| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward S1 probability distribution question requiring only routine application of standard formulas: sum of probabilities equals 1, basic probability calculations, cumulative distribution function evaluation, expectation and variance properties. All parts are textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\alpha + 0.6 = 1 \Rightarrow \alpha = 0.2\) | M1A1 | linear function of \(\alpha = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(-1 \leq X < 2) = P(-1) + P(0) + P(1) = 0.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(F(0.6) = 0.8\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X) = (-3 \times 0.2) + \ldots + (2 \times 0.1) = -0.9\) | M1A1 | \(\sum xP(X=x)\); -0.9 |
| \(aE(X) + 3 = 1.2 \Rightarrow a(-0.9) = -1.8\) | M1 | \(aE(X) + 3\) |
| \(a = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X^2) = (-3^2 \times 0.2) + \ldots + (2^2 \times 0.1) = 3.3\) | M1A1 | \(\sum x^2P(X=x)\); 3.3 |
| \(\text{Var}(X) = 3.3 - (-0.9)^2 = 2.49\) | M1A1 | \(\sum x^2P(X=x) - (E(X))^2\); 2.49 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{Var}(3X-2) = 9\text{Var}(X)\) | M1 | |
| \(= 9 \times 2.49 = 22.41\) | A1 |
## Question 4:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $2\alpha + 0.6 = 1 \Rightarrow \alpha = 0.2$ | M1A1 | linear function of $\alpha = 1$ |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(-1 \leq X < 2) = P(-1) + P(0) + P(1) = 0.5$ | B1 | |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $F(0.6) = 0.8$ | B1 | |
### Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X) = (-3 \times 0.2) + \ldots + (2 \times 0.1) = -0.9$ | M1A1 | $\sum xP(X=x)$; -0.9 |
| $aE(X) + 3 = 1.2 \Rightarrow a(-0.9) = -1.8$ | M1 | $aE(X) + 3$ |
| $a = 2$ | A1 | |
### Part (e):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X^2) = (-3^2 \times 0.2) + \ldots + (2^2 \times 0.1) = 3.3$ | M1A1 | $\sum x^2P(X=x)$; 3.3 |
| $\text{Var}(X) = 3.3 - (-0.9)^2 = 2.49$ | M1A1 | $\sum x^2P(X=x) - (E(X))^2$; 2.49 |
### Part (f):
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{Var}(3X-2) = 9\text{Var}(X)$ | M1 | |
| $= 9 \times 2.49 = 22.41$ | A1 | |
---4. The discrete random variable $X$ has probability function
$$\mathrm { P } ( X = x ) = \begin{array} { l l }
0.2 , & x = - 3 , - 2 \\
\alpha , & x = - 1,0 \\
0.1 , & x = 1,2 .
\end{array}$$
Find
\begin{enumerate}[label=(\alph*)]
\item $\alpha$,
\item $\mathrm { P } ( - 1 \leq X < 2 )$,
\item $\mathrm { F } ( 0.6 )$,
\item the value of $a$ such that $\mathrm { E } ( a X + 3 ) = 1.2$,
\item $\operatorname { Var } ( X )$,
\item $\operatorname { Var } ( 3 X - 2 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q4 [14]}}