| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| 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 mechanical application of standard formulas: converting CDF to PMF by subtraction, computing E(X) and Var(X) using definitions, and applying linear transformation rules. All steps are routine with no problem-solving or conceptual challenges beyond basic recall. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(X\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| \(\mathrm {~F} ( x )\) | 0.1 | 0.2 | 0.25 | 0.4 | 0.5 | 0.6 | 0.75 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probability distribution table with \(x = 1,2,3,4,5,6,7,8\) and \(P(X=x) = 0.1, 0.1, 0.05, 0.15, 0.1, 0.1, 0.15, 0.25\) | M1, A2 | A2 (−1 each error) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (1 \times 0.1) + (2 \times 0.1) + \ldots + (8 \times 0.25) = 5.2\) | M1, A1 | |
| \(E(X^2) = (1^2 \times 0.1) + (2^2 \times 0.1) + \ldots + (8^2 \times 0.25) = 32.8\) | M1, A1 | |
| \(\text{Var}(X) = E(X^2) - \{E(X)\}^2 = 32.8 - (5.2)^2 = 5.76\) | M1, A1 | Answer given on paper |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Y) = 2E(X) + 3 = 13.4\) | B1 | |
| \(\text{Var}(Y) = 2^2 \text{Var}(X)\) | M1 | |
| \(= 4 \times 5.76 = 23.04\) | A1 |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probability distribution table with $x = 1,2,3,4,5,6,7,8$ and $P(X=x) = 0.1, 0.1, 0.05, 0.15, 0.1, 0.1, 0.15, 0.25$ | M1, A2 | A2 (−1 each error) |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = (1 \times 0.1) + (2 \times 0.1) + \ldots + (8 \times 0.25) = 5.2$ | M1, A1 | |
| $E(X^2) = (1^2 \times 0.1) + (2^2 \times 0.1) + \ldots + (8^2 \times 0.25) = 32.8$ | M1, A1 | |
| $\text{Var}(X) = E(X^2) - \{E(X)\}^2 = 32.8 - (5.2)^2 = 5.76$ | M1, A1 | Answer given on paper |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Y) = 2E(X) + 3 = 13.4$ | B1 | |
| $\text{Var}(Y) = 2^2 \text{Var}(X)$ | M1 | |
| $= 4 \times 5.76 = 23.04$ | A1 | |
---4. A discrete random variable $X$ takes only positive integer values. It has a cumulative distribution function $\mathrm { F } ( x ) = \mathrm { P } ( X \leq x )$ defined in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$X$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
$\mathrm {~F} ( x )$ & 0.1 & 0.2 & 0.25 & 0.4 & 0.5 & 0.6 & 0.75 & 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Determine the probability function, $\mathrm { P } ( X = x )$, of $X$.
\item Calculate $\mathrm { E } ( X )$ and show that $\operatorname { Var } ( X ) = 5.76$.
\item Given that $Y = 2 X + 3$, find the mean and variance of $Y$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2002 Q4 [12]}}