| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| 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 a quadratic equation in part (c), the setup is straightforward with clear steps: verify probabilities sum to 1, apply E(X) formula, find range constraints, then use Var(X) = E(X²) - [E(X)]² to solve for a. This is slightly easier than average as it's a textbook-style multi-part question with well-signposted steps and no conceptual surprises. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(a\) | 0.4 | \(0.6 - a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 2a + 3\times0.4 + 4(0.6-a) = 3.6 - 2a\) | M1 A1 | For attempt with 2 out of 3 products correct; for \(2a+1.2+4(0.6-a)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 < a < 0.6\) | B1 | Allow \(\leq\) or \(\geq\) for \(<\) or \(>\); may be written in words |
| \(2\times0.6+3\times0.4 = 2.4\) or \(3.6-2\times0.6=2.4\) and \(3\times0.4+4\times0.6=3.6\) or \(3.6-2\times0=3.6\) | M1 | For correct method for finding lower and upper bounds; may be implied by 2.4 and 3.6 |
| \(2.4 < E(X) < 3.6\) | A1 | Allow e.g. \(2.4\),, \(3.6-2a\),, \(3.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X^2) = 4a + 3.6 + 9.6 - 16a = 13.2 - 12a\) | M1 A1 | Attempt at \(E(X^2)\) with 2 terms correct |
| \(\text{Var}(X) = (13.2-12a) - (3.6-2a)^2\) | M1 | Use of \(\text{Var}(X) = E(X^2) - E(X)^2\) ft their values |
| \(-4a^2 + 2.4a - 0.32 = 0\) | A1 | Correct 3TQ e.g. \(25a^2 - 15a + 2 = 0\) |
| \(a = \frac{-2.4 \pm \sqrt{2.4^2 - 4\times(-4)\times(-0.32)}}{2\times(-4)}\) | M1 | Correct method for solving their 3TQ |
| \(a = \frac{1}{5}\) and \(a = \frac{2}{5}\) | A1 | oe; allow any letter for \(a\) |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 2a + 3\times0.4 + 4(0.6-a) = 3.6 - 2a$ | M1 A1 | For attempt with 2 out of 3 products correct; for $2a+1.2+4(0.6-a)$ oe |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 < a < 0.6$ | B1 | Allow $\leq$ or $\geq$ for $<$ or $>$; may be written in words |
| $2\times0.6+3\times0.4 = 2.4$ or $3.6-2\times0.6=2.4$ and $3\times0.4+4\times0.6=3.6$ or $3.6-2\times0=3.6$ | M1 | For correct method for finding lower and upper bounds; may be implied by 2.4 and 3.6 |
| $2.4 < E(X) < 3.6$ | A1 | Allow e.g. $2.4$,, $3.6-2a$,, $3.6$ |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X^2) = 4a + 3.6 + 9.6 - 16a = 13.2 - 12a$ | M1 A1 | Attempt at $E(X^2)$ with 2 terms correct |
| $\text{Var}(X) = (13.2-12a) - (3.6-2a)^2$ | M1 | Use of $\text{Var}(X) = E(X^2) - E(X)^2$ ft their values |
| $-4a^2 + 2.4a - 0.32 = 0$ | A1 | Correct 3TQ e.g. $25a^2 - 15a + 2 = 0$ |
| $a = \frac{-2.4 \pm \sqrt{2.4^2 - 4\times(-4)\times(-0.32)}}{2\times(-4)}$ | M1 | Correct method for solving their 3TQ |
| $a = \frac{1}{5}$ and $a = \frac{2}{5}$ | A1 | oe; allow any letter for $a$ |
---\begin{enumerate}
\item The probability distribution of the discrete random variable $X$ is given by
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $a$ & 0.4 & $0.6 - a$ \\
\hline
\end{tabular}
\end{center}
where $a$ is a constant.\\
(a) Find, in terms of $a , \mathrm { E } ( X )$\\
(b) Find the range of the possible values of $\mathrm { E } ( X )$
Given that $\operatorname { Var } ( X ) = 0.56$\\
(c) find the possible values of $a$
\hfill \mbox{\textit{Edexcel S1 2023 Q3 [11]}}