Question
A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.
Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{ .}}\]
(i) Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .
(ii) Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .
State and justify an appropriate test procedure giving the null and alternate hypotheses.
What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?
If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?
Answer/Explanation
Markscheme
\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) M1A1
\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) A1
[3 marks]
(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) M1
\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) A1
\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} – \ldots } \right)\) R1A1
(ii) \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} – \ldots } \right)\) M1
\( = 0.38292 = 0.383\) A1
[6 marks]
this is a two tailed test of the sample mean \(\overline x \)
we use the central limit theorem to justify assuming that R1
\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1A1
\({{\rm{H}}_0}:\mu = 28\) A1
\({{\rm{H}}_1}:\mu \ne 28\) A1
[5 marks]
since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) (M1)A1
and (\(\overline x \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) (M1)(A1)(A1)
\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\)
so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) A1A1
[7 marks]
if \(\mu = 28.1\)
\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1
\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)
\( = 0.884\) A1
Note: Depending on the degree of accuracy used for the critical region the answer for part (c) can be anywhere from \(0.8146\) to \(0.879\).
[2 marks]
Question
The continuous random variable \(X\) takes values only in the interval [\(a\), \(b\)] and \(F\) denotes its cumulative distribution function. Using integration by parts, show that:\[E(X) = b – \int_a^b {F(x){\rm{d}}x}. \]
The continuous random variable \(Y\) has probability density function \(f\) given by:\[\begin{array}{*{20}{c}}
{f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\
{f(y) = 0,}&{{\text{elsewhere}}{\text{.}}}
\end{array}\]
(i) Obtain an expression for the cumulative distribution function of \(Y\) , valid for \(0 \le y \le \frac{\pi }{2}\) . Use the result in (a) to determine \(E(Y)\) .
(ii) The random variable \(U\) is defined by \(U = {Y^n}\) , where \(n \in {\mathbb{Z}^ + }\) . Obtain an expression for the cumulative distribution function of \(U\) valid for \(0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}\) .
(iii) The medians of \(U\) and \(Y\) are denoted respectively by \({m_u}\) and \({m_y}\) . Show that \({m_u} = m_y^n\) .
Answer/Explanation
Markscheme
\(E(X) = \int_a^b {xf(x){\rm{d}}x} \) M1
\( = \left[ {xF(x)} \right]_a^b – \int_a^b {F(x){\rm{d}}x} \) A1
\( = bF(b) – aF(a) – \int_a^b {F(x){\rm{d}}x} \) A1
\( = b – \int_a^b {F(x){\rm{d}}x} \) because \(F(a) = 0\) and \(F(b) = 1\) A1
[4 marks]
(i) let \(G\) denote the cumulative distribution function of \(Y\)
\(G(y) = \int_0^y {\cos t{\rm{d}}t} \) M1
\( = \left[ {\sin t} \right]_0^y\) (A1)
\( = \sin y\) A1
\(E(Y) = \frac{\pi }{2} – \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y} \) M1
\( = \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}\) A1
\( = \frac{\pi }{2} – 1\) A1
(ii) CDF of \(U = P(U \le u)\) M1
\( = P({Y^n} \le u)\) A1
\( = P({Y^{}} \le {u^{\frac{1}{n}}})\) A1
\( = G({u^{\frac{1}{n}}})\) (A1)
\( = \sin \left( {{u^{\frac{1}{n}}}} \right)\) A1
(iii) \({m_y}\) satisfies the equation \(\sin {m_y} = \frac{1}{2}\) A1
\({m_u}\) satisfies the equation \(\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}\) A1
therefore \({m_y} = m_u^{\frac{1}{n}}\) A1
\({m_u} = m_y^n\) AG
[14 marks]
Question
A random variable \(X\) has probability density function \(f\) given by:\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{\lambda {e^{ – \lambda x}},}&{{\text{for }}x \geqslant 0{\text{ where }}\lambda > 0} \\
{0,}&{{\text{for }}x < 0.}
\end{array}} \right.\]
(i) Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .
A chicken crosses a road. It is known that cars pass the chicken’s crossing route, with intervals between cars measured in seconds, according to the random variable \(X\) , with \(\lambda = 0.03\) . The chicken, which takes \(10\) seconds to cross the road, starts to cross just as one car passes.
(ii) Find the probability that the chicken will reach the other side of the road before the next car arrives.
Later, the chicken crosses the road again just after a car has passed.
(iii) Show that the probability that the chicken completes both crossings is greater than \(0.5\).
A rifleman shoots at a circular target. The distance in centimetres from the centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if \(5 < X \le 10\) and no points if \(X > 10\) .
(i) Find the expected score when one bullet is fired at the target.
A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his value of \(\lambda \) .
(ii) Given that his expected score is \(6.5\), find his value of \(\lambda \) .
Answer/Explanation
Markscheme
(i) \({\rm{P}}(X > a) = \int_a^\infty {\lambda {e^{ – \lambda x}}{\rm{d}}x} \) M1
\(\left[ { – {e^{ – \lambda x}}} \right]_a^\infty \) A1
\( = {e^{ – \lambda a}}\) A1
(ii) \({\rm{P}}(X > 10) = {e^{ – 0.3}}( = 0.74 \ldots )\) (M1)A1
(iii) probability of a safe double crossing \( = {e^{ – 0.6}}\) \(( = {0.74^2})\) \( = 0.55\) A1
which is greater than \(0.5\) AG
[6 marks]
(i) \({\rm{P}}(X \le 1) = 0.3296 \ldots \) (A1)
\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \) (A1)
\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \) (A1)
\({\rm{E(score)}} = 10 \times 0.3296 \ldots + 5 \times 0.5349 \ldots + 1 \times 0.1170 \ldots \) M1A1
\( = 6.09\) A1
Note: Accept probabilities in exponential form until the final decimal answer.
(ii) \({\rm{E(score)}}\) for X with unknown parameter can be expressed as \(10 \times (1 – {e^{ – \lambda }}) + 5 \times ({e^{ – \lambda }} – {e^{ – 5\lambda }}) + ({e^{ – 5\lambda }} – {e^{ – 10\lambda }})\) (M1)(A1)
attempt to solve \({\rm{E(score)}} = 6.5\) (M1)
obtain \(\lambda = 0.473\) A1
[10 marks]
Examiners report
This question was generally well done.
This question was generally well done.
Question
The discrete random variable X follows a geometric distribution Geo(\(p\)) where
\({\text{P}}\left( {X = x} \right) = \left\{ {\begin{array}{*{20}{c}}
{p{q^{x – 1}},\,{\text{for}}\,x = 1,\,2 \ldots } \\
{0,\,{\text{otherwise}}}
\end{array}} \right.\)
Show that the probability generating function of X is given by
\[G\left( t \right) = \frac{{pt}}{{1 – qt}}.\]
Deduce that \({\text{E}}\left( X \right) = \frac{1}{p}\).
Two friends A and B play a ball game with the following rules.
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 \({q_A} = 1 – {p_A}\).
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 \({q_B} = 1 – {p_B}\).
Show that, after the first 6 serves, the probability that each player has 3 points is
\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 – x}}{\left( {{q_B}} \right)^{3 – x}}\).
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 N be the number of further serves required before the game ends. Given that \({p_A}\) = 0.7 and \({p_A}\) = 0.6 find P(N = 2).
Let \(M = \frac{1}{2}N\). Show that M has a geometric distribution and hence find the value of E(N).
Answer/Explanation
Markscheme
\({\text{P}}\left( {X = x} \right) = p{q^{x – 1}},\,{\text{for}}\,x = 1,\,2 \ldots \)
\(G\left( t \right) = \sum\limits_{x = 1}^\infty {{t^x}} p{q^{x – 1}}\) M1
\( = pt\sum\limits_{x = 1}^\infty {{{\left( {tq} \right)}^{x – 1}}} \) A1
\( = pt\left( {1 + tq + {{\left( {tq} \right)}^2} \ldots } \right)\) M1
\( = \frac{{pt}}{{1 – tq}}\) AG
[3 marks]
\(G’\left( t \right) = \frac{{\left( {1 – tq} \right)p – pt\left( { – q} \right)}}{{{{\left( {1 – tq} \right)}^2}}}\) M1A1
\({\text{E}}\left( X \right) = G’\left( 1 \right)\) M1
\( = \frac{{\left( {1 – q} \right)p + pq}}{{{{\left( {1 – q} \right)}^2}}}\) A1
\( = \frac{1}{p}\) AG
[4 marks]
after 6 serves (3 serves each) we have ABBAAB
A serves B serves
3 wins 0 losses \({p_1} = {}^3{C_3}p_A^3q_A^0{}^3{C_0}p_B^3q_B^0\) M1A1
2 wins 1 loss \({p_2} = {}^3{C_2}p_A^2q_A^1{}^3{C_1}p_B^2q_B^1\) A1
1 win 2 losses \({p_3} = {}^3{C_1}p_A^1q_A^2{}^3{C_2}p_B^1q_B^2\) A1
0 wins 3 losses \({p_4} = {}^3{C_0}p_A^0q_A^3{}^3{C_3}p_B^0q_B^3\) A1
since \({}^3{C_0} = {}^3{C_3},\,\,{}^3{C_1} = {}^3{C_2}\)
\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 – x}}{\left( {{q_B}} \right)^{3 – x}}\) AG
[5 marks]
for N = 2 serves are B, A respectively
P(N = 2) = P(B wins twice) + P(A wins twice) (M1)
= 0.6 × 0.3 + 0.4 × 0.7 A1
= 0.46 A1
[3 marks]
for \(M = \frac{1}{2}N\)
\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M}\) M1
\({\text{P}}\left( {M = 2} \right) = {\text{P}}\left( {N = 4} \right)\)
\( = {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game does not end after}}} \\
{{\text{first two serves}}}
\end{array}} \right) \times {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game ends after}}} \\
{{\text{next two serves}}}
\end{array}} \right) = \left( {1 – {p_M}} \right){p_M}\) A1
similarly \({\text{P}}\left( {M = 3} \right) = {\left( {1 – {p_M}} \right)^2}{p_M}\) (A1)
hence \({\text{P}}\left( {M = r} \right) = {\left( {1 – {p_M}} \right)^{r – 1}}{p_M}\) A1
hence M has a geometric distribution AG
\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M} = 0.46\) A1
hence \({\text{E}}\left( M \right) = \frac{1}{p} = \frac{1}{{0.46}} = 2.174\)
\({\text{E}}\left( N \right) = {\text{E}}\left( {2M} \right) = 2{\text{E}}\left( M \right)\) M1
= 4.35 A1
[7 marks]
Question
The random variable \(X\) has cumulative distribution function\[F(x) = \left\{ {\begin{array}{*{20}{c}}
0&{x < 0} \\
{{{\left( {\frac{x}{a}} \right)}^3}}&{0 \leqslant x \leqslant a} \\
1&{x > a}
\end{array}} \right.\]where \(a\) is an unknown parameter. You are given that the mean and variance of \(X\) are \(\frac{{3a}}{4}\) and \(\frac{{3{a^2}}}{{80}}\) respectively. To estimate the value of \(a\) , a random sample of \(n\) independent observations, \({X_1},{X_2}, \ldots {X_n}\) is taken from the distribution of \(X\) .
(i) Find an expression for \(c\) in terms of \(n\) such that \(U = c\sum\limits_{i = 1}^n {{X_i}} \) is an unbiased estimator for \(a\) .
(ii) Determine an expression for \({\text{Var}}(U)\) in this case.
(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 probability density function of \(Y\) .
(ii) Find \({\rm{E}}(Y)\) .
(iii) Show that \({\rm{Var}}(Y) = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) .
(iv) Find an expression for \(d\) in terms of \(n\) such that \(V = dY\) is an unbiased estimator for \(a\) .
(v) Determine an expression for \({\rm{Var}}(V)\) in this case.
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 the more efficient estimator for \(a\) .
Answer/Explanation
Markscheme
(i) \({\rm{E}}(U) = c \times n \times \frac{{3a}}{4} = a \Rightarrow c = \frac{4}{{3n}}\) M1A1
(ii) \(Var(U) = \frac{{16}}{{9{n^2}}} \times n \times \frac{{3{a^2}}}{{80}} = \frac{{{a^2}}}{{15n}}\) M1A1
[4 marks]
(i) \({\rm{P}}(Y \le y) = {\rm{P}}({\text{all }}Xs \le y)\) M1
\( = \left[ {\rm{P}} \right.{\left. {(X \le y)} \right]^n}\) (A1)
\( = {\left( {{{\left( {\frac{y}{a}} \right)}^3}} \right)^n}\) (A1)
Note: Only one of the two A1 marks above may be implied.
\( = {\left( {\frac{y}{a}} \right)^{3n}}\) AG
\(g(y) = \frac{{\rm{d}}}{{{\rm{d}}y}}{\left( {\frac{y}{a}} \right)^{3n}} = \frac{{3n{y^{3n – 1}}}}{{{a^{3n}}}},(0 < y < a)\) M1A1
(\(g(y) = 0\) otherwise)
(ii) \({\rm{E}}(Y) = \int_0^a {\frac{{3n{y^{3n}}}}{{{a^{3n}}}}} {\rm{d}}y\) M1
\( = \left[ {\frac{{3n{y^{3n + 1}}}}{{(3n + 1){a^{3n}}}}} \right]_0^a\) A1
\( = \frac{{3na}}{{3n + 1}}\) A1
(iii) \({\rm{Var}}(Y) = \int_0^a {\frac{{3n{y^{3n + 1}}}}{{{a^{3n}}}}} {\rm{d}}y – {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) M1
\( = \left[ {\frac{{3n{y^{3n + 2}}}}{{(3n + 2){a^{3n}}}}} \right]_0^a – {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) A1
\( = \frac{{3n{a^2}}}{{3n + 2}} – \frac{{9{n^2}{a^2}}}{{{{(3n + 1)}^2}}}\) M1
\( = \frac{{3n{a^2}(9{n^2} + 6n + 1) – 9{n^2}{a^2}(3n + 2)}}{{(3n + 2)(3n + 1)}}\) A1
\( = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) AG
(iv) \({\rm{E}}(V) = d \times \frac{{3na}}{{3n + 1}} = a \Rightarrow d = \frac{{3n + 1}}{{3n}}\) M1A1
(v) \(Var(V) = {\left( {\frac{{3n + 1}}{{3n}}} \right)^2} \times \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) M1
\( = \frac{{{a^2}}}{{3n(3n + 2)}}\) A1
[16 marks]
\(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{\frac{{{a^2}}}{{15n}}}}{{\frac{{{a^2}}}{{3n(3n + 2)}}}}\) A1
\( = \frac{{3n + 2}}{5}\) AG
\(V\) is the more efficient estimator because \(3n + 2 > 5\) (for \(n > 1\) ) R1
[2 marks]