A continuous random variable $X$ has the probability density function

$f\left(x\right)=\left\{\begin{array}{cc}\frac{2}{\left(b-a\right)\left(c-a\right)}\left(x-a\right),& a\le x\le c\\ \frac{2}{\left(b-a\right)\left(b-c\right)}\left(b-x\right),& c<x\le b\\ 0,& \text{otherwise}\end{array}\right.$.

The following diagram shows the graph of $y=f\left(x\right)$ for $a\le x\le b$.

Given that $c\ge \frac{a+b}{2}$, find an expression for the median of $X$ in terms of $a, b$ and $c$.

let $m$ be the median

EITHER

attempts to find the area of the required triangle          M1

base is $\left(m-a\right)$          (A1)

and height is $\frac{2}{\left(b-a\right)\left(c-a\right)}\left(m-a\right)$

area $=\frac{1}{2}\left(m-a\right)\times \frac{2}{\left(b-a\right)\left(c-a\right)}\left(m-a\right) \left(=\frac{{\left(m-a\right)}^2}{\left(b-a\right)\left(c-a\right)}\right)$         A1

OR

attempts to integrate the correct function          M1

$\underset{a}{\overset{m}{\int }}\frac{2}{\left(b-a\right)\left(c-a\right)}\left(x-a\right) dx$

$=\frac{2}{\left(b-a\right)\left(c-a\right)}{\left[\frac{1}{2}{\left(x-a\right)}^2\right]}_a^m$  OR  $\frac{2}{\left(b-a\right)\left(c-a\right)}{\left[\frac{x^2}{2}-ax\right]}_a^m$         A1A1

Note: Award A1 for correct integration and A1 for correct limits.

THEN

sets up (their) $\underset{a}{\overset{m}{\int }}\frac{2}{\left(b-a\right)\left(c-a\right)}\left(x-a\right) dx$ or area $=\frac{1}{2}$         M1

Note: Award M0A0A0M1A0A0 if candidates conclude that $m>c$ and set up their area or sum of integrals $=\frac{1}{2}$.

$\frac{{\left(m-a\right)}^2}{\left(b-a\right)\left(c-a\right)}=\frac{1}{2}$

$m=a\pm \sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$         (A1)

as $m>a$, rejects $m=a-\sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$

so $m=a+\sqrt{\frac{\left(b-a\right)\left(c-a\right)}{2}}$         A1

[6 marks]

Show that the probability that Chloe wins the game is $\frac{3}{8}$.

Determine the mean of X.

Determine the variance of X.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

number of possible “deals” $=4!=24$     A1

consider ways of achieving “no matches” (Chloe winning):

Selena could deal B, C, D (ie, 3 possibilities)

as her first card     R1

for each of these matches, there are only 3 possible combinations for the remaining 3 cards     R1

so no. ways achieving no matches $=3\times 3=9$     M1A1

so probability Chloe wins $=\frac{9}{23}=\frac{3}{8}$     A1AG

METHOD 2

number of possible “deals” $=4!=24$     A1

consider ways of achieving a match (Selena winning)

Selena card A can match with Chloe card A, giving 6 possibilities for this happening     R1

if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D     R1

so no. ways achieving one match is $=6+3+3+3=15$     M1A1

so probability Chloe wins $=1-\frac{15}{24}=\frac{3}{8}$     A1AG

METHOD 3

systematic attempt to find number of outcomes where Chloe wins (no matches)

(using tree diag. or otherwise)     M1

9 found     A1

each has probability $\frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}\times 1$     M1

$=\frac{1}{24}$     A1

their 9 multiplied by their $\frac{1}{24}$     M1A1

$=\frac{3}{8}$     AG

[6 marks]

$X\sim \text{B}\left(50,\frac{3}{8}\right)$     (M1)

$\mu =np=50\times \frac{3}{8}=\frac{150}{8}\left(=\frac{75}{4}\right)(=18.75)$     (M1)A1

[3 marks]

${\sigma }^2=np(1-p)=50\times \frac{3}{8}\times \frac{5}{8}=\frac{750}{64}\left(=\frac{375}{32}\right)(=11.7)$     (M1)A1

[2 marks]

State the mode of $X$.

Find $\int \text{arcsin}x\text{d}x$.

Hence show that $k=\frac{2}{2+\pi }$.

$\frac{\text{d}y}{\text{d}x}=x\text{arcsin}x$.

$\text{E}\left(X\right)=\frac{3\pi }{4\left(\pi +2\right)}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

mode is 0    A1

[1 mark]

attempt at integration by parts      (M1)

$\frac{\text{d}u}{\text{d}x}=\frac{1}{\sqrt{1-x^2}}\text{,}\text{d}v=\text{d}x$

$=x\text{arcsin}x-\int \frac{x\text{d}x}{\sqrt{1-x^2}}$    A1

$=x\text{arcsin}x+\sqrt{1-x^2}\left(+c\right)$    A1

[3 marks]

$\underset{0}{\overset{1}{\int }}\left(\pi -\text{arcsin}x\right)\text{d}x={\left[\pi x-x\text{arcsin}x-\sqrt{1-x^2}\right]}_0^1$   A1

$=\left(\pi -\frac{\pi }{2}-0\right)-\left(0-0-1\right)=\frac{\pi }{2}+1$

$=\frac{\pi +2}{2}$    A1

$\underset{0}{\overset{1}{\int }}k\left(\pi -\text{arcsin}x\right)\text{d}x=1$   (M1)

Note: This line can be seen (or implied) anywhere.

Note: Do not allow FT A marks from bi to bii.

$k\left(\frac{\pi +2}{2}\right)=1$

$\Rightarrow k=\frac{2}{2+\pi }$    AG

[3 marks]

attempt to use product rule to differentiate    M1

$\frac{\text{d}y}{\text{d}x}=x\text{arcsin}x+\frac{x^2}{2\sqrt{1-x^2}}-\frac{1}{4\sqrt{1-x^2}}-\frac{x^2}{4\sqrt{1-x^2}}+\frac{\sqrt{1-x^2}}{4}$  A2

Note: Award A2 for all terms correct, A1 for 4 correct terms.

$=x\text{arcsin}x+\frac{2x^2}{4\sqrt{1-x^2}}-\frac{1}{4\sqrt{1-x^2}}-\frac{x^2}{4\sqrt{1-x^2}}+\frac{1-x^2}{4\sqrt{1-x^2}}$    A1

Note: Award A1 for equivalent combination of correct terms over a common denominator.

$=x\text{arcsin}x$    AG

[4 marks]

$\text{E}\left(X\right)=k\underset{0}{\overset{1}{\int }}x\left(\pi -\text{arcsin}x\right)\text{d}x$    M1

$=k\underset{0}{\overset{1}{\int }}\left(\pi x-x\text{arcsin}x\right)\text{d}x$

$=k{\left[\frac{\pi x^2}{2}-\frac{x^2}{2}\text{arcsin}x+\frac{1}{4}\text{arcsin}x-\frac{x}{4}\sqrt{1-x^2}\right]}_0^1$      A1A1

Note: Award A1 for first term, A1 for next 3 terms.

$=k\left[\left(\frac{\pi }{2}-\frac{\pi }{4}+\frac{\pi }{8}\right)-\left(0\right)\right]$      A1

$=\left(\frac{2}{2+\pi }\right)\frac{3\pi }{8}$      A1

$=\frac{3\pi }{4\left(\pi +2\right)}$    AG

[5 marks]

Find the value of $k$.

By considering the graph of f write down the mean of $X$;

By considering the graph of f write down the median of $X$;

By considering the graph of f write down the mode of $X$.

Show that $P(0⩽X⩽2)=\frac{1}{4}$.

Hence state the interquartile range of $X$.

Calculate $P(X⩽4|X⩾3)$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to equate integral to 1 (may appear later)     M1

$k\underset{0}{\overset{6}{\int }}\sin \left(\frac{\pi x}{6}\right)\text{d}x=1$

correct integral     A1

$k{\left[-\frac{6}{\pi }\cos \left(\frac{\pi x}{6}\right)\right]}_0^6=1$

substituting limits     M1

$-\frac{6}{\pi }(-1-1)=\frac{1}{k}$

$k=\frac{\pi }{12}$ A1

[4 marks]

mean $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

median $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

mode $=3$     A1

Note:     Award A1A0A0 for three equal answers in $(0,6)$.

[1 mark]

$\frac{\pi }{12}\underset{0}{\overset{2}{\int }}\sin \left(\frac{\pi x}{6}\right)\text{d}x$    M1

$=\frac{\pi }{12}{\left[-\frac{6}{\pi }\cos \left(\frac{\pi x}{6}\right)\right]}_0^2$     A1

Note:     Accept without the $\frac{\pi }{12}$ at this stage if it is added later.

$\frac{\pi }{12}\left[-\frac{6}{\pi }\left(\cos \frac{\pi }{3}-1\right)\right]$     M1

$=\frac{1}{4}$     AG

[4 marks]

from (c)(i) $Q_1=2$     (A1)

as the graph is symmetrical about the middle value $x=3\Rightarrow Q_3=4$     (A1)

so interquartile range is

$4-2$

$=2$     A1

[3 marks]

$P(X⩽4|X⩾3)=\frac{P(3⩽X⩽4)}{P(X⩾3)}$

$=\frac{\frac{1}{4}}{\frac{1}{2}}$     (M1)

$=\frac{1}{2}$     A1

[2 marks]

Find the value of $p$.

Given that $\text{E}\left(X\right)=10$, find the value of $N$.

$p=1-\frac{1}{2}-\frac{1}{5}-\frac{1}{5}$       (M1)

$=\frac{1}{10}$       A1

[2 marks]

attempt to find $\text{E}\left(X\right)$        (M1)

$\frac{1}{2}+1+2+\frac{N}{10}=10$       A1

$\Rightarrow N=65$       A1

Note: Do not allow FT in part (b) if their $p$ is outside the range .

[3 marks]

Show that $\text{P}(A\cup B)=\text{P}(A)+\text{P}(A^{\prime }\cap B)$.

(i)     show that $\text{P}(A)=\frac{1}{3}$;

(ii)     hence find $\text{P}(B)$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)$    M1

$=\text{P}(A)+\text{P}(A\cap B)+\text{P}(A^{\prime }\cap B)-\text{P}(A\cap B)$    M1A1

$=\text{P}(A)+\text{P}(A^{\prime }\cap B)$    AG

METHOD 2

$\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)$    M1

$=\text{P}(A)+\text{P}(B)-\text{P}(A|B)\times \text{P}(B)$    M1

$=\text{P}(A)+\left(1-\text{P}(A|B)\right)\times \text{P}(B)$

$=\text{P}(A)+\text{P}(A^{\prime }|B)\times \text{P}(B)$    A1

$=\text{P}(A)+\text{P}(A^{\prime }\cap B)$    AG

[3 marks]

(i)     use $\text{P}(A\cup B)=\text{P}(A)+\text{P}(A^{\prime }\cap B)$ and $\text{P}(A^{\prime }\cap B)=\text{P}(B|A^{\prime })\text{P}(A^{\prime })$     (M1)

$\frac{4}{9}=\text{P}(A)+\frac{1}{6}\left(1-\text{P}(A)\right)$    A1

$8=18\text{P}(A)+3\left(1-\text{P}(A)\right)$    M1

$\text{P}(A)=\frac{1}{3}$    AG

(ii)     METHOD 1

$\text{P}(B)=\text{P}(A\cap B)+\text{P}(A^{\prime }\cap B)$    M1

$=\text{P}(B|A)\text{P}(A)+\text{P}(B|A^{\prime })\text{P}(A^{\prime })$    M1

$=\frac{1}{3}\times \frac{1}{3}+\frac{1}{6}\times \frac{2}{3}=\frac{2}{9}$    A1

METHOD 2

$\text{P}(A\cap B)=\text{P}(B|A)\text{P}(A)\Rightarrow \text{P}(A\cap B)=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$    M1

$\text{P}(B)=\text{P}(A\cup B)+\text{P}(A\cap B)-\text{P}(A)$    M1

$\text{P}(B)=\frac{4}{9}+\frac{1}{9}-\frac{1}{3}=\frac{2}{9}$    A1

[6 marks]

Complete the probability distribution table for $X$.

Find the expected value of $X$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

     A1A1

Note:     Award A1 for each correct row.

[2 marks]

$\text{E}(X)=1\times \frac{1}{6}+2\times \frac{1}{3}+4\times \frac{1}{3}+6\times \frac{1}{6}$    (M1)

$=\frac{19}{6}\left(=3\frac{1}{6}\right)$    A1

Note:     If the probabilities in (a) are not values between 0 and 1 or lead to $\text{E}(X)>6$ award M1A0 to correct method using the incorrect probabilities; otherwise allow FT marks.

[2 marks]

Find the value of p.

Find μ, the expected value of X.

Find P(X > μ).

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

equating sum of probabilities to 1 (p + 0.5 − p + 0.25 + 0.125 + p³ = 1)       M1

p³ = 0.125 = $\frac{1}{8}$

p= 0.5      A1

[2 marks]

μ = 0 × 0.5 + 1 × 0 + 2 × 0.25 + 3 × 0.125 + 4 × 0.125       M1

= 1.375 $\left(=\frac{11}{8}\right)$     A1

[2 marks]

P(X > μ) = P(X = 2) + P(X = 3) + P(X = 4)      (M1)

= 0.5       A1

Note: Do not award follow through A marks in (b)(i) from an incorrect value of p.

Note: Award M marks in both (b)(i) and (b)(ii) provided no negative probabilities, and provided a numerical value for μ has been found.

[2 marks]

By drawing a Venn diagram, or otherwise, find $\text{P}\left(A\cup B\right)$.

Show that the events $A$ and $B$ are not independent.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

        (M1)

Note: Award M1 for a Venn diagram with at least one probability in the correct region.

EITHER

$\text{P}\left(A\cap B^{\prime }\right)=0.3$     (A1)

$\text{P}\left(A\cup B\right)=0.3+0.4+0.1=0.8$     A1

OR

$\text{P}\left(B\right)=0.5$     (A1)

$\text{P}\left(A\cup B\right)=0.5+0.4-0.1=0.8$     A1

[3 marks]

METHOD 1

$\text{P}\left(A\right)\text{P}\left(B\right)=0.4\times 0.5$        (M1)

= 0.2      A1

statement that their $\text{P}\left(A\right)\text{P}\left(B\right)\ne \text{P}\left(A\cap B\right)$      R1

Note: Award R1 for correct reasoning from their value.

⇒ $A$, $B$ not independent     AG

METHOD 2

$\text{P}\left(A|B\right)=\frac{\text{P}\left(A\cap B\right)}{\text{P}\left(B\right)}=\frac{0.1}{0.5}$        (M1)

= 0.2      A1

statement that their $\text{P}\left(A|B\right)\ne \text{P}\left(A\right)$      R1

Note: Award R1 for correct reasoning from their value.

⇒ $A$, $B$ not independent     AG

Note: Accept equivalent argument using $\text{P}\left(B|A\right)=0.25$.

[3 marks]

A discrete random variable $X$ has the probability distribution given by the following table.

Given that $\text{E}\left(X\right)=\frac{19}{12}$, determine the value of $p$ and the value of $q$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\text{E}\left(X\right)=\left(0\times p\right)+\left(1\times \frac{1}{4}\right)+\left(2\times \frac{1}{6}\right)+3q\left(=\frac{19}{12}\right)$        (M1)

$\left(\Rightarrow \frac{1}{4}+\frac{1}{3}+3q=\frac{19}{12}\right)$

$q=\frac{1}{3}$        A1

$p+\frac{1}{4}+\frac{1}{6}+q=1$        (M1)

$\left(\Rightarrow p+q=\frac{7}{12}\right)$

$p=\frac{1}{4}$        A1

[4 marks]

A continuous random variable X has the probability density function $f$ given by

.

Find P(0 ≤ X ≤ 3).

attempting integration by parts, eg

$u=\frac{\pi x}{36}\text{,}\text{d}u=\frac{\pi }{36}\text{d}x\text{,}\text{d}v=\text{sin}\left(\frac{\pi x}{6}\right)\text{d}x\text{,}v=-\frac{6}{\pi }\text{cos}\left(\frac{\pi x}{6}\right)$               (M1)

P(0 ≤ X ≤ 3) $=\frac{\pi }{36}\left({\left[-\frac{6x}{\pi }\text{cos}\left(\frac{\pi x}{6}\right)\right]}_0^3+\frac{6}{\pi }\underset{0}{\overset{3}{\int }}\text{cos}\left(\frac{\pi x}{6}\right)\text{d}x\right)$ (or equivalent)      A1A1

Note: Award A1 for a correct $uv$ and A1 for a correct $\int v\text{d}u$.

attempting to substitute limits       M1

$\frac{\pi }{36}{\left[-\frac{6x}{\pi }\text{cos}\left(\frac{\pi x}{6}\right)\right]}_0^3=0$       (A1)

so P(0 ≤ X ≤ 3) $=\frac{1}{\pi }{\left[\text{sin}\left(\frac{\pi x}{6}\right)\right]}_0^3$ (or equivalent)       A1

$=\frac{1}{\pi }$      A1

[7 marks]

$\text{P}\left(X>\mu +5\right)$.

.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

use of symmetry eg diagram       (M1)

$\text{P}\left(X>\mu +5\right)=0.2$       A1

[2 marks]

EITHER

       (M1)

             (A1)

      $=\frac{0.6}{0.8}$      A1A1

Note: A1 for denominator is independent of the previous A marks.

OR

use of diagram       (M1)

Note: Only award (M1) if the region  is indicated and used.

$\text{P}\left(X>\mu -5\right)=0.8$             (A1)

Note: Probabilities can be shown on the diagram.

$=\frac{0.6}{0.8}$      M1A1

THEN

$=\frac{3}{4}=\left(0.75\right)$      A1

[5 marks]

Find the value of $k$.

Find $\text{E}\left(X\right)$.

attempt to integrate $\frac{k}{\sqrt{4-3x^2}}$            (M1)

$=k⌊\frac{1}{\sqrt{3}}\text{arcsin}\left(\frac{\sqrt{3}}{2}x\right)⌋$            A1

Note: Award (M1)A0 for $\text{arcsin}\left(\frac{\sqrt{3}}{2}x\right)$.
Condone absence of $k$ up to this stage.

equating their integrand to $1$             M1

$k{\left[\frac{1}{\sqrt{3}}\text{arcsin}\left(\frac{\sqrt{3}}{2}x\right)\right]}_0^1=1$

$k=\frac{3\sqrt{3}}{\mathrm{\pi }}$            A1

[4 marks]

$\text{E}\left(X\right)=\frac{3\sqrt{3}}{\mathrm{\pi }}{\int }_0^1\frac{x}{\sqrt{4-3x^2}}dx$            A1

Note: Condone absence of limits if seen at a later stage.

EITHER

attempt to integrate by inspection            (M1)

$=\frac{3\sqrt{3}}{\mathrm{\pi }}\times -\frac{1}{6}\int -6x{\left(4-3x^2\right)}^{-\frac{1}{2}}dx$

$=\frac{3\sqrt{3}}{\mathrm{\pi }}{\left[-\frac{1}{3}\sqrt{4-3x^2}\right]}_0^1$            A1

Note: Condone the use of $k$ up to this stage.

OR

for example, $u=4-3x^2\Rightarrow \frac{du}{dx}=-6x$

Note: Other substitutions may be used. For example $u=-3x^2$.

$=-\frac{\sqrt{3}}{2\mathrm{\pi }}{\int }_4^1{u}^{-\frac{1}{2}}du$            M1

Note: Condone absence of limits up to this stage.

$=-\frac{\sqrt{3}}{2\mathrm{\pi }}{\left[2\sqrt{u}\right]}_4^1$            A1

Note: Condone the use of $k$ up to this stage.

THEN

$=\frac{\sqrt{3}}{\mathrm{\pi }}$            A1

Note: Award A0M1A1A0 for their $k\left[-\frac{1}{3}\sqrt{4-3x^2}\right]$ or $k\left[-2\sqrt{u}\right]$ for working with incorrect or no limits.

[4 marks]

Most candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x² was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.

Find the value of a and the value of b.

Find the expected value of T.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$a=\frac{3}{16}$ and $b=\frac{5}{16}$     (M1)A1A1

[3 marks]

Note: Award M1 for consideration of the possible outcomes when rolling the two dice.

$\text{E}\left(T\right)=\frac{1+6+15+28}{16}=\frac{25}{8}\left(=3.125\right)$     (M1)A1

Note: Allow follow through from part (a) even if probabilities do not add up to 1.

[2 marks]