| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Standard +0.8 This S2 question requires multiple techniques: using continuity conditions at x=5 to get one equation, differentiating the CDF to find the PDF, computing E(X²) by integration and setting equal to 6.25 for a second equation, then solving the simultaneous system. While methodical, it involves more algebraic manipulation and integration than typical S2 questions, plus the constraint-solving aspect adds complexity beyond standard CDF exercises. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(f(x) = \frac{1}{100}(3ax^2 + 2bx + 15)\) \(E(X^2) = \int x^2 f(x)dx\) \(\frac{1}{100}[\frac{2}{5}ax^5 + \frac{1}{2}bx^4 + 5x^3]_0^6 = 6.25\) \(1875a + 312.5b + 625 = 625\) \(6a + b = 0\) * | M1 A1 M1 dM1 A1 cso (5) |
| (b) | \(F(5) = 1\) \(\frac{1}{100}(125a + 25b + 75) = 1\) Solving simultaneously \(\begin{cases} 5a + b = 1 \\ 6a + b = 0 \end{cases}\) \(a = -1\) and \(b = 6\) | M1 M1 A1 A1 (4) |
| (c) | \([P(3 < X \leq 7)] = \text{F}(5) - \text{F}(3)\) or \(1 - \text{F}(3)\) \(1 - \frac{1}{100}(27a + 9b + 45)\) | M1 A1 (2) |
(a) | $f(x) = \frac{1}{100}(3ax^2 + 2bx + 15)$ $E(X^2) = \int x^2 f(x)dx$ $\frac{1}{100}[\frac{2}{5}ax^5 + \frac{1}{2}bx^4 + 5x^3]_0^6 = 6.25$ $1875a + 312.5b + 625 = 625$ $6a + b = 0$ * | M1 A1 M1 dM1 A1 cso (5) | **Mark parts (a) and (b) together as may see use of F(5) = 1 in part (a)** / 1st M1 for use of $\frac{d}{dx}[F(x)]$ to find f(x) ($x^n \to x^{n+1}$) / 1st A1 for correct differentiation / 2nd M1 for attempt to integrate $\int x^2 f(x)dx$ (ignore limits) ($x^n \to x^{n+1}$) / Should see $x^5$ as the highest power of x here / 3rd dM1 for $\int x^2 f(x)dx = 6.25$ and substitution of $x = 5$ to obtain an equation in terms of a and b only (dependent upon 2nd M1) / 2nd A1 cso for $6a + b = 0$ from correct working
(b) | $F(5) = 1$ $\frac{1}{100}(125a + 25b + 75) = 1$ Solving simultaneously $\begin{cases} 5a + b = 1 \\ 6a + b = 0 \end{cases}$ $a = -1$ and $b = 6$ | M1 M1 A1 A1 (4) | 1st M1 for use of F(5) = 1 / 2nd M1 for solving simultaneously leading to $a = \ldots$ / 1st A1 for $a = -1$ / 2nd A1 for $b = 6$
(c) | $[P(3 < X \leq 7)] = \text{F}(5) - \text{F}(3)$ or $1 - \text{F}(3)$ $1 - \frac{1}{100}(27a + 9b + 45)$ | M1 A1 (2) | M1 for $1 - \text{F}(3)$ or $\int_3^5 f(x)dx$ attempt to integrate with use of correct limits / A1 0.28oe Total 11 | **0.28** |
---5. The random variable $X$ has cumulative distribution function given by
$$F ( x ) = \left\{ \begin{array} { l r }
0 & x < 0 \\
\frac { 1 } { 100 } \left( a x ^ { 3 } + b x ^ { 2 } + 15 x \right) & 0 \leqslant x \leqslant 5 \\
1 & x > 5
\end{array} \right.$$
Given that $\mathrm { E } \left( X ^ { 2 } \right) = 6.25$
\begin{enumerate}[label=(\alph*)]
\item show that $6 a + b = 0$
\item find the value of $a$ and the value of $b$
\item find $\mathrm { P } ( 3 \leqslant X \leqslant 7 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q5 [11]}}