| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard CDF properties: using F(5)=1 to find k, evaluating the CDF at a point, and differentiating to find the pdf. All parts are routine applications of definitions with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(F(5) = 1\); \(k(95 - 25 - 34) = 1; 36k = 1 \therefore k = \frac{1}{36}\) | M1 A1 | |
| (b) \(P(X > 4) = 1 - F(4) = 1 - \frac{5}{36}(76 - 16 - 34) = \frac{5}{18}\) | M1 A1 | |
| (c) \(f(x) = F'(x) = \frac{1}{36}(19 - 2x)\); \(\therefore f(x) = \begin{cases} \frac{1}{36}(19 - 2x), & 2 \le x \le 5, \\ 0, & \text{otherwise} \end{cases}\) | M1 A1 A1 | (7 marks total) |
(a) $F(5) = 1$; $k(95 - 25 - 34) = 1; 36k = 1 \therefore k = \frac{1}{36}$ | M1 A1 |
(b) $P(X > 4) = 1 - F(4) = 1 - \frac{5}{36}(76 - 16 - 34) = \frac{5}{18}$ | M1 A1 |
(c) $f(x) = F'(x) = \frac{1}{36}(19 - 2x)$; $\therefore f(x) = \begin{cases} \frac{1}{36}(19 - 2x), & 2 \le x \le 5, \\ 0, & \text{otherwise} \end{cases}$ | M1 A1 A1 | (7 marks total)
---\begin{enumerate}
\item The continuous random variable $X$ has the following cumulative distribution function:
\end{enumerate}
$$F ( x ) = \begin{cases} 0 , & x < 2 \\ k \left( 19 x - x ^ { 2 } - 34 \right) , & 2 \leq x \leq 5 \\ 1 , & x > 5 \end{cases}$$
(a) Show that $k = \frac { 1 } { 36 }$.\\
(b) Find $\mathrm { P } ( X > 4 )$.\\
(c) Find and specify fully the probability density function $\mathrm { f } ( x )$ of $X$.\\
\hfill \mbox{\textit{Edexcel S2 Q1 [7]}}