| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for mean and variance from a discrete probability distribution, followed by routine use of linear transformation properties. Both parts require only direct calculation with no problem-solving insight, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(\boldsymbol { x }\) | 40 | 45 | 55 | 74 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.30 | 0.24 | 0.36 | 0.10 |
| Answer | Marks | Guidance |
|---|---|---|
| \[E(X) = \sum_{\text{all } x} x \cdot P(X=x) = 50\] | B1 | |
| \[E(X^2) = \sum_{\text{all } x} x^2 P(X=x) = 2602.6(0)\] | M1 | |
| \[\text{Var}(X) = E(X^2) - [E(X)]^2 = 2602.6 - 50^2 = 102.6(0)\] | M1 | |
| \[\therefore \text{standard deviation}(X) = 10.13\] | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \[E(Y) = \mu = E(10X + 250) = 10 \times E(X) + 250 = 750\] | B1√ | |
| \[\text{s.d}(Y) = 10 \times 10.1 = 101\] | B1√ | 2 marks |
**5(a)**
$$E(X) = \sum_{\text{all } x} x \cdot P(X=x) = 50$$ | B1 | | (cao)
$$E(X^2) = \sum_{\text{all } x} x^2 P(X=x) = 2602.6(0)$$ | M1 | |
$$\text{Var}(X) = E(X^2) - [E(X)]^2 = 2602.6 - 50^2 = 102.6(0)$$ | M1 | |
$$\therefore \text{standard deviation}(X) = 10.13$$ | A1 | 4 marks | (to nearest 1p)
**5(b)**
$$E(Y) = \mu = E(10X + 250) = 10 \times E(X) + 250 = 750$$ | B1√ | | (on their E(X))
$$\text{s.d}(Y) = 10 \times 10.1 = 101$$ | B1√ | 2 marks | (on their sd(X))
**Question 5 Total: 6 marks**
---5 The Globe Express agency organises trips to the theatre. The cost, $\pounds X$, of these trips can be modelled by the following probability distribution:
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 40 & 45 & 55 & 74 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.30 & 0.24 & 0.36 & 0.10 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean and standard deviation of $X$.
\item For special celebrity charity performances, Globe Express increases the cost of the trips to $\pounds Y$, where
$$Y = 10 X + 250$$
Determine the mean and standard deviation of $Y$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q5 [6]}}