| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with marginal/conditional probabilities |
| Difficulty | Moderate -0.3 This is a standard S3 bivariate distribution question requiring routine application of formulas for covariance, correlation, and linear combinations of random variables. All necessary values are given, requiring only substitution into standard formulas (Cov(X,Y) = E(XY) - E(X)E(Y), ρ = Cov/√(Var(X)Var(Y)), Var(aX+bY) = a²Var(X) + b²Var(Y) + 2abCov(X,Y)). The table reading and verification in part (a)(i) is trivial. This is slightly easier than average as it's purely computational with no problem-solving or interpretation required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \multirow{2}{*}{} | Number of morning operations ( \(\boldsymbol { X }\) ) | ||||||
| 2 | 3 | 4 | 5 | 6 | \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\) | ||
| \multirow{3}{*}{Number of afternoon operations ( \(\boldsymbol { Y }\) )} | 3 | 0.00 | 0.05 | 0.20 | 0.20 | 0.05 | 0.50 |
| 4 | 0.00 | 0.15 | 0.10 | 0.05 | 0.00 | 0.30 | |
| 5 | 0.05 | 0.05 | 0.10 | 0.00 | 0.00 | 0.20 | |
| \(\mathrm { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.05 | 0.25 | 0.40 | 0.25 | 0.05 | 1.00 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The distribution of \(X\) is symmetric about \(x = 4\) | B1 | Accept equivalent statement about equal probabilities either side |
| \(E(X^2) = 2^2(0.05) + 3^2(0.25) + 4^2(0.40) + 5^2(0.25) + 6^2(0.05)\) | M1 | Attempt \(E(X^2)\) using correct probabilities |
| \(= 0.20 + 2.25 + 6.40 + 6.25 + 1.80 = 16.9\) | A1 | Correct value |
| \(\text{Var}(X) = 16.9 - 4^2 = 0.9\) | A1 | Correct answer shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Cov}(X,Y) = E(XY) - E(X)E(Y)\) | M1 | Correct formula used |
| \(= 14.4 - 4 \times 3.7 = 14.4 - 14.8 = -0.4\) | A1 | Correct value |
| \(\rho_{XY} = \dfrac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X)\cdot\text{Var}(Y)}}\) | M1 | Correct formula used |
| \(= \dfrac{-0.4}{\sqrt{0.9 \times 0.61}} = \dfrac{-0.4}{\sqrt{0.549}} = -0.540\) | A1 | Accept \(-0.54\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T) = E(X) + E(Y) = 4 + 3.7 = 7.7\) | B1 | Correct mean |
| \(\text{Var}(T) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)\) | M1 | Correct formula with covariance term |
| \(= 0.9 + 0.61 + 2(-0.4) = 0.71\) | A1 | Correct variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(D) = E(X) - E(Y) = 4 - 3.7 = 0.3\) | B1 | Correct mean |
| \(\text{Var}(D) = \text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y)\) | ||
| \(= 0.9 + 0.61 - 2(-0.4) = 2.31\) | A1 | Correct variance, follow through on Cov |
# Question 5:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The distribution of $X$ is symmetric about $x = 4$ | B1 | Accept equivalent statement about equal probabilities either side |
| $E(X^2) = 2^2(0.05) + 3^2(0.25) + 4^2(0.40) + 5^2(0.25) + 6^2(0.05)$ | M1 | Attempt $E(X^2)$ using correct probabilities |
| $= 0.20 + 2.25 + 6.40 + 6.25 + 1.80 = 16.9$ | A1 | Correct value |
| $\text{Var}(X) = 16.9 - 4^2 = 0.9$ | A1 | Correct answer shown |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Cov}(X,Y) = E(XY) - E(X)E(Y)$ | M1 | Correct formula used |
| $= 14.4 - 4 \times 3.7 = 14.4 - 14.8 = -0.4$ | A1 | Correct value |
| $\rho_{XY} = \dfrac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X)\cdot\text{Var}(Y)}}$ | M1 | Correct formula used |
| $= \dfrac{-0.4}{\sqrt{0.9 \times 0.61}} = \dfrac{-0.4}{\sqrt{0.549}} = -0.540$ | A1 | Accept $-0.54$ |
## Part (b)(i): $T = X + Y$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T) = E(X) + E(Y) = 4 + 3.7 = 7.7$ | B1 | Correct mean |
| $\text{Var}(T) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)$ | M1 | Correct formula with covariance term |
| $= 0.9 + 0.61 + 2(-0.4) = 0.71$ | A1 | Correct variance |
## Part (b)(ii): $D = X - Y$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(D) = E(X) - E(Y) = 4 - 3.7 = 0.3$ | B1 | Correct mean |
| $\text{Var}(D) = \text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y)$ | | |
| $= 0.9 + 0.61 - 2(-0.4) = 2.31$ | A1 | Correct variance, follow through on Cov |
---5 The numbers of daily morning operations, $X$, and daily afternoon operations, $Y$, in an operating theatre of a small private hospital can be modelled by the following bivariate probability distribution.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{5}{|c|}{Number of morning operations ( $\boldsymbol { X }$ )} & \\
\hline
& & 2 & 3 & 4 & 5 & 6 & $\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )$ \\
\hline
\multirow{3}{*}{Number of afternoon operations ( $\boldsymbol { Y }$ )} & 3 & 0.00 & 0.05 & 0.20 & 0.20 & 0.05 & 0.50 \\
\hline
& 4 & 0.00 & 0.15 & 0.10 & 0.05 & 0.00 & 0.30 \\
\hline
& 5 & 0.05 & 0.05 & 0.10 & 0.00 & 0.00 & 0.20 \\
\hline
& $\mathrm { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.05 & 0.25 & 0.40 & 0.25 & 0.05 & 1.00 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State why $\mathrm { E } ( X ) = 4$ and show that $\operatorname { Var } ( X ) = 0.9$.
\item Given that
$$\mathrm { E } ( Y ) = 3.7 , \operatorname { Var } ( Y ) = 0.61 \text { and } \mathrm { E } ( X Y ) = 14.4$$
calculate values for $\operatorname { Cov } ( X , Y )$ and $\rho _ { X Y }$.
\end{enumerate}\item Calculate values for the mean and the variance of:
\begin{enumerate}[label=(\roman*)]
\item $T = X + Y$;
\item $\quad D = X - Y$.\\[0pt]
[4 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2014 Q5 [4]}}