| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Standard +0.8 This Further Statistics question requires setting up and solving a system of two equations (probabilities sum to 1, and E(X) = 3.8) to find k and m, then computing variance. It involves algebraic manipulation with fractions and systematic calculation across multiple values, going beyond routine single-parameter distribution problems but following a clear methodological path. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\Sigma p = 1 \Rightarrow k + \frac{k}{2} + \frac{k}{3} + \frac{m}{12} + \frac{m}{18} = 1\) and \(\Sigma px = 3.8 \Rightarrow k + \frac{k}{2}(2) + \frac{k}{3}(3) + \frac{m}{12}(6) + \frac{m}{18}(9) = 3.8\) | M1 | Attempt at both equations with at least one term in \(k\) and one in \(m\) correct |
| \(\frac{11k}{6} + \frac{5m}{36} = 1\ [= 66k + 5m = 36]\) | A1 | Correct equation using \(\Sigma p = 1\) |
| \(3k + m = 3.8\) | A1 | Correct equation using \(\Sigma px = 3.8\) |
| Solving simultaneously to eliminate one variable | dM1 | dep on 1st M1; may be implied by one correct value |
| \(k = \frac{1}{3}\) and \(m = \frac{14}{5}\) | A1 | Both values correct |
| \(E(X^2) = 1^2 \times k + 2^2 \times \frac{k}{2} + 3^2 \times \frac{k}{3} + 6^2 \times \frac{m}{12} + 9^2 \times \frac{m}{18}\ [=23]\) | M1 | Attempt to find \(E(X^2)\) using their \(k\) and \(m\) with at least 3 correct products. Note: \(E(X^2) = 6k + 7.5m\) |
| \(\text{Var}(X) = 23 - 3.8^2 =\) 8.56 | A1 | 8.56 cao |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Sigma p = 1 \Rightarrow k + \frac{k}{2} + \frac{k}{3} + \frac{m}{12} + \frac{m}{18} = 1$ and $\Sigma px = 3.8 \Rightarrow k + \frac{k}{2}(2) + \frac{k}{3}(3) + \frac{m}{12}(6) + \frac{m}{18}(9) = 3.8$ | M1 | Attempt at both equations with at least one term in $k$ and one in $m$ correct |
| $\frac{11k}{6} + \frac{5m}{36} = 1\ [= 66k + 5m = 36]$ | A1 | Correct equation using $\Sigma p = 1$ |
| $3k + m = 3.8$ | A1 | Correct equation using $\Sigma px = 3.8$ |
| Solving simultaneously to eliminate one variable | dM1 | dep on 1st M1; may be implied by one correct value |
| $k = \frac{1}{3}$ and $m = \frac{14}{5}$ | A1 | Both values correct |
| $E(X^2) = 1^2 \times k + 2^2 \times \frac{k}{2} + 3^2 \times \frac{k}{3} + 6^2 \times \frac{m}{12} + 9^2 \times \frac{m}{18}\ [=23]$ | M1 | Attempt to find $E(X^2)$ using their $k$ and $m$ with at least 3 correct products. Note: $E(X^2) = 6k + 7.5m$ |
| $\text{Var}(X) = 23 - 3.8^2 =$ **8.56** | A1 | 8.56 cao |
---\begin{enumerate}
\item The probability distribution of the discrete random variable $X$ is
\end{enumerate}
$$P ( X = x ) = \begin{cases} \frac { k } { x } & \text { for } x = 1,2 \text { and } 3 \\ \frac { m } { 2 x } & \text { for } x = 6 \text { and } 9 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ and $m$ are positive constants.\\
Given that $\mathrm { E } ( X ) = 3.8$, find $\operatorname { Var } ( X )$
\hfill \mbox{\textit{Edexcel FS1 AS 2020 Q3 [7]}}