| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Linear combinations of independent variables |
| Difficulty | Moderate -0.3 This is a standard S1 question on combining independent discrete random variables. Part (a) requires systematic enumeration of outcomes (straightforward but slightly tedious), while parts (b)-(d) apply standard formulas for expectation and variance. The calculations are routine with no conceptual challenges beyond knowing that E(X+Y)=E(X)+E(Y) and Var(aZ+b)=a²Var(Z). Slightly easier than average due to small probability distributions and direct application of learned techniques. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | 0 | 1 | 2 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 3 } { 5 }\) | \(\frac { 3 } { 10 }\) | \(\frac { 1 } { 10 }\) |
| \(y\) | 0 | 1 |
| \(\mathrm { P } ( Y = y )\) | \(\frac { 5 } { 8 }\) | \(\frac { 3 } { 8 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(z\) | \(0\) |
| \(P(Z = z)\) | \(\frac{3}{8}\) | \(\frac{33}{80}\) |
| M1 A1 A1 A1 A1 | ||
| (b) \(E(Z) = \frac{7}{8}\) | M1 M1 A1 | |
| (c) (i) \(E(Z^2) = \frac{29}{20}\) | M1 A1 M1 A1 | |
| (ii) Var\((Z) = \frac{29}{20} - \frac{49}{64} = \frac{219}{320}\) or \(0.684\) | ||
| (d) Var\((3Z - 4) = \text{Var}(3Z) = 9 \text{ Var}(Z) = 6\frac{51}{320}\) or \(6.16\) | M1 A1 | Total: 14 marks |
(a) | $z$ | $0$ | $1$ | $2$ | $3$ |
| --- | --- | --- | --- | --- |
| $P(Z = z)$ | $\frac{3}{8}$ | $\frac{33}{80}$ | $\frac{7}{40}$ | $\frac{3}{80}$ |
| M1 A1 A1 A1 A1 |
(b) $E(Z) = \frac{7}{8}$ | M1 M1 A1 |
(c) (i) $E(Z^2) = \frac{29}{20}$ | M1 A1 M1 A1 |
(ii) Var$(Z) = \frac{29}{20} - \frac{49}{64} = \frac{219}{320}$ or $0.684$ |
(d) Var$(3Z - 4) = \text{Var}(3Z) = 9 \text{ Var}(Z) = 6\frac{51}{320}$ or $6.16$ | M1 A1 | **Total: 14 marks**6. The distributions of two independent discrete random variables $X$ and $Y$ are given in the tables:
\begin{center}
\begin{tabular}{ | c | | c | c | c | }
\hline
$x$ & 0 & 1 & 2 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 3 } { 5 }$ & $\frac { 3 } { 10 }$ & $\frac { 1 } { 10 }$ \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ | c | | c | c | }
\hline
$y$ & 0 & 1 \\
\hline
$\mathrm { P } ( Y = y )$ & $\frac { 5 } { 8 }$ & $\frac { 3 } { 8 }$ \\
\hline
\end{tabular}
\end{center}
The random variable $Z$ is defined to be the sum of one observation from $X$ and one from $Y$.
\begin{enumerate}[label=(\alph*)]
\item Tabulate the probability distribution for $Z$.
\item Calculate $\mathrm { E } ( Z )$.
\item Calculate (i) $\mathrm { E } \left( Z ^ { 2 } \right)$, (ii) $\operatorname { Var } ( Z )$.
\item Calculate Var (3Z-4).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [14]}}