| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| 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 standard CDF properties: using F(2)=1 to find k, differentiating to get the pdf, and evaluating P(Y>1)=1-F(1). All steps 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 cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(F(2)=1\) gives: \(\frac{1}{4}(2^3 - 4\times2^2 + 2k)=1\) | M1 | For an attempt to use \(F(2)=1\); clear attempt to form a linear equation for \(k\) |
| \(k=6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(f(y) = \frac{d}{dy}(F(y)) = \frac{1}{4}(3y^2 - 8y + \text{"6"})\) | M1A1ft | M1 for some correct differentiation \(y^n \to y^{n-1}\); 1st A1ft for \(3y^2-8y+\text{"6"}\), follow through their value of \(k\) or even \(k\) as a letter |
| \(f(y) = \begin{cases} \frac{1}{4}(3y^2-8y+6) & 0 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases}\) | A1 | 2nd A1 for fully correct solution including the 0 otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(P(Y>1) = 1 - F(1) = 1 - \frac{1}{4}(1^3 - 4\times1^2 + k)\) | M1 | For clear use of \(1-F(y)\) or attempt at integrating \(f(y)\); at least one correct term with correct coefficient, using limits of 1 and 2 |
| \(= \frac{1}{4}\) (o.e.) | A1 | For \(\frac{1}{4}\) or any exact equivalent |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $F(2)=1$ gives: $\frac{1}{4}(2^3 - 4\times2^2 + 2k)=1$ | M1 | For an attempt to use $F(2)=1$; clear attempt to form a linear equation for $k$ |
| $k=6$ | A1 | |
### Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $f(y) = \frac{d}{dy}(F(y)) = \frac{1}{4}(3y^2 - 8y + \text{"6"})$ | M1A1ft | M1 for some correct differentiation $y^n \to y^{n-1}$; 1st A1ft for $3y^2-8y+\text{"6"}$, follow through their value of $k$ or even $k$ as a letter |
| $f(y) = \begin{cases} \frac{1}{4}(3y^2-8y+6) & 0 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases}$ | A1 | 2nd A1 for fully correct solution including the 0 otherwise |
### Part (c)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $P(Y>1) = 1 - F(1) = 1 - \frac{1}{4}(1^3 - 4\times1^2 + k)$ | M1 | For clear use of $1-F(y)$ or attempt at integrating $f(y)$; at least one correct term with correct coefficient, using limits of 1 and 2 |
| $= \frac{1}{4}$ (o.e.) | A1 | For $\frac{1}{4}$ or any exact equivalent |2. The continuous random variable $Y$ has cumulative distribution function
$$\mathrm { F } ( y ) = \left\{ \begin{array} { c c }
0 & y < 0 \\
\frac { 1 } { 4 } \left( y ^ { 3 } - 4 y ^ { 2 } + k y \right) & 0 \leqslant y \leqslant 2 \\
1 & y > 2
\end{array} \right.$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the probability density function of $Y$, specifying it for all values of $y$.
\item Find $\mathrm { P } ( Y > 1 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q2 [7]}}