| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Moderate -0.3 This is a straightforward S2 question requiring standard probability distribution calculations: completing a table using a given formula, finding probabilities, calculating E(R) and Var(R), and applying linear transformations. All steps are routine applications of formulas with no conceptual challenges, though the multi-part nature and arithmetic (especially the geometric series calculations) require care. Slightly easier than average due to being purely procedural. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(\boldsymbol { r }\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\) | 0.5 | 0.0296 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\) | 1 | 2 |
| \(p\) | .5 | .24 |
| \(0.4 \times 0.6 = 0.24\) | B2,1 | 2 |
| \(0.24 \times 0.6 = 0.144\) | ||
| \(0.144 \times 0.6 = 0.0864\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(\text{fewer than 3 bedrooms are rented}) = \mathbb{P}(R = 1,2)\) | Alternative: \(\mathbb{P}(\text{fewer than 3 not rented}) = \mathbb{P}(0, 1 \text{ or 2 not rented}) = \mathbb{P}(5, 4 \text{ or 3 are rented}) = \mathbb{P}(R = 3, 4, 5)\) M1 | |
| \(\mathbb{P}(\text{fewer than 3 bedrooms not rented}) = 1 - \mathbb{P}(R = 1,2)\) | M1 | \(p = 0.4 \times 0.6^2 + 0.4 \times 0.6^3 + 0.0296 = 0.144 + 0.0864 + 0.0296\) m1 |
| \(= 1 - \mathbb{P}(1 \text{ or 2 rooms are rented})\) | [or their \(p(3) + p(4) \leq 0.4704\) value from table used] | |
| \(= 1 - (0.5 + \mathbf{0.24})\) | m1 | |
| [their \(0 < p(2) \leq 0.4704\) value from table used] | ||
| \(= 1 - 0.74 = 0.26\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(R) = 0.5 \times 1 + 0.4 \times 0.6 \times 2\) | M1 | \(\sum r_i \times \mathbb{P}(R = r)\) from their table |
| \(+ 0.4 \times 0.6^2 \times 3 + 0.4 \times 0.6^3 \times 4\) | ||
| \(+ 0.0296 \times 5\) | ||
| \(= 0.5 + 0.48 + 0.432 + 0.3456 + 0.148\) | ||
| \(\left[= \frac{1}{2} + \frac{12}{25} + \frac{54}{125} + \frac{216}{625} + \frac{37}{250}\right]\) | ||
| \(\therefore E(R) = 1.9056\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(R^2) = 0.5 \times 1^2 + 0.4 \times 0.6 \times 2^2\) | B1 | AG |
| \(+ 0.4 \times 0.6^2 \times 3^2 + 0.4 \times 0.6^3 \times 4^2\) | ||
| \(+ 0.0296 \times 5^2\) | ||
| \(E(R^2) = 4.8784\) | ||
| \(\text{Var}(R) = 4.8784 - 1.9056^2\) | M1 | 4.8784 – their \(E^2(R)\) |
| \((= 1.24708864)\) | ||
| \(= 1.25 (3sf)\) | A1 | 3 |
| 13 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(M) = 1250E(R) - 282 = 1250 \times 1.9056 - 282 = 2100\) | B1 | cao |
| \(\text{Var}(M) = 1250^2 \times [4.8784 - 1.9056^2]\) | M1 | \(1250^2 \times\) their Var\((R) > 0\) in (c)(ii) (1 948 473 to 1 953 125) |
| \(\text{sd}(M) = 1250 \times \sqrt{1.24708864} = 1395.91\) | A1 | 3 |
| 13 |
### 4(a)
| $r$ | 1 | 2 | 3 | 4 | 5 |
| --- | --- | --- | --- | --- | --- |
| $p$ | .5 | .24 | **.144** | .0864 | 0.0296 |
| | | | | | |
| $0.4 \times 0.6 = 0.24$ | B2,1 | 2 | B1 for any 1 correct (unsimplified) (B1) B2 all correct and simplified |
| $0.24 \times 0.6 = 0.144$ | | | |
| $0.144 \times 0.6 = 0.0864$ | | | |
### 4(b)
| $\mathbb{P}(\text{fewer than 3 bedrooms are rented}) = \mathbb{P}(R = 1,2)$ | | **Alternative:** $\mathbb{P}(\text{fewer than 3 not rented}) = \mathbb{P}(0, 1 \text{ or 2 not rented}) = \mathbb{P}(5, 4 \text{ or 3 are rented}) = \mathbb{P}(R = 3, 4, 5)$ **M1** |
| --- | --- | --- |
| $\mathbb{P}(\text{fewer than 3 bedrooms not rented}) = 1 - \mathbb{P}(R = 1,2)$ | M1 | $p = 0.4 \times 0.6^2 + 0.4 \times 0.6^3 + 0.0296 = 0.144 + 0.0864 + 0.0296$ **m1** |
| $= 1 - \mathbb{P}(1 \text{ or 2 rooms are rented})$ | | [or their $p(3) + p(4) \leq 0.4704$ value from table used] |
| $= 1 - (0.5 + \mathbf{0.24})$ | m1 | |
| [their $0 < p(2) \leq 0.4704$ value from table used] | | |
| $= 1 - 0.74 = 0.26$ | A1 | 3 | = 0.26 (cao) **A1** [SC 0.74 for B1] |
### 4(c)(i)
| $E(R) = 0.5 \times 1 + 0.4 \times 0.6 \times 2$ | M1 | $\sum r_i \times \mathbb{P}(R = r)$ from their table |
| --- | --- | --- |
| $+ 0.4 \times 0.6^2 \times 3 + 0.4 \times 0.6^3 \times 4$ | | |
| $+ 0.0296 \times 5$ | | |
| $= 0.5 + 0.48 + 0.432 + 0.3456 + 0.148$ | | |
| $\left[= \frac{1}{2} + \frac{12}{25} + \frac{54}{125} + \frac{216}{625} + \frac{37}{250}\right]$ | | |
| $\therefore E(R) = 1.9056$ | A1 | 2 | [awfw 1.9 to 1.91] $\frac{566}{625}$ |
### 4(c)(ii)
| $E(R^2) = 0.5 \times 1^2 + 0.4 \times 0.6 \times 2^2$ | B1 | AG |
| --- | --- | --- |
| $+ 0.4 \times 0.6^2 \times 3^2 + 0.4 \times 0.6^3 \times 4^2$ | | |
| $+ 0.0296 \times 5^2$ | | |
| $E(R^2) = 4.8784$ | | |
| $\text{Var}(R) = 4.8784 - 1.9056^2$ | M1 | 4.8784 – their $E^2(R)$ |
| $(= 1.24708864)$ | | |
| $= 1.25 (3sf)$ | A1 | 3 | (awfw 1.23 to 1.25) |
| | | **13** | |
## Question 4(d)
| $E(M) = 1250E(R) - 282 = 1250 \times 1.9056 - 282 = 2100$ | B1 | cao |
| --- | --- | --- |
| $\text{Var}(M) = 1250^2 \times [4.8784 - 1.9056^2]$ | M1 | $1250^2 \times$ their Var$(R) > 0$ in (c)(ii) (1 948 473 to 1 953 125) |
| $\text{sd}(M) = 1250 \times \sqrt{1.24708864} = 1395.91$ | A1 | 3 | $\text{sd}(M) = \sqrt{1948437} = 1395.9$ ($\sqrt{1953125} = 1397.5$) (awfw 1395 to 1400) |
| | | **13** | |4 A house has a total of five bedrooms, at least one of which is always rented.\\
The probability distribution for $R$, the number of bedrooms that are rented at any given time, is given by
$$\mathrm { P } ( R = r ) = \begin{cases} 0.5 & r = 1 \\ 0.4 ( 0.6 ) ^ { r - 1 } & r = 2,3,4 \\ 0.0296 & r = 5 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below.
\item Find the probability that fewer than 3 bedrooms are not rented at any given time.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\mathrm { E } ( R )$.
\item Show that $\mathrm { E } \left( R ^ { 2 } \right) = 4.8784$ and hence find the value of $\operatorname { Var } ( R )$.
\end{enumerate}\item Bedrooms are rented on a monthly basis.
The monthly income, $\pounds M$, from renting bedrooms in the house may be modelled by
$$M = 1250 R - 282$$
Find the mean and the standard deviation of $M$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$\boldsymbol { r }$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )$ & 0.5 & & & & 0.0296 \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q4 [13]}}