AQA Further AS Paper 2 Statistics Specimen — Question 5 5 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Statistics (Further AS Paper 2 Statistics)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Random Variables
TypeMixed continuous-discrete problems
DifficultyStandard +0.8 This question requires finding E(T) where T = L² + ½S with L continuous and S discrete. Students must find the constant w using integration, calculate E(L²) using the transformation formula for continuous variables, calculate E(S) for the discrete variable, then apply linearity of expectation and independence. The multi-step nature, mixing continuous and discrete distributions, and requiring integration of l³ makes this moderately challenging for Further AS level.
Spec5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY)

5 Participants in a school jumping competition gain a total score for each jump based on the length, \(L\) metres, jumped beyond a fixed point and a mark, \(S\), for style. \(L\) may be regarded as a continuous random variable with probability density function $$\mathrm { f } ( l ) = \left\{ \begin{array} { c c } w l & 0 \leq l \leq 15 \\ 0 & \text { otherwise } \end{array} \right.$$ where \(w\) is a constant. \(S\) may be regarded as a discrete random variable with probability function $$\mathrm { P } ( S = s ) = \left\{ \begin{array} { c l } \frac { 1 } { 15 } s & s = 1,2,3,4,5 \\ 0 & \text { otherwise } \end{array} \right.$$ Assume that \(L\) and \(S\) are independent. The total score for a participant in this competition, \(T\), is given by \(T = L ^ { 2 } + \frac { 1 } { 2 } S\) Show that the expected total score for a participant is \(114 \frac { 1 } { 3 }\)

Question 5:
AnswerMarks Guidance
Part 1: \(\int_0^{15} wl \, dl = 1 \Rightarrow w\left[\frac{l^2}{2}\right]_0^{15} = \frac{225w}{2} = 1 \Rightarrow w = \frac{2}{225}\)M1 Allow one error if method correct
Part 2: \(E(L^2) = \int_0^{15} l^2 \times \frac{2}{225} l \, dl = \frac{2}{225}\left[\frac{l^4}{4}\right]_0^{15} = \frac{225}{2}\)M1 FT 'their' value for \(w\)
Part 3: \(E(S) = \frac{1}{15}(0\times0 + 1\times1 + 2\times2 + \ldots + 5\times5) = \frac{55}{15} = \frac{11}{3}\)M1 Allow one error if method correct
Part 4: Uses \(E(T) = E(L^2) + \frac{1}{2}E(S)\)M1
Part 5: \(E(T) = \frac{225}{2} + \frac{11}{6} = \frac{343}{3} = 114\frac{1}{3}\) AGR1 Must have completely correct, clear, easy to follow solution with no slips
## Question 5:

**Part 1:** $\int_0^{15} wl \, dl = 1 \Rightarrow w\left[\frac{l^2}{2}\right]_0^{15} = \frac{225w}{2} = 1 \Rightarrow w = \frac{2}{225}$ | M1 | Allow one error if method correct

**Part 2:** $E(L^2) = \int_0^{15} l^2 \times \frac{2}{225} l \, dl = \frac{2}{225}\left[\frac{l^4}{4}\right]_0^{15} = \frac{225}{2}$ | M1 | FT 'their' value for $w$

**Part 3:** $E(S) = \frac{1}{15}(0\times0 + 1\times1 + 2\times2 + \ldots + 5\times5) = \frac{55}{15} = \frac{11}{3}$ | M1 | Allow one error if method correct

**Part 4:** Uses $E(T) = E(L^2) + \frac{1}{2}E(S)$ | M1 |

**Part 5:** $E(T) = \frac{225}{2} + \frac{11}{6} = \frac{343}{3} = 114\frac{1}{3}$ **AG** | R1 | Must have completely correct, clear, easy to follow solution with no slips

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5 Participants in a school jumping competition gain a total score for each jump based on the length, $L$ metres, jumped beyond a fixed point and a mark, $S$, for style.\\
$L$ may be regarded as a continuous random variable with probability density function

$$\mathrm { f } ( l ) = \left\{ \begin{array} { c c } 
w l & 0 \leq l \leq 15 \\
0 & \text { otherwise }
\end{array} \right.$$

where $w$ is a constant.\\
$S$ may be regarded as a discrete random variable with probability function

$$\mathrm { P } ( S = s ) = \left\{ \begin{array} { c l } 
\frac { 1 } { 15 } s & s = 1,2,3,4,5 \\
0 & \text { otherwise }
\end{array} \right.$$

Assume that $L$ and $S$ are independent.

The total score for a participant in this competition, $T$, is given by $T = L ^ { 2 } + \frac { 1 } { 2 } S$

Show that the expected total score for a participant is $114 \frac { 1 } { 3 }$

\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics  Q5 [5]}}