| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring only basic probability distribution knowledge: calculating E(X) using the formula Σxp(x), then solving a simple linear equation. No problem-solving insight needed, just routine application of definitions with arithmetic. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | \(k\) | \(k + 4\) | \(2 k\) |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 8 }\) | \(\frac { 3 } { 8 }\) | \(\frac { 1 } { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum xP(x) = \frac{1}{k} + \frac{3}{2}(k+4) + \frac{1}{2}(2k) = \frac{3}{4}(k+1)\) | M2 A1 | |
| \(\frac{3}{2}(k+1) = 9; k = 5\) | M1 A1 | (5) |
$\sum xP(x) = \frac{1}{k} + \frac{3}{2}(k+4) + \frac{1}{2}(2k) = \frac{3}{4}(k+1)$ | M2 A1 |
$\frac{3}{2}(k+1) = 9; k = 5$ | M1 A1 | (5)\begin{enumerate}
\item The discrete random variable $X$ has the following probability distribution.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & $k$ & $k + 4$ & $2 k$ \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 8 }$ & $\frac { 3 } { 8 }$ & $\frac { 1 } { 2 }$ \\
\hline
\end{tabular}
\end{center}
(a) Find and simplify an expression in terms of $k$ for $\mathrm { E } ( X )$.
Given that $\mathrm { E } ( X ) = 9$,\\
(b) find the value of $k$.\\
\hfill \mbox{\textit{Edexcel S1 Q1 [5]}}