| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing basic CDF properties: using boundary conditions to find k, evaluating the CDF at a point, and differentiating to find the pdf. All steps are routine applications of standard techniques with no conceptual challenges beyond knowing F(2)=1 and f(y)=F'(y). |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(K(2^4 + 2^2 - 2) = 1 \Rightarrow K = \frac{1}{18}\) | M1, A1 |
| (b) | \(1 - F(1.5) = 1 - \frac{1}{18}(1.5^4 + 1.5^2 - 2) = 0.705\) or \(\frac{203}{288}\) | M1, A1 |
| (c) | \(f(y) = \begin{cases} \dfrac{1}{9}(2y^3 + y) & 1 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases}\) | M1, A1, B1 |
## Question 4:
**(a)** | $K(2^4 + 2^2 - 2) = 1 \Rightarrow K = \frac{1}{18}$ | M1, A1 | M1: putting $F(2)=1$ or $F(2)-F(1)=1$. A1: cso, must show substituting $y=2$ and obtaining $\frac{1}{18}$.
**(b)** | $1 - F(1.5) = 1 - \frac{1}{18}(1.5^4 + 1.5^2 - 2) = 0.705$ or $\frac{203}{288}$ | M1, A1 | M1: either attempting $1-F(1.5)$, or using $F(2)-F(1.5)$. A1: awrt 0.705.
**(c)** | $f(y) = \begin{cases} \dfrac{1}{9}(2y^3 + y) & 1 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases}$ | M1, A1, B1 | M1: attempting to differentiate; must see $y^n \to y^{n-1}$ at least once. A1: $\frac{1}{9}(2y^3+y)$, allow $1<y<2$. B1: for the $0$ *otherwise* (allow $0$ for $y<1$ and $0$ for $y>2$). Allow any letter.\begin{enumerate}
\item The continuous random variable $Y$ has cumulative distribution function $\mathrm { F } ( y )$ given by
\end{enumerate}
$$\mathrm { F } ( y ) = \left\{ \begin{array} { c l }
0 & y < 1 \\
k \left( y ^ { 4 } + y ^ { 2 } - 2 \right) & 1 \leqslant y \leqslant 2 \\
1 & y > 2
\end{array} \right.$$
(a) Show that $k = \frac { 1 } { 18 }$.\\
(b) Find $\mathrm { P } ( Y > 1.5 )$.\\
(c) Specify fully the probability density function f(y).
\hfill \mbox{\textit{Edexcel S2 2008 Q4 [7]}}