IB DP Maths Topic 7.1 Cumulative distribution functions for both discrete and continuous distributions HL Paper 3

 

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Question

The random variable X represents the lifetime in hours of a battery. The lifetime may be assumed to be a continuous random variable X with a probability density function given by \(f(x) = \lambda {{\text{e}}^{ – \lambda x}}\), where \(x \geqslant 0\).

Find the cumulative distribution function, \(F(x)\), of X.

[3]
b.

Find the probability that the lifetime of a particular battery is more than twice the mean.

[2]
c.

Find the median of X in terms of \(\lambda \).

[3]
d.

Find the probability that the lifetime of a particular battery lies between the median and the mean.

[2]
e.
Answer/Explanation

Markscheme

\(\int {\lambda {{\text{e}}^{ – \lambda t}}{\text{d}}t = – {{\text{e}}^{ – \lambda t}}{\text{ }}( + c)} \)     A1

\( \Rightarrow F(x) = \left[ { – {{\text{e}}^{ – \lambda t}}} \right]_0^x\)     (M1)

\( = 1 – {{\text{e}}^{ – \lambda t}}{\text{ }}(x \geqslant 0)\)     A1

[3 marks]

b.

\(1 – F\left( {\frac{2}{\lambda }} \right)\)     M1

\( = {{\text{e}}^{ – 2}}\,\,\,\,\,( = 0.135)\)     A1

[2 marks]

c.

\(F(m) = \frac{1}{2}\)     (M1)

\( \Rightarrow {{\text{e}}^{ – \lambda m}} = \frac{1}{2}\)     A1

\( \Rightarrow – \lambda m = \ln \frac{1}{2}\)

\( \Rightarrow m = \frac{1}{\lambda }\ln 2\)     A1

[3 marks]

d.

\(F\left( {\frac{1}{\lambda }} \right) – F\left( {\frac{{\ln 2}}{\lambda }} \right)\)     M1

\( = \frac{1}{2} – {{\text{e}}^{ – 1}}\,\,\,\,\,( = 0.132)\)     A1

[2 marks]

e.

Examiners report

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ – \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

b.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ – \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

c.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ – \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

d.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ – \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

e.

Question

The continuous random variable X has probability density function f given by

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {2x,}&{0 \leqslant x \leqslant 0.5,} \\
  {\frac{4}{3} – \frac{2}{3}x,}&{0.5 \leqslant x \leqslant 2} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Sketch the function f and show that the lower quartile is 0.5.

[3]
a.

(i)     Determine E(X ).

(ii)     Determine \({\text{E}}({X^2})\).

[4]
b.

Two independent observations are made from X and the values are added.

The resulting random variable is denoted Y .

(i)     Determine \({\text{E}}(Y – 2X)\) .

(ii)     Determine \({\text{Var}}\,(Y – 2X)\).

[5]
c.

(i)     Find the cumulative distribution function for X .

(ii)     Hence, or otherwise, find the median of the distribution.

[7]
d.
Answer/Explanation

Markscheme

piecewise linear graph

 

correct shape     A1

with vertices (0, 0), (0.5, 1) and (2, 0)     A1

LQ: x = 0.5 , because the area of the triangle is 0.25     R1

[3 marks]

a.

(i)     \({\text{E}}(X) = \int_0^{0.5} {x \times 2x{\text{d}}x + \int_{0.5}^2 {x \times \left( {\frac{4}{3} – \frac{2}{3}x} \right){\text{d}}x = \frac{5}{6}{\text{ }}( = 0.833…)} } \)     (M1)A1

 

(ii)     \({\text{E}}({X^2}) = \int_0^{0.5} {{x^2} \times 2x{\text{d}}x + \int_{0.5}^2 {{x^2} \times \left( {\frac{4}{3} – \frac{2}{3}x} \right){\text{d}}x = \frac{7}{8}{\text{ }}( = 0.875)} } \)     (M1)A1

[4 marks]

b.

(i)     \({\text{E}}(Y – 2X) = 2{\text{E}}(X) – 2{\text{E}}(X) = 0\)     A1

 

(ii)     \({\text{Var}}\,(X) = \left( {{\text{E}}({X^2}) – {\text{E}}{{(X)}^2}} \right) = \frac{{13}}{{72}}\)     A1

\(Y = {X_1} + {X_2} \Rightarrow {\text{Var}}\,(Y) = 2{\text{Var }}(X)\)     (M1)

\({\text{Var}}\,(Y – 2X) = 2{\text{Var}}\,(X) + 4{\text{Var}}\,(X) = \frac{{13}}{{12}}\)     M1A1

[5 marks]

c.

(i)     attempt to use \(cf(x) = \int {f(u){\text{d}}u} \)     M1

 

obtain \(cf(x) = \left\{ {\begin{array}{*{20}{c}}
  {{x^2},}&{0 \leqslant x \leqslant 0.5,} \\
  {\frac{{4x}}{3} – \frac{1}{3}{x^2} – \frac{1}{3},}&{0.5 \leqslant x \leqslant 2,}
\end{array}} \right.\)     \(\begin{array}{*{20}{c}}
  {{\boldsymbol{A1}}} \\
  {{\boldsymbol{A2}}}
\end{array}\)

 

(ii)     attempt to solve \(cf(x) = 0.5\)     M1

\(\frac{{4x}}{3} – \frac{1}{3}{x^2} – \frac{1}{3} = 0.5\)     (A1)

obtain 0.775     A1 

 

Note: Accept attempts in the form of an integral with upper limit the unknown median.

Note: Accept exact answer \(2 – \sqrt {1.5} \) .

 

[7 marks]

d.

Examiners report

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

a.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

b.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

c.

There was a curious issue about the lower quartile in part (a): The LQ coincides with a quarter of the range of the distribution \(\frac{2}{4} = 0.5\). Sadly this is wrong reasoning – the correct reasoning involves a consideration of areas.

In part (b) many candidates used hand calculation rather than their GDC.

The random variable Y was not well understood, and that followed into incorrect calculations involving Y – 2X.

d.

Question

The continuous random variable X has probability density function f given by

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{3{x^2} + 2x}}{{10}},}&{{\text{for }}1 \leqslant x \leqslant 2} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

(i)     Determine an expression for \(F(x)\), valid for \(1 \leqslant x \leqslant 2\), where F denotes the cumulative distribution function of X.

(ii)     Hence, or otherwise, determine the median of X.

[6]
a.

(i)     State the central limit theorem.

(ii)     A random sample of 150 observations is taken from the distribution of X and \(\bar X\) denotes the sample mean. Use the central limit theorem to find, approximately, the probability that \(\bar X\) is greater than 1.6.

[8]
b.
Answer/Explanation

Markscheme

(i)     \(F(x) = \int_1^x {\frac{{3{u^2} + 2u}}{{10}}{\text{d}}u} \)     (M1)

\( = \left[ {\frac{{{u^3} + {u^2}}}{{10}}} \right]_1^x\)     A1

Note: Do not penalise missing or wrong limits at this stage.

Accept the use of x in the integrand.

 

\( = \frac{{{x^3} + {x^2} – 2}}{{10}}\)     A1

 

(ii)     the median m satisfies the equation \(F(m) = \frac{1}{2}\) so     (M1)

\({m^3} + {m^2} – 7 = 0\)     (A1)

Note: Do not FT from an incorrect \(F(x)\).

 

\(m = 1.63\)     A1

Note: Accept any answer that rounds to 1.6.

 

[6 marks]

a.

(i)     the mean of a large sample from any distribution is approximately 

normal     A1

Note: This is the minimum acceptable explanation.

 

(ii)     we require the mean \(\mu \) and variance \({\sigma ^2}\) of X

\(\mu  = \int_1^2 {\left( {\frac{{3{x^3} + 2{x^2}}}{{10}}} \right){\text{d}}x} \)     (M1)

\( = \frac{{191}}{{120}}{\text{ }}(1.591666 \ldots )\)     A1

\({\sigma ^2} = \int_1^2 {\left( {\frac{{3{x^4} + 2{x^3}}}{{10}}} \right){\text{d}}x – {\mu ^2}} \)     (M1)

\( = 0.07659722 \ldots \)     A1

the central limit theorem states that

\(\bar X \approx N\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right),\) i.e. \(N(1.591666 \ldots ,{\text{ }}0.0005106481 \ldots )\)     M1A1

\({\text{P}}(\bar X > 1.6) = 0.356\)     A1

Note: Accept any answer that rounds to 0.36.

 

[8 marks]

b.

Examiners report

Solutions to (a)(i) were disappointing in general, suggesting that many candidates are unfamiliar with the concept of the cumulative distribution function. Many candidates knew that it was something to do with the integral of the probability density function but some thought it was \(\int\limits_1^2 {f(x){\text{d}}x} \) which they then evaluated as \(1\) while others thought it was just \(\int {f(x){\text{d}}x}  = \frac{{\left( {{x^2} + {x^3}} \right)}}{{10}}\) which is not, in general, a valid method. However, most candidates solved (a)(ii) correctly, usually by integrating the probability density function from \(1\) to \(m\).

a.

In (b)(i), the statement of the central limit theorem was often quite dreadful. The term ‘sample mean’ was often not mentioned and a common misconception appears to be that the actual distribution rather than the sample mean tends to normality as the sample size increases. Solutions to (b)(ii) often failed to go beyond finding the mean and variance of \(X\) . In calculating the variance, some candidates rounded the mean from \(1.5916666..\) to \(1.59\) which resulted in an incorrect value for the variance. It is important to note that calculating a variance usually involves a small difference of two large numbers so that full accuracy must be maintained.

b.

Question

A random variable \(X\) has probability density function

\(f(x) = \left\{ {\begin{array}{*{20}{c}} 0&{x < 0} \\ {\frac{1}{2}}&{0 \le x < 1} \\ {\frac{1}{4}}&{1 \le x < 3} \\ 0&{x \ge 3} \end{array}} \right.\)

Sketch the graph of \(y = f(x)\).

[1]
a.

Find the cumulative distribution function for \(X\).

[5]
b.

Find the interquartile range for \(X\).

[3]
c.
Answer/Explanation

Markscheme

         A1

Note:     Ignore open / closed endpoints and vertical lines.

Note:     Award A1 for a correct graph with scales on both axes and a clear indication of the relevant values.

[1 mark]

a.

\(F(x) = \left\{ {\begin{array}{*{20}{c}} 0&{x < 0} \\ {\frac{x}{2}}&{0 \le x < 1} \\ {\frac{x}{4} + \frac{1}{4}}&{1 \le x < 3} \\ 1&{x \ge 3} \end{array}} \right.\)

considering the areas in their sketch or using integration     (M1)

\(F(x) = 0,{\text{ }}x < 0,{\text{ }}F(x) = 1,{\text{ }}x \ge 3\)     A1

\(F(x) = \frac{x}{2},{\text{ }}0 \le x < 1\)     A1

\(F(x) = \frac{x}{4} + \frac{1}{4},{\text{ }}1 \le x < 3\)     A1A1

Note:     Accept \( < \) for \( \le \) in all places and also \( > \) for \( \ge \) first A1.

[5 marks]

b.

\({Q_3} = 2,{\text{ }}{Q_1} = 0.5\)     A1A1

\({\text{IQR is }}2 – 0.5 = 1.5\)     A1

[3 marks]

Total [9 marks]

c.

Examiners report

Part (a) was correctly answered by most candidates. Some graphs were difficult to mark because candidates drew their lines on top of the ruled lines in the answer book. Candidates should be advised not to do this. Candidates should also be aware that the command term ‘sketch’ requires relevant values to be indicated.

a.

In (b), most candidates realised that the cumulative distribution function had to be found by integration but the limits were sometimes incorrect.

b.

In (c), candidates who found the upper and lower quartiles correctly sometimes gave the interquartile range as \([0.5,{\text{ }}2]\). It is important for candidates to realise that that the word range has a different meaning in statistics compared with other branches of mathematics.

c.

Question

A discrete random variable \(U\) follows a geometric distribution with \(p = \frac{1}{4}\).

Find \(F(u)\), the cumulative distribution function of \(U\), for \(u = 1,{\text{ }}2,{\text{ }}3 \ldots \)

[3]
a.

Hence, or otherwise, find the value of \(P(U > 20)\).

[2]
b.

Prove that the probability generating function of \(U\) is given by \({G_u}(t) = \frac{t}{{4 – 3t}}\).

[4]
c.

Given that \({U_i} \sim {\text{Geo}}\left( {\frac{1}{4}} \right),{\text{ }}i = 1,{\text{ }}2,{\text{ }}3\), and that \(V = {U_1} + {U_2} + {U_3}\), find

(i)     \({\text{E}}(V)\);

(ii)     \({\text{Var}}(V)\);

(iii)     \({G_v}(t)\), the probability generating function of \(V\).

[6]
d.

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 – 3t)}^3}}}\).

By differentiating \({G_w}(t)\), find \({\text{E}}(W)\).

[4]
e.

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 – 3t)}^3}}}\).

Prove that \(V = W + 3\).

[3]
f.
Answer/Explanation

Markscheme

METHOD 1

\({\text{P}}(U = u) = \frac{1}{4}{\left( {\frac{3}{4}} \right)^{u – 1}}\)     (M1)

\(F(u) = {\text{P}}(U \le u) = \sum\limits_{r = 1}^u {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r – 1}}\;\;\;} \)(or equivalent)

\( = \frac{{\frac{1}{4}\left( {1 – {{\left( {\frac{3}{4}} \right)}^u}} \right)}}{{1 – \frac{3}{4}}}\)     (M1)

\( = 1 – {\left( {\frac{3}{4}} \right)^u}\)     A1

METHOD 2

\({\text{P}}(U \le u) = 1 – {\text{P}}(U > u)\)     (M1)

\({\text{P}}(U > u) = \) probability of \(u\) consecutive failures     (M1)

\({\text{P}}(U \le u) = 1 – {\left( {\frac{3}{4}} \right)^u}\)     A1

[3 marks]

a.

\({\text{P}}(U > 20) = 1 – {\text{P}}(U \le 20)\)     (M1)

\( = {\left( {\frac{3}{4}} \right)^{20}}\;\;\;( = 0.00317)\)     A1

[2 marks]

b.

\({G_U}(t) = \sum\limits_{r = 1}^\infty  {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r – 1}}{t^r}\;\;\;} \)(or equivalent)     M1A1

\( = \sum\limits_{r = 1}^\infty  {\frac{1}{3}{{\left( {\frac{3}{4}t} \right)}^r}} \)     (M1)

\( = \frac{{\frac{1}{3}\left( {\frac{3}{4}t} \right)}}{{1 – \frac{3}{4}t}}\;\;\;\left( { = \frac{{\frac{1}{4}t}}{{1 – \frac{3}{4}t}}} \right)\)     A1

\( = \frac{t}{{4 – 3t}}\)     AG

[4 marks]

c.

(i)     \(E(U) = \frac{1}{{\frac{1}{4}}} = 4\)     (A1)

\(E({U_1} + {U_2} + {U_3}{\text{)}} = 4 + 4 + 4 = 12\)     A1

(ii)     \({\text{Var}}(U) = \frac{{\frac{3}{4}}}{{{{\left( {\frac{1}{4}} \right)}^2}}}=12\)     A1

\({\text{Var(}}{U_1} + {U_2} + {U_3}) = 12 + 12 + 12 = 36\)     A1

(iii)     \({G_v}(t) = {\left( {{G_U}(t)} \right)^3}\)     (M1)

\( = {\left( {\frac{t}{{4 – 3t}}} \right)^3}\)     A1

[6 marks]

d.

\({G_W}^\prime (t) =  – 3{(4 – 3t)^{ – 4}}( – 3)\;\;\;\left( { = \frac{9}{{{{(4 – 3t)}^4}}}} \right)\)     (M1)(A1)

\(E(W) = {G_W}^\prime (1) = 9\)     (M1)A1

Note:     Allow the use of the calculator to perform the differentiation.

[4 marks]

e.

EITHER

probability generating function of the constant 3 is \({t^3}\)     A1

OR

\({G_{W – 3}}(t) = E({t^{W + 3}}) = E({t^W})E({t^3})\)     A1

THEN

\(W + 3\) has generating function \({G_{W + 3}} = \frac{1}{{{{(4 – 3t)}^3}}} \times {t^3} = {G_V}(t)\)     M1

as the generating functions are the same \(V = W + 3\)     R1AG

[3 marks]

Total [22 marks]

f.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

Question

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

\[f(x) = \left\{ {\begin{array}{*{20}{c}} {{{\text{e}}^{ – x}}}&{x \geqslant 0} \\ 0&{x < 0} \end{array}.} \right.\]

The discrete random variable \(Y\) is defined as the integer part of \(X\), that is the largest integer less than or equal to \(X\).

Show that the probability distribution of \(Y\) is given by \({\text{P}}(Y = y) = {{\text{e}}^{ – y}}(1 – {{\text{e}}^{ – 1}}),{\text{ }}y \in \mathbb{N}\).

[4]
a.

(i)     Show that \(G(t)\), the probability generating function of \(Y\), is given by \(G(t) = \frac{{1 – {{\text{e}}^{ – 1}}}}{{1 – {{\text{e}}^{ – 1}}t}}\).

(ii)     Hence determine the value of \({\text{E}}(Y)\) correct to three significant figures.

[8]
b.
Answer/Explanation

Markscheme

\({\text{P}}(Y = y) = \int_y^{y + 1} {{{\text{e}}^{ – x}}{\text{d}}x} \)    M1A1

\( = {[ – {{\text{e}}^{ – x}}]^{y + 1}}y\)    A1

\( =  – {{\text{e}}^{ – (y + 1)}} + {{\text{e}}^{ – y}}\)    A1

\( = {{\text{e}}^{ – y}}(1 – {{\text{e}}^{ – 1}})\)    AG

[4 marks]

a.

(i)     attempt to use \(G(t) = \sum {{\text{P}}(Y = y){t^y}} \)     (M1)

\( = \sum\limits_{y = 0}^\infty  {{{\text{e}}^{ – y}}(1 – {{\text{e}}^{ – 1}}){t^y}} \)    A1

Note:     Accept a listing of terms without the use of \(\Sigma \).

this is an infinite geometric series with first term \(1 – {{\text{e}}^{ – 1}}\) and common ratio \({{\text{e}}^{ – 1}}t\)     M1

\(G(t) = \frac{{1 – {{\text{e}}^{ – 1}}}}{{1 – {{\text{e}}^{ – 1}}t}}\)    AG

(ii)     \({\text{E}}(Y) = G'(1)\)     M1

\(G'(t) = \frac{{1 – {{\text{e}}^{ – 1}}}}{{{{(1 – {{\text{e}}^{ – 1}}t)}^2}}} \times {{\text{e}}^{ – 1}}\)     (M1)(A1)

\({\text{E}}(Y) = \frac{{{{\text{e}}^{ – 1}}}}{{(1 – {{\text{e}}^{ – 1}})}}\)    (A1)

\( = 0.582\)    A1

Note:     Allow the use of GDC to determine \(G'(1)\).

[8 marks]

b.

Examiners report

In (a), it was disappointing to find that very few candidates realised that \({\text{P}}(Y = y)\) could be found by integrating \(f(x)\) from \(y\) to \(y + 1\). Candidates who simply integrated \(f(x)\) to find the cumulative distribution function of \(X\) were given no credit unless they attempted to use their result to find the probability distribution of \(Y\).

a.

Solutions to (b)(i) were generally good although marks were lost due to not including the \(y = 0\) term.

Part (b)(ii) was also well answered in general with the majority of candidates using the GDC to evaluate \(G'(1)\).

Candidates who tried to differentiate \(G(t)\) algebraically often made errors.

b.

Question

Two independent discrete random variables \(X\) and \(Y\) have probability generating functions \(G(t)\) and \(H(t)\) respectively. Let \(Z = X + Y\) have probability generating function \(J(t)\).

Write down an expression for \(J(t)\) in terms of \(G(t)\) and \(H(t)\).

[1]
a.

By differentiating \(J(t)\), prove that

(i)     \({\text{E}}(Z) = {\text{E}}(X) + {\text{E}}(Y)\);

(ii)     \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\).

[10]
b.
Answer/Explanation

Markscheme

\(J(t) = G(t)H(t)\)    A1

[1 mark]

a.

(i)     \(J'(t) = G'(t)H(t) + G(t)H'(t)\)     M1A1

\(J'(1) = G'(1)H(1) + G(1)H'(1)\)    M1

\(J'(1) = G'(1) + H'(1)\)    A1

so \(E(Z) = E(X) + E(Y)\)     AG

(ii)     \(J”(t) = G”(t)H(t) + G'(t)H'(t) + G'(t)H'(t) + G(t)H”(t)\)     M1A1

\(J”(1) = G”(1)H(1) + 2G'(1)H'(1) + G(1)H”(1)\)

\( = G”(1) + 2G'(1)H'(1) + H”(1)\)    A1

\({\text{Var}}(Z) = J”(1) + J'(1) – {\left( {J'(1)} \right)^2}\)    M1

\( = G”(1) + 2G'(1)H'(1) + H”(1) + G'(1) + H'(1) – {\left( {G'(1) + H'(1)} \right)^2}\)    A1

\( = G”(1) + G'(1) – {\left( {G'(1)} \right)^2} + H”(1) + H'(1) – {\left( {H'(1)} \right)^2}\)    A1

so \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\)     AG

Note: If addition is wrongly used instead of multiplication in (a) it is inappropriate to give FT apart from the second M marks in each part, as the working is too simple.

[10 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

Two independent discrete random variables \(X\) and \(Y\) have probability generating functions \(G(t)\) and \(H(t)\) respectively. Let \(Z = X + Y\) have probability generating function \(J(t)\).

Write down an expression for \(J(t)\) in terms of \(G(t)\) and \(H(t)\).

[1]
a.

By differentiating \(J(t)\), prove that

(i)     \({\text{E}}(Z) = {\text{E}}(X) + {\text{E}}(Y)\);

(ii)     \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\).

[10]
b.
Answer/Explanation

Markscheme

\(J(t) = G(t)H(t)\)    A1

[1 mark]

a.

(i)     \(J'(t) = G'(t)H(t) + G(t)H'(t)\)     M1A1

\(J'(1) = G'(1)H(1) + G(1)H'(1)\)    M1

\(J'(1) = G'(1) + H'(1)\)    A1

so \(E(Z) = E(X) + E(Y)\)     AG

(ii)     \(J”(t) = G”(t)H(t) + G'(t)H'(t) + G'(t)H'(t) + G(t)H”(t)\)     M1A1

\(J”(1) = G”(1)H(1) + 2G'(1)H'(1) + G(1)H”(1)\)

\( = G”(1) + 2G'(1)H'(1) + H”(1)\)    A1

\({\text{Var}}(Z) = J”(1) + J'(1) – {\left( {J'(1)} \right)^2}\)    M1

\( = G”(1) + 2G'(1)H'(1) + H”(1) + G'(1) + H'(1) – {\left( {G'(1) + H'(1)} \right)^2}\)    A1

\( = G”(1) + G'(1) – {\left( {G'(1)} \right)^2} + H”(1) + H'(1) – {\left( {H'(1)} \right)^2}\)    A1

so \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\)     AG

Note: If addition is wrongly used instead of multiplication in (a) it is inappropriate to give FT apart from the second M marks in each part, as the working is too simple.

[10 marks]

b.

Examiners report

[N/A]

a.

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b.

Question

A continuous random variable \(T\) has a probability density function defined by

\(f(t) = \left\{ {\begin{array}{*{20}{c}} {\frac{{t(4 – {t^2})}}{4}}&{0 \leqslant t \leqslant 2} \\ {0,}&{{\text{otherwise}}} \end{array}} \right.\).

Find the cumulative distribution function \(F(t)\), for \(0 \leqslant t \leqslant 2\).

[3]
a.

Sketch the graph of \(F(t)\) for \(0 \leqslant t \leqslant 2\), clearly indicating the coordinates of the endpoints.

[2]
b.i.

Given that \(P(T < a) = 0.75\), find the value of \(a\).

[2]
b.ii.
Answer/Explanation

Markscheme

\(F(t) = \int_0^t {\left( {x – \frac{{{x^3}}}{4}} \right){\text{d}}x{\text{ }}\left( { = \int_0^t {\frac{{x(4 – {x^2})}}{4}{\text{d}}x} } \right)} \)     M1

\( = \left[ {\frac{{{x^2}}}{2} – \frac{{{x^4}}}{{16}}} \right]_0^t{\text{ }}\left( { = \left[ {\frac{{{x^2}(8 – {x^2})}}{{16}}} \right]_0^t} \right){\text{ }}\left( { = \left[ {\frac{{ – 4 – {x^2}{)^2}}}{{16}}} \right]_0^t} \right)\)     A1

\( = \frac{{{t^2}}}{2} – \frac{{{t^4}}}{{16}}{\text{ }}\left( { = \frac{{{t^2}(8 – {t^2})}}{{16}}} \right){\text{ }}\left( { = 1 – \frac{{{{(4 – {t^2})}^2}}}{{16}}} \right)\)     A1

Note:     Condone integration involving \(t\) only.

Note:     Award M1A0A0 for integration without limits eg, \(\int {\frac{{t(4 – {t^2})}}{4}{\text{d}}t = \frac{{{t^2}}}{2} – \frac{{{t^4}}}{{16}}} \) or equivalent.

Note:     But allow integration \( + \) \(C\) then showing \(C = 0\) or even integration without \(C\) if \(F(0) = 0\) or \(F(2) = 1\) is confirmed.

[3 marks]

a.

correct shape including correct concavity     A1

clearly indicating starts at origin and ends at \((2,{\text{ }}1)\)     A1

Note:     Condone the absence of \((0,{\text{ }}0)\).

Note:     Accept 2 on the \(x\)-axis and 1 on the \(y\)-axis correctly placed.

[2 marks]

b.i.

attempt to solve \(\frac{{{a^2}}}{2} – \frac{{{a^4}}}{{16}} = 0.75\) (or equivalent) for \(a\)     (M1)

\(a = 1.41{\text{ }}( = \sqrt 2 )\)     A1

Note:     Accept any answer that rounds to 1.4.

[2 marks]

b.ii.

Examiners report

[N/A]

a.

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b.i.

[N/A]

b.ii.

Question

The random variable \(X\) follows a Poisson distribution with mean \(\lambda \). The probability generating function of \(X\) is given by \({G_X}(t) = {{\text{e}}^{\lambda (t – 1)}}\).

The random variable \(Y\), independent of \(X\), follows a Poisson distribution with mean \(\mu \).

Find expressions for \({G’_X}(t)\) and \({G’’_X}(t)\).

[2]
a.i.

Hence show that \({\text{Var}}(X) = \lambda \).

[3]
a.ii.

By considering the probability generating function, \({G_{X + Y}}(t)\), of \(X + Y\), show that \(X + Y\) follows a Poisson distribution with mean \(\lambda  + \mu \).

[3]
b.

Show that \({\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 – \frac{\lambda }{{\lambda + \mu }}} \right)^{n – x}}\), where \(n\), \(x\) are non-negative integers and \(n \geqslant x\).

[5]
c.i.

Identify the probability distribution given in part (c)(i) and state its parameters.

[2]
c.ii.
Answer/Explanation

Markscheme

\({G’_X}(t) = \lambda {{\text{e}}^{\lambda (t – 1)}}\)     A1

\({G’’_X}(t) = {\lambda ^2}{{\text{e}}^{\lambda (t – 1)}}\)     A1

[2 marks]

a.i.

\({\text{Var}}(X) = {G”_X}(1) + {G’_X}(1) – {\left( {{{G’}_X}(1)} \right)^2}\)     (M1)

\({G’_X}(1) = \lambda \) and \({G’’_X}(1) = {\lambda ^2}\)     (A1)

\({\text{Var}}(X) = {\lambda ^2} + \lambda  – {\lambda ^2}\)     A1

\( = \lambda \)     AG

[3 marks]

a.ii.

\({G_{X + Y}}(t) = {{\text{e}}^{\lambda (t – 1)}} \times {{\text{e}}^{\mu (t – 1)}}\)     M1

Note:     The M1 is for knowing to multiply pgfs.

\( = {{\text{e}}^{(\lambda  + \mu )(t – 1)}}\)     A1

which is the pgf for a Poisson distribution with mean \(\lambda  + \mu \)     R1AG

Note:     Line 3 identifying the Poisson pgf must be seen.

[3 marks]

b.

\({\text{P}}(X = x|X + Y = n) = \frac{{{\text{P}}(X = x \cap Y = n – x)}}{{{\text{P}}(X + Y = n)}}\)     (M1)

\( = \left( {\frac{{{{\text{e}}^{ – \lambda }}{\lambda ^x}}}{{x!}}} \right)\left( {\frac{{{{\text{e}}^{ – \mu }}{\mu ^{n – x}}}}{{(n – x)!}}} \right)\left( {\frac{{n!}}{{{{\text{e}}^{ – (\lambda  + \mu )}}{{(\lambda  + \mu )}^n}}}} \right)\) (or equivalent)     M1A1

\( = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right)\frac{{{\lambda ^x}{\mu ^{n – x}}}}{{{{(\lambda + \mu )}^n}}}\)     A1

\( = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {\frac{\mu }{{\lambda + \mu }}} \right)^{n – x}}\)     A1

leading to \({\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 – \frac{\lambda }{{\lambda + \mu }}} \right)^{n – x}}\)     AG

[5 marks]

c.i.

\({\text{B}}\left( {n,{\text{ }}\frac{\lambda }{{\lambda  + \mu }}} \right)\)     A1A1

Note:     Award A1 for stating binomial and A1 for stating correct parameters.

[2 marks]

c.ii.

Examiners report

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a.i.

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a.ii.

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b.

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c.i.

[N/A]

c.ii.

Question

Consider an unbiased tetrahedral (four-sided) die with faces labelled 1, 2, 3 and 4 respectively.

The random variable X represents the number of throws required to obtain a 1.

State the distribution of X.

[1]
a.

Show that the probability generating function, \(G\left( t \right)\), for X is given by \(G\left( t \right) = \frac{t}{{4 – 3t}}\).

[4]
b.

Find \(G’\left( t \right)\).

[2]
c.

Determine the mean number of throws required to obtain a 1.

[1]
d.
Answer/Explanation

Markscheme

X is geometric (or negative binomial)      A1

[1 mark]

a.

\(G\left( t \right) = \frac{1}{4}t + \frac{1}{4}\left( {\frac{3}{4}} \right){t^2} + \frac{1}{4}{\left( {\frac{3}{4}} \right)^2}{t^3} +  \ldots \)     M1A1

recognition of GP \(\left( {{u_1} = \frac{1}{4}t,\,\,r = \frac{3}{4}t} \right)\)     (M1)

\( = \frac{{\frac{1}{4}t}}{{1 – \frac{3}{4}t}}\)     A1

leading to \(G\left( t \right) = \frac{t}{{4 – 3t}}\)     AG

[4 marks]

b.

attempt to use product or quotient rule      M1

\(G’\left( t \right) = \frac{4}{{{{\left( {4 – 3t} \right)}^2}}}\)     A1

[2 marks]

c.

4      A1

Note: Award A1FT to a candidate that correctly calculates the value of \(G’\left( 1 \right)\) from their \(G’\left( t \right)\).

[1 mark]

d.

Examiners report

[N/A]

a.

[N/A]

b.

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c.

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d.

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