| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.8 This is a standard S1 textbook exercise requiring routine application of probability axioms (sum=1) and expectation formula to find two unknowns via simultaneous equations. Parts (b)-(e) are direct formula applications with no problem-solving required. Easier than average A-level due to being purely procedural. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.2 | 0.3 | \(b\) | \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.5+b+a=1\) | M1, A1 | Use of \(\Sigma P(X=x)=1\) |
| \(0.3+2b+3a=1.7\) | M1, A1 | Use of \(E(x)=\Sigma x P(X=x)\) |
| \(\therefore a=0.4\) | ||
| \(b=0.1\) | B1 | \(a=0.4\), \(b=0.1\) (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
\(P(0| B1 |
(1 mark) |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(2X-3)=2E(X)-3\) | M1 | Use of \(E(aX+b)\) |
| \(=2\times 1.7-3=0.4\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X)=(1^2\times 0.3)+(2^2\times 0.1)+(3^2\times 0.4)-1.7^2\) | M1 | Use of \(E(x^2)-\{E(X)\}^2\) |
| \(=1.41\) \((*)\) | A1 ft | |
| A1 | cso (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(2X-3)=2^2\,\text{Var}(X)\) | M1 | Use of Var |
| \(=4\times 1.41=5.64\) | A1 | (2 marks) |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.5+b+a=1$ | M1, A1 | Use of $\Sigma P(X=x)=1$ |
| $0.3+2b+3a=1.7$ | M1, A1 | Use of $E(x)=\Sigma x P(X=x)$ |
| $\therefore a=0.4$ | | |
| $b=0.1$ | B1 | $a=0.4$, $b=0.1$ **(5 marks)** |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(0<X<1.5)=P(X=1)=0.3$ | B1 | **(1 mark)** |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(2X-3)=2E(X)-3$ | M1 | Use of $E(aX+b)$ |
| $=2\times 1.7-3=0.4$ | A1 | **(2 marks)** |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X)=(1^2\times 0.3)+(2^2\times 0.1)+(3^2\times 0.4)-1.7^2$ | M1 | Use of $E(x^2)-\{E(X)\}^2$ |
| $=1.41$ $(*)$ | A1 ft | |
| | A1 | cso **(3 marks)** |
### Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(2X-3)=2^2\,\text{Var}(X)$ | M1 | Use of Var |
| $=4\times 1.41=5.64$ | A1 | **(2 marks)** |
**(13 marks total)**
---3. A discrete random variable $X$ has a probability function as shown in the table below, where $a$ and $b$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.2 & 0.3 & $b$ & $a$ \\
\hline
\end{tabular}
\end{center}
Given that $\mathrm { E } ( X ) = 1.7$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$ and the value of $b$.
Find
\item $\mathrm { P } ( 0 < X < 1.5 )$,
\item $\mathrm { E } ( 2 X - 3 )$.
\item Show that $\operatorname { Var } ( X ) = 1.41$.
\item Evaluate $\operatorname { Var } ( 2 X - 3 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q3 [13]}}