-size: medium;">is to be \(3.5\%\)?</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">If the mean weight of the bags is actually \(28\).1 kg, what would be the probability </span><span style="font-family: times new roman,times; font-size: medium;">of a Type II error?</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = 1 + \left( { - \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) <strong><em>M1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} - \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></strong></em></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 - \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} - \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) <strong><em>M1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} - \ldots } \right)\) <strong><em>R1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{P}}( - 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 - \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} - \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} - \ldots } \right)\) <em><strong>M1</strong> </em></span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.38292 = 0.383\) <em><strong>A1 </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks] </span></strong></em></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">this is a two tailed test of the sample mean \(\overline x \) </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">we use the central limit theorem to justify assuming that <em><strong>R1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) <strong><em>R1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{H}}_0}:\mu = 28\) <strong><em>A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{H}}_1}:\mu \ne 28\) <em><strong>A1 </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[5 marks]</strong> </span></em></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) <strong><em>(M1)A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">and (\(\overline x \le 28 - 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) <em><strong>(M1)(A1)(A1)</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\) </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) <strong><em>A1A1</em> </strong></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">if \(\mu = 28.1\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) <strong><em>R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)</span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.884\) <em><strong>A1</strong> </em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Depending on the degree of accuracy used for the critical region the answer </span><span style="font-family: times new roman,times; font-size: medium;">for part (c) can be anywhere from \(0.8146\) to \(0.879\). </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></strong></em></p>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The derivation of a series from a given one by substitution seems not to be well known. This made finding the required series from \(({{\rm{e}}^x})\) in part (a) to be much more difficult than it need have been. The fact that this part was worth only 3 marks was a clear hint that an easy derivation was possible. </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In part (b)(i) the \(0.5\) was usually missing which meant that this part came out incorrectly. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The conditions required in part (a) were rarely stated correctly and some candidates were unable to state the hypotheses precisely. There was some confusion with "less than" and "less than or equal to". </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">There was some confusion with "less than" and "less than or equal to". </span></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Levels of accuracy in the body of the question varied wildly leading to a wide range of answers to part (c). </span></p>
<div class="question_part_label">B.c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The weights, \(X\) grams, of tomatoes may be assumed to be normally distributed </span><span style="font-family: times new roman,times; font-size: medium;">with mean \(\mu \) grams and standard deviation \(\sigma \) grams. Barry weighs \(21\) tomatoes </span><span style="font-family: times new roman,times; font-size: medium;">selected at random and calculates the following statistics.\[\sum {x = 1071} \) ; \(\sum {{x^2} = 54705} \]</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Determine a \(95\%\) confidence interval for \(\mu \) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The random variable \(Y\) has variance \({\sigma ^2}\) , where \({\sigma ^2} > 0\) . A random sample </span><span style="font-family: times new roman,times; font-size: medium;">of \(n\) observations of \(Y\) is taken and \(S_{n - 1}^2\)</span><span style="font-family: times new roman,times; font-size: medium;"> denotes the unbiased estimator for \({\sigma ^2}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">By considering the expression </span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{Var}}({S_{n - 1}}) = {\rm{E}}(S_{n - 1}^2) - {\left\{ {E\left. {({S_{n - 1}})} \right\}} \right.^2}\)</span><span style="font-family: times new roman,times; font-size: medium;"> ,</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">show that \(S_{n - 1}^{}\) is not an unbiased estimator for \(\sigma \) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \(\overline x = \frac{{1071}}{{21}} = 51\) <strong><em> A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S_{n - 1}^2 = \frac{{54705}}{{20}} - \frac{{{{1071}^2}}}{{20 \times 21}} = 4.2\) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) degrees of freedom \( = 20\) ; \(t\)-value \( = 2.086\) <strong><em>(A1)(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(95\%\) confidence limits are </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(51 \pm 2.086\sqrt {\frac{{4.2}}{{21}}} \) <strong><em>(M1)(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">leading to \(\left[ {50.1,51.9} \right]\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 marks] </span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{Var}}({S_{n - 1}}) > 0\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(E(S_{n - 1}^2) = {\sigma ^2}\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting in the given equation, </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\sigma ^2} - E(S_{n - 1}^{}) > 0\) <strong><em>M1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it follows that </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(E(S_{n - 1}^{}) < \sigma \) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">this shows that \({S_{n - 1}}\) is not an unbiased estimator for \(\sigma \) since that would require = instead of \( < \) <strong><em>R1 </em></strong></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[5 marks]</em> </span></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates attempted (a) although some used the normal distribution instead of the \(t\)-distribution. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Many candidates were unable even to start (b) and many of those who did filled several pages of algebra with factors such as \(n\) / \((n - 1)\) prominent. Few candidates realised that the solution required only a few lines. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The weights of apples, in grams, produced on a farm may be assumed to be normally </span><span style="font-family: times new roman,times; font-size: medium;">distributed with mean \(\mu \) and variance \({\sigma ^2}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The farm manager selects a random sample of \(10\) apples and weighs them with </span><span style="font-family: times new roman,times; font-size: medium;">the following results, given in grams.\[82, 98, 102, 96, 111, 95, 90, 89, 99, 101\]</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Determine a \(95\%\) confidence interval for \(\mu \) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The farm manager claims that the mean weight of apples is \(100\) grams but </span><span style="font-family: times new roman,times; font-size: medium;">the buyer from the local supermarket claims that the mean is less than this. </span><span style="font-family: times new roman,times; font-size: medium;">To test these claims, they select a random sample of \(100\) apples and weigh them. </span><span style="font-family: times new roman,times; font-size: medium;">Their results are summarized as follows, where \(x\) is the weight of an apple </span><span style="font-family: times new roman,times; font-size: medium;">in grams.\[\sum {x = 9831;\sum {{x^2} = 972578} } \]</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) State suitable hypotheses for testing these claims.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Determine the \(p\)-value for this test.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) At the \(1\%\) significance level, state which claim you accept and justify </span><span style="font-family: times new roman,times; font-size: medium;">your answer.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) from the GDC,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">unbiased estimate for \(\mu = 96.3\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">unbiased estimate for \({\sigma ^2} = 8.028{ \ldots ^2} = 64.5\) <strong><em>(M1)A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(95\%\) confidence interval is [\(90.6\), \(102\)] <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept \(102.0\) as the upper limit.</span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \({H_0}:\mu = 100;{H_1}:\mu < 100\) <strong><em> A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overline x = 98.31,{S_{n - 1}} = 7.8446 \ldots \) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(p\)-value \( = 0.0168\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) the farm manager’s claim is accepted because \(0.0168 > 0.01\) <em><strong>A1R1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>The discrete random variable <em>X</em> follows a geometric distribution Geo(\(p\)) where</p>
<p style="text-align: center;">\({\text{P}}\left( {X = x} \right) = \left\{ {\begin{array}{*{20}{c}}<br> {p{q^{x - 1}},\,{\text{for}}\,x = 1,\,2 \ldots } \\ <br> {0,\,{\text{otherwise}}} <br>\end{array}} \right.\)</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the probability generating function of <em>X</em> is given by</p>
<p>\[G\left( t \right) = \frac{{pt}}{{1 - qt}}.\]</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Deduce that \({\text{E}}\left( X \right) = \frac{1}{p}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Two friends A and B play a ball game with the following rules.</p>
<p>Each player starts with zero points. Player A serves first and then the players have alternate pairs of serves so that the service order is A, B, B, A, A, … When player A serves, the probability of her scoring 1 point is \({p_A}\) and the probability of B scoring 1 point is \({q_A}\), where \({q_A} = 1 - {p_A}\).</p>
<p>When player B serves, the probability of her scoring 1 point is \({p_B}\) and the probability of A scoring 1 point is \({q_B}\), where \({q_B} = 1 - {p_B}\).</p>
<p>Show that, after the first 6 serves, the probability that each player has 3 points is</p>
<p>\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}<br> 3 \hfill \\<br> x \hfill \\ <br>\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\).</p>
<p> </p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>After 6 serves the score is 3 points each. Play continues and the game ends when one player has scored two more points than the other player. Let <em>N</em> be the number of further serves required before the game ends. Given that \({p_A}\) = 0.7 and \({p_A}\) = 0.6 find P(<em>N</em> = 2).</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let \(M = \frac{1}{2}N\). Show that <em>M</em> has a geometric distribution and hence find the value of E(<em>N</em>).</p>
<div class="marks">[7]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\({\text{P}}\left( {X = x} \right) = p{q^{x - 1}},\,{\text{for}}\,x = 1,\,2 \ldots \)</p>
<p>\(G\left( t \right) = \sum\limits_{x = 1}^\infty {{t^x}} p{q^{x - 1}}\) <em><strong>M1</strong></em></p>
<p>\( = pt\sum\limits_{x = 1}^\infty {{{\left( {tq} \right)}^{x - 1}}} \) <em><strong>A1</strong></em></p>
<p>\( = pt\left( {1 + tq + {{\left( {tq} \right)}^2} \ldots } \right)\) <em><strong>M1</strong></em></p>
<p>\( = \frac{{pt}}{{1 - tq}}\) <em><strong> AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(G'\left( t \right) = \frac{{\left( {1 - tq} \right)p - pt\left( { - q} \right)}}{{{{\left( {1 - tq} \right)}^2}}}\) <em><strong>M1A1</strong></em></p>
<p>\({\text{E}}\left( X \right) = G'\left( 1 \right)\) <em><strong>M1</strong></em></p>
<p>\( = \frac{{\left( {1 - q} \right)p + pq}}{{{{\left( {1 - q} \right)}^2}}}\) <em><strong>A1</strong></em></p>
<p>\( = \frac{1}{p}\) <em><strong> AG</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>after 6 serves (3 serves each) we have <em>ABBAAB</em></p>
<p><em>A</em> serves <em>B</em> serves</p>
<p>3 wins 0 losses \({p_1} = {}^3{C_3}p_A^3q_A^0{}^3{C_0}p_B^3q_B^0\) <em><strong>M1A1</strong></em></p>
<p>2 wins 1 loss \({p_2} = {}^3{C_2}p_A^2q_A^1{}^3{C_1}p_B^2q_B^1\) <em><strong>A1</strong></em></p>
<p>1 win 2 losses \({p_3} = {}^3{C_1}p_A^1q_A^2{}^3{C_2}p_B^1q_B^2\) <em><strong>A1</strong></em></p>
<p>0 wins 3 losses \({p_4} = {}^3{C_0}p_A^0q_A^3{}^3{C_3}p_B^0q_B^3\)<em><strong> A1</strong></em></p>
<p>since \({}^3{C_0} = {}^3{C_3},\,\,{}^3{C_1} = {}^3{C_2}\)</p>
<p>\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}<br> 3 \hfill \\<br> x \hfill \\ <br>\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\) <em><strong>AG</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>for <em>N</em> = 2 serves are <em>B</em>, <em>A</em> respectively</p>
<p>P(<em>N</em> = 2) = P(<em>B</em> wins twice) + P(<em>A</em> wins twice) <em><strong> (M1)</strong></em></p>
<p>= 0.6 × 0.3 + 0.4 × 0.7 <em><strong>A1</strong></em></p>
<p>= 0.46 <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>for \(M = \frac{1}{2}N\)</p>
<p>\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M}\) <em><strong>M1</strong></em></p>
<p>\({\text{P}}\left( {M = 2} \right) = {\text{P}}\left( {N = 4} \right)\)</p>
<p>\( = {\text{P}}\left( {\begin{array}{*{20}{c}}<br> {{\text{game does not end after}}} \\ <br> {{\text{first two serves}}} <br>\end{array}} \right) \times {\text{P}}\left( {\begin{array}{*{20}{c}}<br> {{\text{game ends after}}} \\ <br> {{\text{next two serves}}} <br>\end{array}} \right) = \left( {1 - {p_M}} \right){p_M}\) <em><strong>A1</strong></em></p>
<p>similarly \({\text{P}}\left( {M = 3} \right) = {\left( {1 - {p_M}} \right)^2}{p_M}\) <em><strong>(A1)</strong></em></p>
<p>hence \({\text{P}}\left( {M = r} \right) = {\left( {1 - {p_M}} \right)^{r - 1}}{p_M}\) <em><strong>A1</strong></em></p>
<p>hence <em>M</em> has a geometric distribution <em><strong>AG</strong></em></p>
<p>\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M} = 0.46\) <em><strong>A1</strong></em></p>
<p>hence \({\text{E}}\left( M \right) = \frac{1}{p} = \frac{1}{{0.46}} = 2.174\)</p>
<p>\({\text{E}}\left( N \right) = {\text{E}}\left( {2M} \right) = 2{\text{E}}\left( M \right)\) <em><strong>M1</strong></em></p>
<p>= 4.35 <em><strong>A1</strong></em></p>
<p><em><strong>[7 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p>The independent random variables <em>X</em> and <em>Y</em> are given by <em>X</em> ~ N\(\left( {{\mu _1},\,\sigma _1^2} \right)\) and <em>Y</em> ~ N\(\left( {{\mu _2},\,\sigma _2^2} \right)\).</p>
</div>
<div class="specification">
<p>Two independent random variables <em>X</em><sub>1</sub> and <em>X</em><sub>2</sub> each have a normal distribution with a mean 3 and a variance 9. Four independent random variables <em>Y</em><sub>1</sub>, <em>Y</em><sub>2</sub>, <em>Y</em><sub>3</sub>, <em>Y</em><sub>4</sub> each have a normal distribution with mean 2 and variance 25. Each of the variables <em>Y</em><sub>1</sub>, <em>Y</em><sub>2</sub>, <em>Y</em><sub>3</sub>, <em>Y</em><sub>4</sub> is independent of each of the variables <em>X</em><sub>1</sub>, <em>X</em><sub>2</sub>. Find</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down the distribution of <em>aX</em> + <em>bY</em> where <em>a</em>, <em>b </em>\( \in \mathbb{R}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>P(<em>X</em><sub>1</sub> + <em>Y</em><sub>1</sub> < 11).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>P(3<em>X</em><sub>1</sub> + 4<em>Y</em><sub>1</sub> > 15).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>P(<em>X</em><sub>1</sub> + <em>X</em><sub>2</sub> + <em>Y</em><sub>1</sub> + <em>Y</em><sub>2</sub> + <em>Y</em><sub>3</sub> + <em>Y</em><sub>4</sub> < 30).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that \({\bar X}\) and \({\bar Y}\) are the respective sample means, find \({\text{P}}\left( {\bar X > \bar Y} \right)\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><em>aX</em> + <em>bY</em> ~ N\(\left( {a{\mu _1} + b{\mu _2},\,\,{a^2}\sigma _1^2 + {b^2}\sigma _2^2} \right)\) <em><strong>A1A1</strong></em></p>
<p><strong>Note:</strong> <em><strong>A1</strong> </em>for N and the mean, <em><strong>A1</strong> </em>for the variance.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><em>X</em><sub>1</sub> + <em>Y</em><sub>1 </sub>∼ N(5,34) <em><strong>(A1)(A1)</strong></em></p>
<p>⇒ P(<em>X</em><sub>1</sub> + <em>Y</em><sub>1</sub> < 11) = 0.848 <em><strong> A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>3<em>X</em><sub>1</sub> + 4<em>Y</em><sub>1</sub><sub> </sub>∼ N(9 + 8, 9 × 9 + 16 × 25) <em><strong>(A1)(M1)(A1)</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>(A1)</strong></em> for correct expectation, <em><strong>(M1)(A1)</strong></em> for correct variance.</p>
<p>∼ N(17, 481)</p>
<p>⇒ P(3<em>X</em><sub>1</sub> + 4<em>Y</em><sub>1</sub> > 15) = 0.536 <em><strong> A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><em>X</em><sub>1</sub> + <em>X</em><sub>2</sub> + <em>Y</em><sub>1</sub> + <em>Y</em><sub>2</sub> + <em>Y</em><sub>3</sub> + <em>Y</em><sub>4</sub><sub> </sub>∼ N(6 + 8, 2 × 9 + 4 × 25) <em><strong>(A1)(A1)</strong></em></p>
<p>∼ N(14, 118)</p>
<p>⇒ P(<em>X</em><sub>1</sub> + <em>X</em><sub>2</sub> + <em>Y</em><sub>1</sub> + <em>Y</em><sub>2</sub> + <em>Y</em><sub>3</sub> + <em>Y</em><sub>4</sub> < 30) = 0.930 <em><strong> A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider \(\bar X - \bar Y\) <em><strong>(M1)</strong></em></p>
<p>\({\text{E}}\left( {\bar X - \bar Y} \right) = 3 - 2 = 1\) <em><strong> A1</strong></em></p>
<p>\({\text{Var}}\left( {\bar X - \bar Y} \right) = \frac{9}{2} + \frac{{25}}{4}\left( { = 10.75} \right)\) <em><strong>(M1)A1</strong></em></p>
<p>\( \Rightarrow {\text{P}}\left( {\bar X - \bar Y > 0} \right) = 0.620\) <em><strong> A1</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">An automatic machine is used to fill bottles of water. The amount delivered, \(</span><span style="font-family: times new roman,times; font-size: medium;">Y\) ml , may be assumed to be normally distributed with mean \(\mu \) ml and standard </span><span style="font-family: times new roman,times; font-size: medium;">deviation \(8\) ml . Initially, the machine is adjusted so that the value of \(\mu \) is \(500\). </span><span style="font-family: times new roman,times; font-size: medium;">In order to check that the value of \(\mu \) remains equal to \(500\), a random sample </span><span style="font-family: times new roman,times; font-size: medium;">of 10 bottles is selected at regular intervals, and the mean amount of water, \(\overline y \) , </span><span style="font-family: times new roman,times; font-size: medium;">in these bottles is calculated. The following hypotheses are set up.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{H}}_0}:\mu = 500\) ; \({{\rm{H}}_1}:\mu \ne 500\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The critical region is defined to be \(\left( {\overline y < 495} \right) \cup \left( {\overline y > 505} \right)\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the significance level of this procedure.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Some time later, the actual value of \(\mu \) is \(503\). Find the probability of a </span><span style="font-family: times new roman,times; font-size: medium;">Type II error.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Under \({{\rm{H}}_0}\) , the distribution of \({\overline y }\) is N(500, 6.4) . <strong><em> (A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Significance level \( = {\rm{P}}\overline y < 495\) or \( > 505|{{\rm{H}}_0}\) <em><strong>M2</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = 2 \times 0.02405\) <strong><em>(A1)</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.0481\) <em><strong>A1 N5</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Using tables, answer is \(0.0478\).</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) The distribution of \(\overline y \) is now N(\(503\), \(6.4\)) . <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">P(Type ΙΙ error) \( = {\rm{P}}(495 < \overline y < 505)\) <strong><em> (M1)</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.785\) <em><strong>A1 N3</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Using tables, answer is \(0.784\).</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The function \(f\) is defined by \(f(x) = \ln (1 + \sin x)\) .</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">When a scientist measures the concentration \(\mu \) of a solution, the measurement </span><span style="font-family: times new roman,times; font-size: medium;">obtained may be assumed to be a normally distributed random variable with mean </span><span style="font-family: times new roman,times; font-size: medium;">\(\mu \) and standard deviation \(1.6\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(f''(x) = \frac{{ - 1}}{{1 + \sin x}}\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Determine the Maclaurin series for \(f(x)\) as far as the term in \({x^4}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Deduce the Maclaurin series for \(\ln (1 - \sin x)\) as far as the term in \({x^4}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">By combining your two series, show that \(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence, or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">He makes 5 independent measurements of the concentration of a particular </span><span style="font-family: times new roman,times; font-size: medium;">solution and correctly calculates the following confidence interval for \(\mu \) .</span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">[\(22.7\) , \(26.1\)]</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Determine the confidence level of this interval.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">He is now given a different solution and is asked to determine a \(95\%\) confidence </span><span style="font-family: times new roman,times; font-size: medium;">interval for its concentration. The confidence interval is required to have a width </span><span style="font-family: times new roman,times; font-size: medium;">less than \(2\). Find the minimum number of independent measurements required.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{ - \sin x - 1}}{{{{(1 + \sin x)}^2}}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{ - 1}}{{1 + \sin x}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">AG</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [4 marks]</span></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({f^{iv}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(0) = 0\) , \(f'(0) = 1\) , \(f''(0) = - 1\) , \(f'''(0) = 1\) , \({f^{iv}}(0) = - 2\) <strong><em>(A2)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for 2 errors and <em><strong>A0</strong></em> for more than 2 errors.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 + \sin x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>M1A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [6 marks]</span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 - \sin x) = \ln (1 + \sin ( - x)) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) <em><strong>M1A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Adding, <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 - {\sin ^2}x) = \ln {\cos ^2}x\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = - {x^2} - \frac{{{x^4}}}{6} + \ldots \) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln \cos x = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} + \ldots \) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Limit \( = 0\) <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">A.e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Interval width \( = 26.1 - 22.7 = 3.4\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">So \(3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(z = 2.375 \ldots \) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Probability \( = 0.9912\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Confidence level \( = 2 \times 0.4912 = 98.2\% \) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(z\)-value \( = 1.96\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">We require</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Whence \(n > 9.83\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">So we need \(n = 10\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept \( = \) signs throughout.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The random variable \(X\) has probability density function given by</p>
<p class="p1">\[f(x) = \left\{ {\begin{array}{*{20}{l}}<br> {x{{\text{e}}^{ - x}},}&{{\text{for }}x \geqslant 0,} \\ <br> {0,}&{{\text{otherwise}}} <br>\end{array}} \right..\]</p>
</div>
<div class="specification">
<p class="p1">A sample of size <span class="s1">50 </span>is taken from the distribution of \(X\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find \({\text{E}}({X^2})\)<span class="s1">.</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>Show that \({\text{Var}}(X) = 2\).</p>
<div class="marks">[10]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the central limit theorem.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the probability that the sample mean is less than <span class="s1">2.3</span><span class="s2">.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">attempt to apply l’Hôpital’s rule <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)</p>
<p class="p2">then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 0\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ - x}}{\text{d}}x} \)</span> <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">attempt at integration by parts <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">the integral \( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ - x}}{\text{d}}x\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ - x}}{\text{d}}x} \) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ - x}}{\text{d}}x} \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + [ - 6{{\text{e}}^{ - x}}]_0^R\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s2">\( = 6\) </span>when \(R \to \infty \) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> \({\text{E}}(X) = 2\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2} = 6 - {2^2}\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 2\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong><em>[10 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">if a random sample of size \(n\) is taken from <strong><em>any </em></strong>distribution \(X\), <span class="s1">with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), </span>then, for <span class="s1"><strong><em>large n</em></strong></span>, <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\) </span><strong><em>(A1)</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b) the infinite upper limit was rarely treated rigorously.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In answering part (c) many failed to say that the Central Limit Theorem is valid for large samples and for any initial distribution. The parameters of the distribution were often not stated.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">In a large population of sheep, their weights are normally distributed with mean \(\mu \) <span class="s1">kg </span>and standard deviation \(\sigma \) <span class="s1">kg. A random sample of \(100\) </span>sheep is taken from the population.</p>
<p class="p1">The mean weight of the sample is \(\bar X\) <span class="s1">kg</span>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the distribution of \(\bar X\) , giving its mean and standard deviation.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) <span class="s1">kg </span>is the weight of a sheep.</p>
<p class="p2">(i) <span class="Apple-converted-space"> </span>Find unbiased estimates for<span class="s2"> \(\mu \) </span>and<span class="s2"> \({\sigma ^2}\)</span><span class="s3">.</span></p>
<p class="p2"><span class="s3">(ii) <span class="Apple-converted-space"> </span>Find a \(95\%\) </span>confidence interval for<span class="s2"> \(\mu \)</span>.</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Test, at the \(1\%\) level of significance, the null hypothesis \(\mu = 35\) against the alternative hypothesis that \(\mu > 35\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1"><strong>Note: </strong>Award <strong><em>A1 </em></strong>for \(N\), <strong><em>A1 </em></strong>for the parameters.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>\(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(s_{n - 1}^2 = \frac{{155341}}{{99}} - \frac{{{{3782}^2}}}{{9900}} = 124\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>\(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \) <span class="Apple-converted-space"> </span><strong><em>(M1)(A1)</em></strong></p>
<p class="p1">\( = (35.6,{\text{ }}40.0)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">one tailed t-test <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">testing <span class="s1">\(37.82\) <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1"><span class="s1">\(99\) </span>degrees of freedom</p>
<p class="p1">reject if \(t > 2.36\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="s1"><em>t</em></span>-value being tested is <span class="s1">\(2.5294\) <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">one tailed t-test <span class="Apple-converted-space"> </span><strong>(<em>A1)</em></strong></p>
<p class="p1">\(p = 0.00650\) <span class="Apple-converted-space"> </span><strong><em>A3</em></strong></p>
<p class="p1">since \(p{\text{ - value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The random variable \(X\) has the binomial distribution \({\text{B}}(n,{\text{ }}p)\), where \(n > 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\frac{X}{n}\) is an unbiased estimator for \(p\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({\left( {\frac{X}{n}} \right)^2}\) is <strong>not </strong>an unbiased estimator for \({p^2}\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \(\frac{{X(X - 1)}}{{n(n - 1)}}\) is an unbiased estimator for \({p^2}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \({\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}(X)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{n} \times np = p\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore unbiased <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({\text{E}}\left[ {{{\left( {\frac{X}{n}} \right)}^2}} \right] = \frac{1}{{{n^2}}}\left( {{\text{Var}}(X) + {{[{\text{E}}(X)]}^2}} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{{{n^2}}}\left( {np(1 - p) + {n^2}{p^2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \ne {p^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore not unbiased <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({\text{E}}\left[ {\left( {\frac{{X(X - 1)}}{{n(n - 1)}}} \right)} \right] = \frac{{{\text{E}}({X^2}) - {\text{E}}(X)}}{{n(n - 1)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{np(1 - p) + {n^2}{p^2} - np}}{{n(n - 1)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{n{p^2}(n - 1)}}{{n(n - 1)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {p^2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore unbiased <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The continuous random variable \(X\) takes values only in the interval [\(a\), \(b\)] and \(F\) </span><span style="font-family: times new roman,times; font-size: medium;">denotes its cumulative distribution function. Using integration by parts, show that:\[E(X) = b - \int_a^b {F(x){\rm{d}}x}. \]<br></span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The continuous random variable \(Y\) has probability density function \(f\) given by:\[\begin{array}{*{20}{c}}<br> {f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\ <br> {f(y) = 0,}&{{\text{elsewhere}}{\text{.}}} <br>\end{array}\]</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) Obtain an expression for the cumulative distribution function of \(Y\) , valid </span><span style="font-family: times new roman,times; font-size: medium;">for \(0 \le y \le \frac{\pi }{2}\) . Use the result in (a) to determine \(E(Y)\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) The random variable \(U\) is defined by \(U = {Y^n}\) , where \(n \in {\mathbb{Z}^ + }\) . Obtain </span><span style="font-family: times new roman,times; font-size: medium;">an expression for the cumulative distribution function of \(U\) valid for \(0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}\) .</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) The medians of \(U\) and \(Y\) are denoted respectively by \({m_u}\) and \({m_y}\) . </span><span style="font-family: times new roman,times; font-size: medium;">Show that \({m_u} = m_y^n\) .</span></p>
<div class="marks">[14]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(E(X) = \int_a^b {xf(x){\rm{d}}x} \) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = \left[ {xF(x)} \right]_a^b - \int_a^b {F(x){\rm{d}}x} \) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = bF(b) - aF(a) - \int_a^b {F(x){\rm{d}}x} \) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = b - \int_a^b {F(x){\rm{d}}x} \)</span><span style="font-family: times new roman,times; font-size: medium;"> because \(F(a) = 0\) and \(F(b) = 1\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) let \(G\) denote the cumulative distribution function of \(Y\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(G(y) = \int_0^y {\cos t{\rm{d}}t} \) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \left[ {\sin t} \right]_0^y\) <strong><em>(A1)</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \sin y\) <em><strong> A1</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(E(Y) = \frac{\pi }{2} - \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y} \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{\pi }{2} - 1\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) CDF of \(U = P(U \le u)\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = P({Y^n} \le u)\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = P({Y^{}} \le {u^{\frac{1}{n}}})\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = G({u^{\frac{1}{n}}})\) <strong><em>(A1)</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \sin \left( {{u^{\frac{1}{n}}}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) \({m_y}\) satisfies the equation \(\sin {m_y} = \frac{1}{2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({m_u}\) satisfies the equation \(\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore \({m_y} = m_u^{\frac{1}{n}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({m_u} = m_y^n\) <strong><em>AG</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [14 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solutions to (a) were often unconvincing. Candidates were expected to include in their solution the fact that \(F(a) = 0\) and \(F(b) = 1\) . </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In (b) (i), it was not enough to state that \(G(y) = \int {\cos y{\rm{d}}y = \sin y} \) although that, fortuitously, gave the correct answer on this occasion. The correct approach was either to state that \(G(y) = \int_0^y {\cos t{\rm{d}}t} = \sin y\) or that \(G(y) = \int {\cos y{\rm{d}}y} = \sin x + C\) and then show that \(C = 0\) because \(F(0) = 0\) or \(F\left( {\frac{\pi }{2}} \right) = 1\) . Solutions to (b) (ii) and (iii) were often disappointing, giving the impression that many of the candidates were not familiar with dealing with cumulative distribution functions. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A random variable \(X\) has probability density function \(f\) given by:\[f(x) = \left\{ {\begin{array}{*{20}{l}}<br> {\lambda {e^{ - \lambda x}},}&{{\text{for }}x \geqslant 0{\text{ where }}\lambda > 0} \\ <br> {0,}&{{\text{for }}x < 0.} <br>\end{array}} \right.\]</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A chicken crosses a road. It is known that cars pass the chicken’s crossing </span><span style="font-family: times new roman,times; font-size: medium;">route, with intervals between cars measured in seconds, according to the random </span><span style="font-family: times new roman,times; font-size: medium;">variable \(X\) , with \(\lambda = 0.03\) . The chicken, which takes \(10\) seconds to cross the </span><span style="font-family: times new roman,times; font-size: medium;">road, starts to cross just as one car passes.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the probability that the chicken will reach the other side of the road </span><span style="font-family: times new roman,times; font-size: medium;">before the next car arrives.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Later, the chicken crosses the road again just after a car has passed.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Show that the probability that the chicken completes both crossings is </span><span style="font-family: times new roman,times; font-size: medium;">greater than \(0.5\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A rifleman shoots at a circular target. The distance in centimetres from the </span><span style="font-family: times new roman,times; font-size: medium;">centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . </span><span style="font-family: times new roman,times; font-size: medium;">The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if </span><span style="font-family: times new roman,times; font-size: medium;">\(5 < X \le 10\) and no points if \(X > 10\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the expected score when one bullet is fired at the target.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his </span><span style="font-family: times new roman,times; font-size: medium;">value of \(\lambda \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Given that his expected score is \(6.5\), find his value of \(\lambda \) .</span></p>
<div class="marks">[10]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({\rm{P}}(X > a) = \int_a^\infty {\lambda {e^{ - \lambda x}}{\rm{d}}x} \) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ { - {e^{ - \lambda x}}} \right]_a^\infty \) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {e^{ - \lambda a}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{P}}(X > 10) = {e^{ - 0.3}}( = 0.74 \ldots )\) <strong><em> (M1)A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) probability of a safe double crossing \( = {e^{ - 0.6}}\) \(( = {0.74^2})\) \( = 0.55\) <em><strong> A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">which is greater than \(0.5\) <em><strong>AG </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong> </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks] </span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({\rm{P}}(X \le 1) = 0.3296 \ldots \) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \) <strong><em>(A1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{E(score)}} = 10 \times 0.3296 \ldots + 5 \times 0.5349 \ldots + 1 \times 0.1170 \ldots \) <strong><em>M1A1</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 6.09\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept probabilities in exponential form until the final decimal </span><span style="font-family: times new roman,times; font-size: medium;">answer. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{E(score)}}\) for <strong><em>X</em></strong> with unknown parameter can be expressed as \(10 \times (1 - {e^{ - \lambda }}) + 5 \times ({e^{ - \lambda }} - {e^{ - 5\lambda }}) + ({e^{ - 5\lambda }} - {e^{ - 10\lambda }})\) <strong><em>(M1)(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve \({\rm{E(score)}} = 6.5\) <strong><em> (M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">obtain \(\lambda = 0.473\) <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[10 marks] </span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was generally well done.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was generally well done.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The discrete random variable \(X\) follows the distribution Geo(\(p\)).</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Arthur tosses a biased coin each morning to decide whether to walk or cycle to school; </span><span style="font-family: times new roman,times; font-size: medium;">he walks if the coin shows a head.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The probability of obtaining a head is \(0.55\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Write down the mode of \(X\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the exact value of \(p\) if \({\rm{Var}}(X) = \frac{{28}}{9}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the smallest value of \(n\) for which the probability of Arthur walking to </span><span style="font-family: times new roman,times; font-size: medium;">school on the next \(n\) days is less than \(0.01\).</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the probability that Arthur cycles to school for the third time on the </span><span style="font-family: times new roman,times; font-size: medium;">last of eight successive days.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the mode is \(1\) <em><strong>A1</strong> </em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) attempt to solve \(\frac{{1 - p}}{{{p^2}}} = \frac{{28}}{9}\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>M1</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">obtain \(p = \frac{3}{7}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1 </span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: </span><span style="font-family: times new roman,times; font-size: medium;">\(p = 0.429\) </span><span style="font-family: times new roman,times; font-size: medium;">is awarded <strong><em>M1A0</em></strong>. </span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) require least \(n\) such that </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({0.55^n} < 0.01\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>(M1)</em> </span></strong></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">EITHER </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">listing values: \(0.55\), \(0.3025\), \(0.166\), \(0.091\), \(0.050\), \(0.028\), \(0.015\), \(0.0084\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">obtain \(n = 8\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1</em> </span></strong></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">OR </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(n > \frac{{\ln 0.01}}{{\ln 0.55}} = 7.70 \ldots \) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>(M1)</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">obtain </span><span style="font-family: times new roman,times; font-size: medium;">\(n = 8\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) recognition of negative binomial <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(X \sim NB(3,0.45)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}(X = 8) = \left( \begin{array}{l}<br>7\\<br>2<br>\end{array} \right) \times {0.45^3} \times {0.55^5}\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>(A1)</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 0.0963\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 </span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If \(0.45\) and \(0.55\) are mixed up, count it as a misread – probability in that case is \(0.0645\). </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[</span><span style="font-family: times new roman,times; font-size: medium;">6 marks] </span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(a)(i) A surprising number of candidates were unaware of the definition of the mode of a distribution.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(a)(ii) Generally well done, although a few candidates gave a decimal answer.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(b) Generally well </span><span style="font-family: times new roman,times; font-size: medium;">done, and it was pleasing that most were familiar with the direct use of the negative binomial </span><span style="font-family: times new roman,times; font-size: medium;">distribution in (ii).</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br>