| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| 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 testing basic CDF/PMF conversions and variance properties. Part (a) uses F(3)=1 to find k (simple algebra), part (b) requires P(X=x)=F(x)-F(x-1) (standard procedure), and part (c) applies the formula Var(aX+b)=a²Var(X) (direct recall). All steps are routine textbook exercises with no problem-solving or insight required. |
| Spec | 2.04a Discrete probability distributions5.02c Linear coding: effects on mean and variance5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F(3)=1\) gives \(\frac{3^3+k}{40}=1\) | M1 | Must see attempt at \(\frac{3^3+k}{40}=1\); \(27+k=40\) without reference to \(F(3)=1\) is M0 |
| \(k=\mathbf{13}\) | A1cso | No incorrect working seen; for Verify: allow M1 for \(\frac{3^3+13}{40}=1\) but A1 requires explicit comment "so \(k=13\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X=1)=\frac{14}{40}\) or \(0.35\) | B1 | Can be labelled \(F(1)\), \(P(X=1)\) or \(p(x)\); associated with \(x=1\) or given in table |
| Use of \(P(X=2)=F(2)-F(1)\) or \(P(X=3)=F(3)-F(2)\) | M1 | Clear method showing how to obtain \(P(X=\ldots)\) from \(F(x)\); implied if either \(P(X=2)\) or \(P(X=3)\) is correct |
| \(P(X=2)=\frac{7}{40}\) or \(0.175\) | A1 | |
| \(P(X=3)=\frac{19}{40}\) or \(0.475\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{Var}(4X-5)=4^2\,\text{Var}(X)\) | M1 | Correct use of variance formula; \(4^2\text{Var}(X)\) alone secures M1; a value for \(\text{Var}(X)\) is not required |
| \(\text{Var}(4X-5)=\frac{259}{20}\) or \(12.95\) | A1 | Any exact equivalent to \(12.95\); correct answer only is 2/2 |
# Question 2:
## Part (a)
| Working | Mark | Guidance |
|---------|------|---------|
| $F(3)=1$ gives $\frac{3^3+k}{40}=1$ | M1 | Must see attempt at $\frac{3^3+k}{40}=1$; $27+k=40$ without reference to $F(3)=1$ is M0 |
| $k=\mathbf{13}$ | A1cso | No incorrect working seen; for Verify: allow M1 for $\frac{3^3+13}{40}=1$ but A1 requires explicit comment "so $k=13$" |
## Part (b)
| Working | Mark | Guidance |
|---------|------|---------|
| $P(X=1)=\frac{14}{40}$ or $0.35$ | B1 | Can be labelled $F(1)$, $P(X=1)$ or $p(x)$; associated with $x=1$ or given in table |
| Use of $P(X=2)=F(2)-F(1)$ or $P(X=3)=F(3)-F(2)$ | M1 | Clear method showing how to obtain $P(X=\ldots)$ from $F(x)$; implied if either $P(X=2)$ or $P(X=3)$ is correct |
| $P(X=2)=\frac{7}{40}$ or $0.175$ | A1 | |
| $P(X=3)=\frac{19}{40}$ or $0.475$ | A1 | |
## Part (c)
| Working | Mark | Guidance |
|---------|------|---------|
| $\text{Var}(4X-5)=4^2\,\text{Var}(X)$ | M1 | Correct use of variance formula; $4^2\text{Var}(X)$ alone secures M1; a value for $\text{Var}(X)$ is not required |
| $\text{Var}(4X-5)=\frac{259}{20}$ or $12.95$ | A1 | Any exact equivalent to $12.95$; correct answer only is 2/2 |
---2. The discrete random variable $X$ can take only the values 1,2 and 3 . For these values the cumulative distribution function is defined by
$$\mathrm { F } ( x ) = \frac { x ^ { 3 } + k } { 40 } \quad x = 1,2,3$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 13$
\item Find the probability distribution of $X$.
Given that $\operatorname { Var } ( X ) = \frac { 259 } { 320 }$
\item find the exact value of $\operatorname { Var } ( 4 X - 5 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2013 Q2 [8]}}