| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Multiple independent observations |
| Difficulty | Standard +0.3 This is a standard S2 continuous probability distribution question requiring routine integration techniques (finding k by normalizing, calculating E(X) and Var(X), finding probabilities). Part (d) adds a simple binomial probability application. While multi-part with 18 marks total, each component follows textbook procedures without requiring novel insight—slightly easier than average due to the symmetric pdf making E(X) immediately apparent and straightforward polynomial integration throughout. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| (a) \(\int_2^{10} f(x) \, dx = 1\), so \(k\int_2^{10} (-x^2 + 12x - 20) \, dx = 1\) | M1 M1 A1 |
| \(k\left[-\frac{1}{3}x^3 + 6x^2 - 20x\right]_2^{10} = 1\), \(\frac{256k}{3} = 1\), \(k = \frac{3}{256}\) | M1 A1 |
| By symmetry of parabola, mean \(= 6\) | B1 |
| (b) \(E(X^2) = \frac{3}{256}\int_2^{10} (-x^4 + 12x^3 - 20x^2) \, dx\) | M1 A1 |
| \(= \frac{3}{256}\left[-\frac{x^5}{5} + 3x^4 - \frac{20x^3}{3}\right]_2^{10} = 39.2\), \(\text{Var}(X) = 3.2\) | M1 A1 A1 |
| Standard deviation \(= \sqrt{3.2} = 1.789\), i.e. £1789 | A1 |
| (c) \(P(X > 8) = \int_8^{10} f(x) \, dx = \frac{3}{256}\left[-\frac{1}{3}x^3 + 6x^2 - 20x\right]_8^{10} = 0.156\) | M1 A1 M1 A1 |
| (d) \(0.156^4 = 0.000592\) | M1 A1 |
(a) $\int_2^{10} f(x) \, dx = 1$, so $k\int_2^{10} (-x^2 + 12x - 20) \, dx = 1$ | M1 M1 A1 |
$k\left[-\frac{1}{3}x^3 + 6x^2 - 20x\right]_2^{10} = 1$, $\frac{256k}{3} = 1$, $k = \frac{3}{256}$ | M1 A1 |
By symmetry of parabola, mean $= 6$ | B1 |
(b) $E(X^2) = \frac{3}{256}\int_2^{10} (-x^4 + 12x^3 - 20x^2) \, dx$ | M1 A1 |
$= \frac{3}{256}\left[-\frac{x^5}{5} + 3x^4 - \frac{20x^3}{3}\right]_2^{10} = 39.2$, $\text{Var}(X) = 3.2$ | M1 A1 A1 |
Standard deviation $= \sqrt{3.2} = 1.789$, i.e. £1789 | A1 |
(c) $P(X > 8) = \int_8^{10} f(x) \, dx = \frac{3}{256}\left[-\frac{1}{3}x^3 + 6x^2 - 20x\right]_8^{10} = 0.156$ | M1 A1 M1 A1 |
(d) $0.156^4 = 0.000592$ | M1 A1 |
**Total: 18 marks**A corner-shop has weekly sales (in thousands of pounds), which can be modelled by the continuous random variable $X$ with probability density function
$$f(x) = k(x-2)(10-x) \quad 2 \leq x \leq 10,$$
$$f(x) = 0 \quad \text{otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{3}{256}$ and write down the mean of $X$. [6 marks]
\item Find the standard deviation of the weekly sales. [6 marks]
\item Find the probability that the sales exceed £8 000 in any particular week. [4 marks]
\end{enumerate}
If the sales exceed £8 000 per week for 4 consecutive weeks, the manager gets a bonus.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the probability that the manager gets a bonus in February. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [18]}}