Question
A continuous random variable X has a probability density function given by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{{(x + 1)}^3}}}{{60}},}&{{\text{for }}1 \leqslant x \leqslant 3} \\
{0,}&{{\text{otherwise}}{\text{.}}}
\end{array},} \right.\]
Find
(a) \({\text{P}}(1.5 \leqslant X \leqslant 2.5)\) ;
(b) E(X) ;
(c) the median of X .
▶️Answer/Explanation
Markscheme
(a) \(\int_{1.5}^{2.5} {\frac{{{{(x + 1)}^3}}}{{60}}{\text{d}}x = 0.4625\,\,\,\,\,{\text{( = 0.463)}}} \) M1A1
(b) \({\text{E}}(X) = \int_1^3 {\frac{{x{{(x + 1)}^3}}}{{60}}{\text{d}}x = 2.31} \) M1A1
(c) \(\int_1^m {\frac{{{{(x + 1)}^3}}}{{60}}{\text{d}}x = 0.5} \) M1
\(\left[ {\frac{{{{(x + 1)}^4}}}{{240}}} \right]_1^m = 0.5\) (A1)
\(m = 2.41\) A1
[7 marks]
Examiners report
Parts (a) and (b) were reasonably well done in general but (c) caused problems for many candidates where several misconceptions regarding the median were seen. The expectation was that candidates would use their GDCs to solve (a) and (b), and possibly even (c), although in the event most candidates did the integrations by hand. Those candidates using their GDCs made fewer mistakes in general than those doing the integrations analytically.
Question
The continuous random variable \(X\) has the probability distribution function \(f(x) = A\sin \left( {\ln (x)} \right),{\text{ }}1 \le x \le 5\).
a.Find the value of \(A\) to three decimal places.[2]
b.Find the mode of \(X\).[2]
c.Find the value \({\text{P}}(X \le 3|X \ge 2)\).[2]
▶️Answer/Explanation
Markscheme
\(A\int_1^5 {\sin (\ln x){\text{d}}x = 1} \) (M1)
\(A = 0.323{\text{ (3 dp only)}}\) A1
[2 marks]
either a graphical approach or \(f'(x) = \frac{{\cos (\ln x)}}{x} = 0\) (M1)
\(x = 4.81\;\;\;\left( { = {{\text{e}}^{\frac{\pi }{2}}}} \right)\) A1
Note: Do not award A1FT for a candidate working in degrees.
[2 marks]
\({\text{P}}(X \le 3|X \ge 2) = \frac{{{\text{P}}(2 \le X \le 3)}}{{{\text{P}}(X \ge 2)}}\;\;\;\left( { = \frac{{\int_2^3 {\sin \left( {\ln (x)} \right){\text{d}}x} }}{{\int_2^5 {\sin \left( {\ln (x)} \right){\text{d}}x} }}} \right)\) (M1)
\( = 0.288\) A1
Note: Do not award A1FT for a candidate working in degrees.
[2 marks]
Total [6 marks]
Question
The probability density function of a continuous random variable \(X\) is given by
\[f(x) = \left\{ {\begin{array}{*{20}{c}} {0,{\text{ }}x < 0} \\ {\frac{{\sin x}}{4},{\text{ }}0 \le x \le \pi } \\ {a(x – \pi ),{\text{ }}\pi < x \le 2\pi } \\ {0,{\text{ }}2\pi < x} \end{array}.} \right.\]
a.Sketch the graph \(y = f(x)\).[2]
b.Find \({\text{P}}(X \le \pi )\).[2]
c.Show that \(a = \frac{1}{{{\pi ^2}}}\).[3]
d.Write down the median of \(X\).[1]
e.Calculate the mean of \(X\).[3]
f.Calculate the variance of \(X\).[3]
g.Find \({\text{P}}\left( {\frac{\pi }{2} \le X \le \frac{{3\pi }}{2}} \right)\).[2]
h.Given that \(\frac{\pi }{2} \le X \le \frac{{3\pi }}{2}\) find the probability that \(\pi \le X \le 2\pi \).[4]
▶️Answer/Explanation
Markscheme
Award A1 for sine curve from \(0\) to \(\pi \), award A1 for straight line from \(\pi \) to \(2\pi \) A1A1
[2 marks]
\(\int_0^\pi {\frac{{\sin x}}{4}{\text{d}}x = \frac{1}{2}} \) (M1)A1
[2 marks]
METHOD 1
require \(\frac{1}{2} + \int_\pi ^{2\pi } {a(x – \pi ){\text{d}}x = 1} \) (M1)
\( \Rightarrow \frac{1}{2} + a\left[ {\frac{{{{(x – \pi )}^2}}}{2}} \right]_\pi ^{2\pi } = 1\;\;\;\left( {{\text{or }}\frac{1}{2} + a\left[ {\frac{{{x^2}}}{2} – \pi x} \right]_\pi ^{2\pi } = 1} \right)\) A1
\( \Rightarrow a\frac{{{\pi ^2}}}{2} = \frac{1}{2}\) A1
\( \Rightarrow a = \frac{1}{{{\pi ^2}}}\) AG
Note: Must obtain the exact value. Do not accept answers obtained with calculator.
METHOD 2
\(0.5 + {\text{ area of triangle }} = 1\) R1
area of triangle \( = \frac{1}{2}\pi \times a\pi = 0.5\) M1A1
Note: Award M1 for correct use of area formula \( = 0.5\), A1 for \(a\pi \).
\(a = \frac{1}{{{\pi ^2}}}\) AG
[3 marks]
median is \(\pi \) A1
[1 mark]
\(\mu = \int_0^\pi {x \cdot \frac{{\sin x}}{4}{\text{d}}x + \int_\pi ^{2\pi } {x \cdot \frac{{x – \pi }}{{{\pi ^2}}}{\text{d}}x} } \) (M1)(A1)
\( = 3.40339 \ldots = 3.40\;\;\;\left( {{\text{or }}\frac{\pi }{4} + \frac{{5\pi }}{6} = \frac{{13}}{{12}}\pi } \right)\) A1
[3 marks]
For \(\mu = 3.40339 \ldots \)
EITHER
\({\sigma ^2} = \int_0^\pi {{x^2} \cdot \frac{{\sin x}}{4}{\text{d}}x + \int_\pi ^{2\pi } {{x^2} \cdot \frac{{x – \pi }}{{{\pi ^2}}}{\text{d}}x – {\mu ^2}} } \) (M1)(A1)
OR
\({\sigma ^2} = \int_0^\pi {{{(x – \mu )}^2} \cdot \frac{{\sin x}}{4}{\text{d}}x + \int_\pi ^{2\pi } {{{(x – \mu )}^2} \cdot \frac{{x – \pi }}{{{\pi ^2}}}{\text{d}}x} } \) (M1)(A1)
THEN
\( = 3.866277 \ldots = 3.87\) A1
[3 marks]
\(\int_{\frac{\pi }{2}}^\pi {\frac{{\sin x}}{4}{\text{d}}x + \int_\pi ^{\frac{{3\pi }}{2}} {\frac{{x – \pi }}{{{\pi ^2}}}{\text{d}}x = 0.375\;\;\;\left( {{\text{or }}\frac{1}{4} + \frac{1}{8} = \frac{3}{8}} \right)} } \) (M1)A1
[2 marks]
\({\text{P}}\left( {\pi \le X \le 2\pi \left| {\frac{\pi }{2} \le X \le \frac{{3\pi }}{2}} \right.} \right) = \frac{{{\text{P}}\left( {\pi \le X \le \frac{{3\pi }}{2}} \right)}}{{{\text{P}}\left( {\frac{\pi }{2} \le X \le \frac{{3\pi }}{2}} \right)}}\) (M1)(A1)
\( = \frac{{\int_\pi ^{\frac{{3\pi }}{2}} {\frac{{(x – \pi )}}{{{\pi ^2}}}{\text{d}}x} }}{{0.375}} = \frac{{0.125}}{{0.375}}\;\;\;\left( {{\text{or }} = \frac{{\frac{1}{8}}}{{\frac{3}{8}}}{\text{ from diagram areas}}} \right)\) (M1)
\( = \frac{1}{3}\;\;\;(0.333)\) A1
[4 marks]
Total [20 marks]
Question
A random variable \(X\) has probability density function
\[f(x) = \left\{ \begin{array}{r}ax + b,\\0,\end{array} \right.\begin{array}{*{20}{c}}{2 \le x \le 3}\\{{\rm{ otherwise}}}\end{array},a,b \in \mathbb{R}\]
(a) Show that \(5a + 2b = 2\).
Let \({\text{E}}(X) = \mu \).
(b) (i) Show that \(a = 12\mu – 30\).
(ii) Find a similar expression for b in terms of \(\mu \).
Let the median of the distribution be 2.3.
(c) (i) Find the value of \(\mu \).
(ii) Find the value of the standard deviation of X.
▶️Answer/Explanation
Markscheme
(a) \(\int_2^3 {(ax + b){\text{d}}x{\text{ }}( = 1)} \) M1A1
\(\left[ {\frac{1}{2}a{x^2} + bx} \right]_2^3{\text{ }}( = 1)\) A1
\(\frac{5}{2}a + b = 1\) M1
\(5a + 2b = 2\) AG
[4 marks]
(b) (i) \(\int_2^3 {\left( {a{x^2} + bx} \right){\text{d}}x{\text{ }}( = \mu )} \) M1A1
\(\left[ {\frac{1}{3}a{x^3} + \frac{1}{2}b{x^2}} \right]_2^3{\text{ }}( = \mu )\) A1
\(\frac{{19}}{3}a + \frac{5}{2}b = \mu \) A1
eliminating b M1
eg
\(\frac{{19}}{3}a + \frac{5}{2}\left( {1 – \frac{5}{2}a} \right) = \mu \) A1
\(\frac{1}{{12}}a + \frac{5}{2} = \mu \)
\(a = 12\mu – 30\) AG
Note: Elimination of b could be at different stages.
(ii) \(b = 1 – \frac{5}{2}(12\mu – 30)\)
\( = 76 – 30\mu \) A1
Note: This solution may be seen in part (i).
[7 marks]
(c) (i) \(\int_2^{2.3} {(ax + b){\text{d}}x{\text{ }}( = 0.5)} \) (M1)(A1)
\(\left[ {\frac{1}{2}a{x^2} + bx} \right]_2^{2.3}{\text{ }}( = 0.5)\)
\(0.645a + 0.3b{\text{ }}( = 0.5)\) (A1)
\(0.645(12\mu – 30) + 0.3(76 – 30\mu ) = 0.5\) M1
\(\mu = 2.34{\text{ }}\left( { = \frac{{295}}{{126}}} \right)\) A1
(ii) \({\text{E}}\left( {{X^2}} \right) = \int_2^3 {{x^2}(ax + b){\text{d}}x} \) (M1)
\(a = 12\mu – 30 = – \frac{{40}}{{21}},{\text{ }}b = 76 – 30\mu = \frac{{121}}{{21}}\) (A1)
\({\text{E}}\left( {{X^2}} \right) = \int_2^3 {{x^2}\left( { – \frac{{40}}{{21}}x + \frac{{121}}{{21}}} \right){\text{d}}x = 5.539 \ldots {\text{ }}\left( { = \frac{{349}}{{63}}} \right)} \) (A1)
\({\text{Var}}(X) = 5.539{\text{K}} – {(2.341{\text{K}})^2} = 0.05813 \ldots \) (M1)
\(\sigma = 0.241\) A1
[10 marks]
Total [21 marks]
Examiners report
Question
A fisherman notices that in any hour of fishing, he is equally likely to catch exactly two fish, as he is to catch less than two fish. Assuming the number of fish caught can be modelled by a Poisson distribution, calculate the expected value of the number of fish caught when he spends four hours fishing.
▶️Answer/Explanation
Markscheme
\(X \sim {\text{Po}}(m)\)
\({\text{P}}(X = 2) = {\text{P}}(X < 2)\) (M1)
\(\frac{1}{2}{m^2}{{\text{e}}^{ – m}} = {{\text{e}}^{ – m}}(1 + m)\) (A1)(A1)
\(m = 2.73 {\text{ }}\left( {1 + \sqrt 3 } \right)\) A1
in four hours the expected value is 10.9\(\,\,\,\,\left( {4 + 4\sqrt 3 } \right)\) A1
Note: Value of m does not need to be rounded.
[5 marks]
Examiners report
Many candidates did not attempt this question and many others did not go beyond setting the equation up. Among the ones who attempted to solve the equation, once again, very few candidates took real advantage of GDC use to obtain the correct answer.
Question
The annual weather-related loss of an insurance company is modelled by a random variable \(X\) with probability density function\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{2.5{{\left( {200} \right)}^{2.5}}}}{{{x^{3.5}}}},}&{x \geqslant 200} \\
{0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]Find the median.
▶️Answer/Explanation
Markscheme
\(\int_{200}^M {\frac{{2.5{{\left( {200} \right)}^{2.5}}}}{{{x^{3.5}}}}{\text{d}}x} = 0.5\) M1A1A1
Note: Award M1 for the integral equal to \(0.5\)
A1A1 for the correct limits.
\(\frac{{ – {{200}^{2.5}}}}{{{M^{2.5}}}}\left( {\frac{{ – {{200}^{2.5}}}}{{{{200}^{2.5}}}}} \right) = 0.5\) M1A1A1
Note: Award M1 for correct integration
A1A1 for correct substitutions.
\(\frac{{ – {{200}^{2.5}}}}{{{M^{2.5}}}} + 1 = 0.5 \Rightarrow {M^{2.5}} = 2{\left( {200} \right)^{2.5}}\) (A1)
\(M = 264\) A1
[8 marks]
Examiners report
Many students used incorrect limits to the integral, although many did correctly let the integral equal to \(0.5\).