| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(aX+b) or Var(aX+b) given distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability distribution question requiring only standard procedures: using the probability sum rule to find p, then applying the linear transformation formulas E(aX+b) and Var(aX+b). All steps are routine calculations with no problem-solving or conceptual challenges beyond basic formula recall. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| \(P(X = x)\) | 0.09 | 0.12 | 0.22 | 0.16 | \(p\) | \(2p\) | 0.2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3p + 0.79 = 1\) → \(3p = 0.21\) → \(p = 0.07\) | M1 A1 | |
| (b) \(E(X) = 3.22\), \(E(X+2) = E(X) + 2 = 5.22\) | M1 A1 M1 A1 | |
| (c) \(\text{Var}(X) = E(X^2) - 3.22^2 = 14.26 - 3.22^2 = 3.892\) | M1 A1 A1 | \(\text{Var}(3X-1) = 9\text{Var}(X) = 35.0\) |
(a) $3p + 0.79 = 1$ → $3p = 0.21$ → $p = 0.07$ | M1 A1 |
(b) $E(X) = 3.22$, $E(X+2) = E(X) + 2 = 5.22$ | M1 A1 M1 A1 |
(c) $\text{Var}(X) = E(X^2) - 3.22^2 = 14.26 - 3.22^2 = 3.892$ | M1 A1 A1 | $\text{Var}(3X-1) = 9\text{Var}(X) = 35.0$ | M1 A1 | Total: 11 marks
---The discrete random variable $X$ has the probability function given by the following table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$P(X = x)$ & 0.09 & 0.12 & 0.22 & 0.16 & $p$ & $2p$ & 0.2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $p = 0.07$ [2 marks]
\item Find the value of $E(X + 2)$. [4 marks]
\item Find the value of $\text{Var}(3X - 1)$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q2 [11]}}