| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from probability function |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability distribution question requiring only standard formula application: summing probabilities to find k, calculating P(X≥2) by addition, computing E(X) and Var(X) using standard formulas, then applying the linear transformation property Var(a+bX)=b²Var(X). All steps are routine textbook exercises with no problem-solving or insight required, making it easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks |
|---|---|
| Marks: M1 A1 cso | (2) |
| Answer | Marks |
|---|---|
| Marks: M1 A1 | (2) |
| Answer | Marks |
|---|---|
| Marks: M1 A1 | (2) |
| Answer | Marks |
|---|---|
| Marks: M1 M1 M1 A1 | (4) |
## Part (a)
**Answer:** $k + 4k + 9k = 1$, $14k = 1$, $k = \frac{1}{14}$ **given**
**Marks:** M1 A1 cso | (2)
**Guidance:**
- M1 for clear attempt to use $\sum p(x) = 1$, full expression needed and the "1" must be clearly seen. This may be seen in a table
- A1cso for no incorrect working seen. The sum and "=" 1" must be explicitly seen somewhere
## Part (b)
**Answer:** $P(X \geq 2) = 1 - P(X = 1)$ or $P(X = 2) + P(X = 3)$, $= 1 - k = \frac{13}{14}$ or 0.92857...
**Marks:** M1 A1 | (2)
**Guidance:**
- M1 for 1- P($X \leq$ 1) or P($X = 2$) + P($X = 3$)
- A1 for awrt 0.929. Answer only scores 2/2
## Part (c)
**Answer:** $E(X) = 1 \times k + 2 \times k \times 4 + 3 \times k \times 9$ or 36k, $= \frac{36}{14} = \frac{18}{7}$ or $2\frac{4}{7}$ (or exact equivalent)
**Marks:** M1 A1 | (2)
**Guidance:**
- M1 for a full expression for E(X) with at least two terms correct
- NB If there is evidence of division (usually by 3) then score M0
- A1 for any exact equivalent - answer only scores 2/2
## Part (d)
**Answer:** $\text{Var}(X) = 1 \times k + 4 \times k \times 4 + 9 \times k \times 9 - \left(\frac{18}{7}\right)^2$, $\text{Var}(1 - X) = \text{Var}(X)$, $= \frac{19}{49}$ or 0.387755...
**Marks:** M1 M1 M1 A1 | (4)
**Guidance:**
- 1st M1 for clear attempt at E($X^2$), need at least 2 terms correct in $1 \times k + 4 \times k \times 4 + 9 \times k \times 9$ or E($X^2$) = 7
- 2nd M1 for their E($X^2$) – (their $\mu$)$^2$
- 3rd M1 for clearly stating that Var(1 – $X$) = Var($X$), wherever seen
- A1 accept awrt 0.388. All 3 M marks are required. Allow 4/4 for correct answer only but must be for Var(1 – $X$)
---The probability function of a discrete random variable $X$ is given by
$$p(x) = kx^2 \quad x = 1, 2, 3$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{1}{14}$ [2]
\end{enumerate}
Find
\begin{enumerate}[label=(\alph*), start=2]
\item P$(X \geq 2)$ [2]
\item E$(X)$ [2]
\item Var$(1-X)$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2010 Q5 [10]}}