| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Moderate -0.3 This is a straightforward S2 uniform distribution question requiring standard formulas and algebraic manipulation. Part (a) uses basic properties (E(X) = (a+b)/2 and probability as length/total length), parts (b-c) apply linearity of expectation and the variance formula, and part (d) is simple probability calculation. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | \(\frac{5-3}{b-a} = \frac{1}{8} \Rightarrow b - a = 16\) | M1 M1 |
| \(\frac{a+b}{2} = 4 \Rightarrow a + b = 8\) | ||
| \(a = -4\) and \(b = 12\) | A1 | both \(a = -4\) and \(b = 12\) |
| Part (b) | \([cE(X) - 2] = [4c - 2 = 0]\) or \(\int_{-4}^{12} \frac{1}{16}(cx - 2) \, dx = 0\) | M1 |
| \(c = \frac{1}{2}\) | A1 | |
| Part (c) | \(\text{Var}(X) = \frac{(12' - ('-4'))^2}{12} \left( = \frac{64}{3} \right)\) or \(\int_{-4}^{12} \frac{1}{16} x^2 \, dx\) | B1ft |
| \(E(X^2) = \left( \frac{64}{3} \right) + 4^2 = \frac{112}{3}\) or \(\left[\frac{'12'^3}{48} - \frac{'-4'^3}{48}\right] = \frac{112}{3}\) | M1 A1 | M1 for substituting \(E(X) = 4\) and their \(\text{Var}(X)\) into a correct expression for \(E(X^2)\) or correct integration with correct use of limits (ft their \(a\) and \(b\)); A1 for \(\frac{112}{3}\) or exact equivalent |
| Part (d) | \(P(2X > a + b) = P(X > 4) = 0.5\) | M1 A1 |
| Total (10) |
| **Part (a)** | $\frac{5-3}{b-a} = \frac{1}{8} \Rightarrow b - a = 16$ | M1 M1 | 1st M1 for $\frac{5-3}{b-a} = \frac{1}{8}$ oe or a sketch of a uniform distribution with a range of 16; 2nd M1 for $\frac{a+b}{2} = 4$ oe or a sketch of a uniform distribution centred on 4 |
| | $\frac{a+b}{2} = 4 \Rightarrow a + b = 8$ | | |
| | $a = -4$ and $b = 12$ | A1 | both $a = -4$ **and** $b = 12$ |
| **Part (b)** | $[cE(X) - 2] = [4c - 2 = 0]$ or $\int_{-4}^{12} \frac{1}{16}(cx - 2) \, dx = 0$ | M1 | for $4c - 2 = 0$ or attempt correct equation for $E(cX - 2) = 0$ using integration |
| | $c = \frac{1}{2}$ | A1 | |
| **Part (c)** | $\text{Var}(X) = \frac{(12' - ('-4'))^2}{12} \left( = \frac{64}{3} \right)$ or $\int_{-4}^{12} \frac{1}{16} x^2 \, dx$ | B1ft | for substituting their $a$ and $b$ into a correct expression for $\text{Var}(X)$ or correct integral with limits (ft their $a$ and $b$) for $E(X^2)$ |
| | $E(X^2) = \left( \frac{64}{3} \right) + 4^2 = \frac{112}{3}$ or $\left[\frac{'12'^3}{48} - \frac{'-4'^3}{48}\right] = \frac{112}{3}$ | M1 A1 | M1 for substituting $E(X) = 4$ and their $\text{Var}(X)$ into a correct expression for $E(X^2)$ or correct integration with correct use of limits (ft their $a$ and $b$); A1 for $\frac{112}{3}$ or exact equivalent |
| **Part (d)** | $P(2X > a + b) = P(X > 4) = 0.5$ | M1 A1 | M1 for correct rearrangement up to $P(X > k)$ where $k = \frac{a+b}{2}$ (may be implied by a correct ft answer) |
| | | | Total (10) |
---The continuous random variable $X$ is uniformly distributed over the interval $[a, b]$
Given that $\mathrm{P}(3 < X < 5) = \frac{1}{8}$ and $\mathrm{E}(X) = 4$
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$ and the value of $b$ [3]
\item find the value of the constant, $c$, such that $\mathrm{E}(cX - 2) = 0$ [2]
\item find the exact value of $\mathrm{E}(X^2)$ [3]
\item find $\mathrm{P}(2X - b > a)$ [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q2 [10]}}