| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.3 This is a standard S1 probability distribution question requiring systematic application of expectation and variance formulas. While it involves solving simultaneous equations with three unknowns, the algebraic manipulation is straightforward and the linear transformation of random variables in parts (d)-(f) follows directly from standard formulas. The symmetry in the distribution (equal probabilities at x=-2 and x=2) simplifies the algebra. This is slightly easier than average as it's a routine multi-part question testing core syllabus content with no novel problem-solving required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | - 2 | 0 | 2 | 4 |
| \(\mathrm { P } ( X = x )\) | \(a\) | \(b\) | \(a\) | \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(-2a + (0) + 2a + 4c = 0.8\) or \(4c = 0.8\) | M1 (2) | |
| \(c = 0.2\) | A1 | |
| (b) \(4a + (0) + 4a + 16c = 5\) or \(8a + 16c = 5\) | M1 (4) | |
| \(8a + 3.2 = 5\) so \(a = 0.225\) or \(\frac{9}{40}\) | A1 | |
| \(2a + b + c = 1\) so \(b = 1 - "0.2" - 2 \times "0.225"\) | M1 | |
| \(b = 0.35\) or \(\frac{7}{20}\) | A1ft | |
| (c) \(\text{Var}(X) = 5 - 0.8^2\) | M1A1 (2) | \(= 4.36\) |
| (d) \([5 - 3E(X) = 5 - 3 \times 0.8]\) \(= 2.6\) | B1 (1) | |
| (e) \(3^2 \text{Var}(X) = 9 \times 4.36\), or \([E(X^2) - (E(X))^2] = 46 - 2.6^2\) \(= 39.24\) awrt \(39.2\) | M1, A1 (2) | |
| (f) \(Y \ge 0 \Rightarrow 5 - 3X \ge 0 \Rightarrow 5 \ge 3X\) | M1 (4) | |
| \(X \le 1\frac{2}{3}\) | A1 | |
| \([P(Y \ge 0) = P(X \le 0) =]\) \(P(X = -2) + P(X = 0)\) or \(a + b\) | M1 | |
| \(= 0.575\) or \(\frac{23}{40}\) | A1ft | [15] |
**(a)** $-2a + (0) + 2a + 4c = 0.8$ or $4c = 0.8$ | M1 (2) |
$c = 0.2$ | A1 |
**(b)** $4a + (0) + 4a + 16c = 5$ or $8a + 16c = 5$ | M1 (4) |
$8a + 3.2 = 5$ so $a = 0.225$ or $\frac{9}{40}$ | A1 |
$2a + b + c = 1$ so $b = 1 - "0.2" - 2 \times "0.225"$ | M1 |
$b = 0.35$ or $\frac{7}{20}$ | A1ft |
**(c)** $\text{Var}(X) = 5 - 0.8^2$ | M1A1 (2) | $= 4.36$ | For correct expression $5 - 0.8^2$ (Division by 4 at any stage is M0)
**(d)** $[5 - 3E(X) = 5 - 3 \times 0.8]$ $= 2.6$ | B1 (1) |
**(e)** $3^2 \text{Var}(X) = 9 \times 4.36$, or $[E(X^2) - (E(X))^2] = 46 - 2.6^2$ $= 39.24$ awrt $39.2$ | M1, A1 (2) |
**(f)** $Y \ge 0 \Rightarrow 5 - 3X \ge 0 \Rightarrow 5 \ge 3X$ | M1 (4) |
$X \le 1\frac{2}{3}$ | A1 |
$[P(Y \ge 0) = P(X \le 0) =]$ $P(X = -2) + P(X = 0)$ or $a + b$ | M1 |
$= 0.575$ or $\frac{23}{40}$ | A1ft | [15]
**Alt-Method:** 1st M1 for attempting to solve the inequality in X as far as $p \ge qX$ with one of $p$ or $q$ correct. 1st A1 for $X \le 1\frac{2}{3}$ or $X \le 0$ or $X \le$ awrt 1.7. 2nd M1 for $P(X=1) + P(Y=5)$ for $P(Y=11) + P(Y=5)$. 2nd A1ft for their $(a+b)$ or 0.575 (where $a$, $b$ and $a+b$ are all probabilities).
---5. The discrete random variable $X$ has the following probability distribution
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & - 2 & 0 & 2 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $a$ & $b$ & $a$ & $c$ \\
\hline
\end{tabular}
\end{center}
where $a$, $b$ and $c$ are probabilities.\\
Given that $\mathrm { E } ( X ) = 0.8$
\begin{enumerate}[label=(\alph*)]
\item find the value of $c$.
Given also that $\mathrm { E } \left( X ^ { 2 } \right) = 5$ find
\item the value of $a$ and the value of $b$,
\item $\operatorname { Var } ( X )$
The random variable $Y = 5 - 3 X$\\
Find
\item $\mathrm { E } ( Y )$
\item $\operatorname { Var } ( Y )$
\item $\mathrm { P } ( Y \geqslant 0 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q5 [15]}}