| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from probability function |
| Difficulty | Moderate -0.3 This is a straightforward S1 probability distribution question requiring standard techniques: finding k using ΣP(X=x)=1, calculating E(X) and Var(X) using standard formulas, and applying the variance transformation rule Var(aX+b)=a²Var(X). All steps are routine textbook exercises with only 3 values to compute, making it slightly easier than average A-level difficulty. |
| 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 |
|---|---|---|
| \(k(16-9) + k(25-9) + k(36-9) = 1\) | M1 A1 | |
| \(\therefore 7k + 16k + 27k = 1 \Rightarrow k = \frac{1}{50}\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| x | 4 | 5 |
| \(P(X=x)\) | \(\frac{7}{50}\) | \(\frac{16}{50}\) |
| \(E(X) = \left(4 \times \frac{7}{50}\right) + \left(5 \times \frac{16}{50}\right) + \left(6 \times \frac{27}{50}\right) = \frac{270}{50} = 5.4\) | M1 A1 | |
| \(E(X^2) = \left(4^2 \times \frac{7}{50}\right) + \left(5^2 \times \frac{16}{50}\right) + \left(6^2 \times \frac{27}{50}\right) = \frac{1484}{50} = 29.68\) | M1 A1 | |
| \(\therefore \text{Var}(X) = 29.68 - 5.4^2\) | M1 A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(2X-3) = 2^2 \text{Var}(X)\) | M1 | |
| \(= 4 \times 0.52 = 2.08\) | A1 | (2 marks) |
| (11 marks) |
**(a)**
| $k(16-9) + k(25-9) + k(36-9) = 1$ | M1 A1 | |
| $\therefore 7k + 16k + 27k = 1 \Rightarrow k = \frac{1}{50}$ | A1 | (3 marks) |
**(b)**
| | x | 4 | 5 | 6 |
|---|---|---|---|---|
| | $P(X=x)$ | $\frac{7}{50}$ | $\frac{16}{50}$ | $\frac{27}{50}$ |
| | | | | |
| $E(X) = \left(4 \times \frac{7}{50}\right) + \left(5 \times \frac{16}{50}\right) + \left(6 \times \frac{27}{50}\right) = \frac{270}{50} = 5.4$ | M1 A1 | |
| $E(X^2) = \left(4^2 \times \frac{7}{50}\right) + \left(5^2 \times \frac{16}{50}\right) + \left(6^2 \times \frac{27}{50}\right) = \frac{1484}{50} = 29.68$ | M1 A1 | |
| $\therefore \text{Var}(X) = 29.68 - 5.4^2$ | M1 A1 | (6 marks) |
**(c)**
| $\text{Var}(2X-3) = 2^2 \text{Var}(X)$ | M1 | |
| $= 4 \times 0.52 = 2.08$ | A1 | (2 marks) |
| | | (11 marks) |
---4. The discrete random variable $X$ has probability function
$$\mathrm { P } ( X = x ) = \begin{array} { l l }
k \left( x ^ { 2 } - 9 \right) , & x = 4,5,6 \\
0 , & \text { otherwise }
\end{array}$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 50 }$.
\item Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\item Find $\operatorname { Var } ( 2 X - 3 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q4 [11]}}