| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sample statistics from uniform |
| Difficulty | Moderate -0.3 This is a straightforward uniform distribution question requiring standard techniques: sketching a rectangular PDF, using symmetry for E(T), computing Var(T) via integration, and applying the Central Limit Theorem for sample means. All steps are routine S2 procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the integration and CLT application. |
| Spec | 5.02e Discrete uniform distribution5.03c Calculate mean/variance: by integration5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram: Box plot with boundaries at 5 and 11 | Marks: M1 | Guidance: Horizontal line; Evidence of truncation [no need for labels] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: μ = 8 | Marks: B1 | Guidance: 8 only, cwd |
| Answer/Working: \(\int_5^{11} \frac{1}{6}t^2 dt = [\frac{t^3}{18}]_5^{11}\) | Marks: M1 | Guidance: Attempt \(\int kt^2 dt\), limits 5 and 11 seen |
| Answer/Working: | = 67 | Marks: B1 |
| Answer/Working: − 8² | Marks: M1 | Guidance: |
| Answer/Working: = 3 | Marks: A1, 5 | Guidance: Answer 3 only, not from MF1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(N(8, 3/48)\) | Marks: M1 | Guidance: Normal stated or implied |
| Answer/Working: Mean 8 | Marks: A1 | Guidance: |
| Answer/Working: \(1 - \Phi\left(\frac{8.3-8}{\sqrt{3/48}}\right) = 1 - \Phi(1.2)\) | Marks: A1 | Guidance: Variance their (non-zero) (ii)/48 |
| Answer/Working: = 1 − 0.8848 | Marks: M1 | Guidance: Standardise, \(\sqrt{n}\), ignore sign or ∇ errors, cc: |
| Answer/Working: = 0.1151 | Marks: A1 | Guidance: M0; Answer, art 0.115 |
| Answer/Working: Normal distribution only approx. | Marks: B1, 6 | Guidance: Any equivalent comment, e.g. CLT used |
### (i)
**Diagram:** Box plot with boundaries at 5 and 11 | **Marks:** M1 | **Guidance:** Horizontal line; Evidence of truncation [no need for labels]
### (ii)
**Answer/Working:** μ = 8 | **Marks:** B1 | **Guidance:** 8 only, cwd
**Answer/Working:** $\int_5^{11} \frac{1}{6}t^2 dt = [\frac{t^3}{18}]_5^{11}$ | **Marks:** M1 | **Guidance:** Attempt $\int kt^2 dt$, limits 5 and 11 seen
**Answer/Working:** | = 67 | **Marks:** B1 | **Guidance:** Subtract their (non-zero) mean²
**Answer/Working:** − 8² | **Marks:** M1 | **Guidance:**
**Answer/Working:** = 3 | **Marks:** A1, 5 | **Guidance:** Answer 3 only, not from MF1
### (iii)
**Answer/Working:** $N(8, 3/48)$ | **Marks:** M1 | **Guidance:** Normal stated or implied
**Answer/Working:** Mean 8 | **Marks:** A1 | **Guidance:**
**Answer/Working:** $1 - \Phi\left(\frac{8.3-8}{\sqrt{3/48}}\right) = 1 - \Phi(1.2)$ | **Marks:** A1 | **Guidance:** Variance their (non-zero) (ii)/48
**Answer/Working:** = 1 − 0.8848 | **Marks:** M1 | **Guidance:** Standardise, $\sqrt{n}$, ignore sign or ∇ errors, cc:
**Answer/Working:** = 0.1151 | **Marks:** A1 | **Guidance:** M0; Answer, art 0.115
**Answer/Working:** Normal distribution only approx. | **Marks:** B1, 6 | **Guidance:** Any equivalent comment, e.g. CLT used
---The continuous random variable $T$ is equally likely to take any value from 5.0 to 11.0 inclusive.
\begin{enumerate}[label=(\roman*)]
\item Sketch the graph of the probability density function of $T$. [2]
\item Write down the value of E($T$) and find by integration the value of Var($T$). [5]
\item A random sample of 48 observations of $T$ is obtained. Find the approximate probability that the mean of the sample is greater than 8.3, and explain why the answer is an approximation. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2010 Q7 [13]}}