| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Discrete CDF to PMF |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring basic understanding that F(4)=1 to find k, then using differences F(x)-F(x-1) to obtain the PMF. It involves simple algebraic manipulation and recall of CDF properties, making it easier than average with no problem-solving insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F(4) = 1\) | M1 | M1 for use of \(F(4) = 1\) only. If \(F(2)=1\) and/or \(F(3)=1\) seen then M0. \(F(2)+F(3)+F(4)=1\) is M0 |
| \((4+k)^2 = 25\) | ||
| \(k = 1\) as \(k > 0\) | A1 | A1 for \(k=1\) and ignore \(k=-9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x = 2\): \(P(X=x) = \dfrac{2}{25}\) | B1ft | B1ft follow through their \(k\) for \(P(X=2)\), either exact or 3sf between 0 and 1 inclusive |
| \(x = 3\): \(P(X=x) = \dfrac{7}{25}\) | B1 | B1 correct answer only or exact equivalent |
| \(x = 4\): \(P(X=x) = \dfrac{9}{25}\) | B1 | B1 correct answer only or exact equivalent |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $F(4) = 1$ | M1 | M1 for use of $F(4) = 1$ only. If $F(2)=1$ and/or $F(3)=1$ seen then M0. $F(2)+F(3)+F(4)=1$ is M0 |
| $(4+k)^2 = 25$ | | |
| $k = 1$ as $k > 0$ | A1 | A1 for $k=1$ and ignore $k=-9$ |
**[2 marks]**
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $x = 2$: $P(X=x) = \dfrac{2}{25}$ | B1ft | B1ft follow through their $k$ for $P(X=2)$, either exact or 3sf between 0 and 1 inclusive |
| $x = 3$: $P(X=x) = \dfrac{7}{25}$ | B1 | B1 correct answer only or exact equivalent |
| $x = 4$: $P(X=x) = \dfrac{9}{25}$ | B1 | B1 correct answer only or exact equivalent |
**[3 marks] — Total: 5**
---6. The discrete random variable $X$ can take only the values 2,3 or 4 . For these values the cumulative distribution function is defined by
$$F ( x ) = \frac { ( x + k ) ^ { 2 } } { 25 } \text { for } x = 2,3,4$$
where $k$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Find $k$.
\item Find the probability distribution of $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q6 [5]}}