AQA Further AS Paper 2 Statistics 2021 June — Question 6 11 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Statistics (Further AS Paper 2 Statistics)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeFind median or percentiles
DifficultyStandard +0.3 This is a straightforward Further Maths statistics question involving standard techniques: integrating a linear pdf to find the median (solving a quadratic), computing a probability via integration, and finding variance of a transformation using E(1/Y²) - [E(1/Y)]². All parts are routine applications of formulas with no conceptual challenges beyond careful algebra.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration
6 The continuous random variable \(X\) has probability density function $$f ( x ) = \begin{cases} \frac { 1 } { 114 } ( 4 x + 7 ) & 0 \leq x \leq 6 \\ 0 & \text { otherwise } \end{cases}$$ 6
  1. Show that the median of \(X\) is 3.87, correct to three significant figures.
    [0pt] [3 marks]
    6
  2. Find the exact value of \(\mathrm { P } ( X > 2 )\)
    6
  3. The continuous random variable \(Y\) has probability density function \(g ( y ) = \begin{cases} \frac { 1 } { 2 } y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 }1 \leq y \leq 3
    0\text { otherwise } \end{cases}\)
    "
    6 (c) (i) Show that \(\operatorname { Var } \left( \frac { 1 } { Y } \right) = \frac { 2 } { 81 }\)
    		\multirow[b]{2}{*}{
    [4 marks]
    [4 marks]
    }
Question 6(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\int_0^m \frac{1}{114}(4x+7)\, dx = 0.5\)M1 Integrates \(\int_0^m \frac{1}{114}(4x+7)\,dx\) and equates with \(0.5\)
\(\frac{1}{114}(2m^2 + 7m) = 0.5\), giving \(\frac{1}{57}m^2 + \frac{7}{114}m - 0.5 = 0\)M1 Obtains correct quadratic equation in terms of the median equal to zero
\(m = 3.8680512\), so \(m = 3.87\) (3 sig fig)R1 Completes rigorous argument; more accurate answer must be seen and other solution from quadratic must be rejected if seen
Question 6(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(X > 2) = \int_2^6 \frac{2}{57}x + \frac{7}{114}\, dx\)M1 Uses an integral of \(f(x)\) with one limit of \(2\)
\(= \frac{46}{57}\)A1 Obtains correct exact value of \(P(X > 2)\)
Question 6(c)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(E\!\left(\frac{1}{Y}\right) = \int_1^3 y^{-1}\!\left(\frac{1}{2}y^2 - \frac{1}{6}y^3\right)dy = \frac{5}{9}\)M1 Uses general formula for \(E(f(Y))\) to obtain \(E\!\left(\frac{1}{Y}\right)\)
\(E\!\left(\frac{1}{Y^2}\right) = \int_1^3 y^{-2}\!\left(\frac{1}{2}y^2 - \frac{1}{6}y^3\right)dy = \frac{1}{3}\)M1 Uses general formula for \(E(f(Y))\) to obtain \(E\!\left(\frac{1}{Y^2}\right)\)
\(\text{Var}\!\left(\frac{1}{Y}\right) = E\!\left(\frac{1}{Y^2}\right) - \left(E\!\left(\frac{1}{Y}\right)\right)^2 = \frac{1}{3} - \left(\frac{5}{9}\right)^2\)M1 Uses variance formula to obtain expression for \(\text{Var}\!\left(\frac{1}{Y}\right)\)
\(= \frac{2}{81}\)A1 Completes rigorous argument to obtain given value of \(\text{Var}\!\left(\frac{1}{Y}\right)\)
Question 6(c)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\text{Var}\!\left(2X - \frac{3}{Y}\right) = 2^2\,\text{Var}(X) + 3^2\,\text{Var}\!\left(\frac{1}{Y}\right) = 4 \times \frac{939}{361} + 9 \times \frac{2}{81}\)M1 Uses correct general formula for sum of linear functions of independent variables; condone sign error
\(= 10.6\)A1 Obtains correct value; AWRT 10.6 
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^m \frac{1}{114}(4x+7)\, dx = 0.5$ | M1 | Integrates $\int_0^m \frac{1}{114}(4x+7)\,dx$ and equates with $0.5$ |
| $\frac{1}{114}(2m^2 + 7m) = 0.5$, giving $\frac{1}{57}m^2 + \frac{7}{114}m - 0.5 = 0$ | M1 | Obtains correct quadratic equation in terms of the median equal to zero |
| $m = 3.8680512$, so $m = 3.87$ (3 sig fig) | R1 | Completes rigorous argument; more accurate answer must be seen and other solution from quadratic must be rejected if seen |

## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 2) = \int_2^6 \frac{2}{57}x + \frac{7}{114}\, dx$ | M1 | Uses an integral of $f(x)$ with one limit of $2$ |
| $= \frac{46}{57}$ | A1 | Obtains correct exact value of $P(X > 2)$ |

## Question 6(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E\!\left(\frac{1}{Y}\right) = \int_1^3 y^{-1}\!\left(\frac{1}{2}y^2 - \frac{1}{6}y^3\right)dy = \frac{5}{9}$ | M1 | Uses general formula for $E(f(Y))$ to obtain $E\!\left(\frac{1}{Y}\right)$ |
| $E\!\left(\frac{1}{Y^2}\right) = \int_1^3 y^{-2}\!\left(\frac{1}{2}y^2 - \frac{1}{6}y^3\right)dy = \frac{1}{3}$ | M1 | Uses general formula for $E(f(Y))$ to obtain $E\!\left(\frac{1}{Y^2}\right)$ |
| $\text{Var}\!\left(\frac{1}{Y}\right) = E\!\left(\frac{1}{Y^2}\right) - \left(E\!\left(\frac{1}{Y}\right)\right)^2 = \frac{1}{3} - \left(\frac{5}{9}\right)^2$ | M1 | Uses variance formula to obtain expression for $\text{Var}\!\left(\frac{1}{Y}\right)$ |
| $= \frac{2}{81}$ | A1 | Completes rigorous argument to obtain given value of $\text{Var}\!\left(\frac{1}{Y}\right)$ |

## Question 6(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}\!\left(2X - \frac{3}{Y}\right) = 2^2\,\text{Var}(X) + 3^2\,\text{Var}\!\left(\frac{1}{Y}\right) = 4 \times \frac{939}{361} + 9 \times \frac{2}{81}$ | M1 | Uses correct general formula for sum of linear functions of independent variables; condone sign error |
| $= 10.6$ | A1 | Obtains correct value; AWRT 10.6 |
6 The continuous random variable $X$ has probability density function

$$f ( x ) = \begin{cases} \frac { 1 } { 114 } ( 4 x + 7 ) & 0 \leq x \leq 6 \\ 0 & \text { otherwise } \end{cases}$$

6
\begin{enumerate}[label=(\alph*)]
\item Show that the median of $X$ is 3.87, correct to three significant figures.\\[0pt]
[3 marks]\\

6
\item Find the exact value of $\mathrm { P } ( X > 2 )$\\

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
 & \begin{tabular}{l}
6
\item \\
The continuous random variable $Y$ has probability density function \(g ( y ) = \begin{cases} \frac { 1 } { 2 } y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 } & 1 \leq y \leq 3 \\ 0 & \text { otherwise } \end{cases}\) \\
" \\
6 (c) (i) Show that $\operatorname { Var } \left( \frac { 1 } { Y } \right) = \frac { 2 } { 81 }$ \\
\end{tabular} & \multirow[b]{2}{*}{\begin{tabular}{l}
[4 marks] \\[0pt]
[4 marks] \\
\end{tabular}} \\
\hline
 &  &  \\
\hline
\end{tabular}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2021 Q6 [11]}}
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