| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Easy -1.2 This is a straightforward application of standard expectation and variance formulas for discrete distributions. Part (a)(i) requires basic E(X) = Σxp(x) calculation, part (a)(ii) uses the linear transformation property E(aX+b) = aE(X)+b, and part (b) uses Var(aX+b) = a²Var(X) with Var(X) = E(X²) - [E(X)]². All are direct formula applications with simple arithmetic, requiring only recall of standard results from S2 with no problem-solving or insight needed. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| \(\boldsymbol { y }\) | 5 | 15 | 25 | 35 |
| \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\) | 0.1 | 0.2 | 0.3 | 0.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(Y) = \sum y P(Y=y)\) | ||
| \(= 5\times0.1 + 15\times0.2 + 25\times0.3 + 35\times0.4 = 25\) | B1 | |
| \(\text{Var}(Y) = E(Y^2) - [E(Y)]^2\) | ||
| \(= 725 - 25^2\) | M1 | |
| \(= 100\) | A1 | CAO |
| Standard deviation \(= 10\) | A1ft | 4 marks total; ft on \(\text{Var}(Y) > 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(C = 10Y + 5\) | ||
| \(E(C) = 10E(Y) + 5 = 10\times25 + 5 = 255\) pence | B1 | 1 mark total; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{Var}(X) = E(X^2) - [E(X)]^2 = 75.25 - 8.35^2 = 75.25 - 69.7225 = 5.5275\) | M1, A1 | AWFW 5.52 to 5.53 |
| \(T = 0.4X + 250\) | ||
| \(\text{Var}(T) = \text{Var}(0.4X + 250) = 0.4^2 \times \text{Var}(X)\) | M1 | \(\text{Var}(X) > 0\) |
| \(= 0.16 \times 5.5275 = 0.8844\) | A1 | 4 marks total; AWFW 0.884 to 0.885 |
## Question 7:
### Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(Y) = \sum y P(Y=y)$ | | |
| $= 5\times0.1 + 15\times0.2 + 25\times0.3 + 35\times0.4 = 25$ | B1 | |
| $\text{Var}(Y) = E(Y^2) - [E(Y)]^2$ | | |
| $= 725 - 25^2$ | M1 | |
| $= 100$ | A1 | CAO |
| Standard deviation $= 10$ | A1ft | 4 marks total; ft on $\text{Var}(Y) > 0$ |
### Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $C = 10Y + 5$ | | |
| $E(C) = 10E(Y) + 5 = 10\times25 + 5 = 255$ pence | B1 | 1 mark total; OE |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{Var}(X) = E(X^2) - [E(X)]^2 = 75.25 - 8.35^2 = 75.25 - 69.7225 = 5.5275$ | M1, A1 | AWFW 5.52 to 5.53 |
| $T = 0.4X + 250$ | | |
| $\text{Var}(T) = \text{Var}(0.4X + 250) = 0.4^2 \times \text{Var}(X)$ | M1 | $\text{Var}(X) > 0$ |
| $= 0.16 \times 5.5275 = 0.8844$ | A1 | 4 marks total; AWFW 0.884 to 0.885 |
---7
\begin{enumerate}[label=(\alph*)]
\item The number of text messages, $N$, sent by Peter each month on his mobile phone never exceeds 40.
When $0 \leqslant N \leqslant 10$, he is charged for 5 messages.\\
When $10 < N \leqslant 20$, he is charged for 15 messages.\\
When $20 < N \leqslant 30$, he is charged for 25 messages.\\
When $30 < N \leqslant 40$, he is charged for 35 messages.\\
The number of text messages, $Y$, that Peter is charged for each month has the following probability distribution:
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$\boldsymbol { y }$ & 5 & 15 & 25 & 35 \\
\hline
$\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )$ & 0.1 & 0.2 & 0.3 & 0.4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Calculate the mean and the standard deviation of $Y$.
\item The Goodtime phone company makes a total charge for text messages, $C$ pence, each month given by
$$C = 10 Y + 5$$
Calculate $\mathrm { E } ( C )$.
\end{enumerate}\item The number of text messages, $X$, sent by Joanne each month on her mobile phone is such that
$$\mathrm { E } ( X ) = 8.35 \quad \text { and } \quad \mathrm { E } \left( X ^ { 2 } \right) = 75.25$$
The Newtime phone company makes a total charge for text messages, $T$ pence, each month given by
$$T = 0.4 X + 250$$
Calculate $\operatorname { Var } ( T )$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2008 Q7 [9]}}