| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Standard +0.3 Part (i) is a standard confidence interval calculation using CLT with given summary statistics. Part (ii) tests conceptual understanding of confidence interval interpretation (a common misconception). Part (iii) applies basic binomial probability. All parts are routine S3 material requiring no novel insight, though slightly easier than average due to straightforward calculations and standard conceptual points. |
| Spec | 5.02c Linear coding: effects on mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s^2 = (1183.65 - 246.6^2/70)/69\) | M1 | AEF |
| Use \(\bar{x} \pm zs/\sqrt{70}\) | M1 | Allow without ft or with \(s^2\); with 70 |
| \(s/\sqrt{70}\) | A1 | Their \(s\) |
| \(1.645\) | A1 | |
| \((3.10, 3.94)\) | A1 5 | A0 if interval not indicated |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Change 90 to around 90 | B1 1 | Or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(4(0.9)^3(0.1) + 0.9^4\) | M1 | Use of bino with \(p=0.9\) or \(0.1\) and 4, and |
| \(= 0.9477\) | A1 2 | Correct terms considered, art 0.948 |
# Question 4:
## Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $s^2 = (1183.65 - 246.6^2/70)/69$ | M1 | AEF |
| Use $\bar{x} \pm zs/\sqrt{70}$ | M1 | Allow without ft or with $s^2$; with 70 |
| $s/\sqrt{70}$ | A1 | Their $s$ |
| $1.645$ | A1 | |
| $(3.10, 3.94)$ | A1 **5** | A0 if interval not indicated |
## Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Change 90 to around 90 | B1 **1** | Or equivalent |
## Part (iii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $4(0.9)^3(0.1) + 0.9^4$ | M1 | Use of bino with $p=0.9$ or $0.1$ and 4, and |
| $= 0.9477$ | A1 **2** | Correct terms considered, art 0.948 |
---4 Part of an ecological study involved measuring the heights of trees in a young forest. In order to obtain an estimate of the mean height of all the trees in the forest, a random sample of 70 trees was selected and their heights measured. These heights, $x$ metres, are summarised by $\Sigma x = 246.6$ and $\Sigma x ^ { 2 } = 1183.65$. The mean height of all trees in the forest is denoted by $\mu$ metres.\\
(i) Calculate a symmetric $90 \%$ confidence interval for $\mu$.\\
(ii) A student was asked to interpret the interval and said,\\
"If 100 independent $90 \%$ confidence intervals were calculated then 90 of them would contain $\mu$."
Explain briefly what is wrong with this statement.\\
(iii) Four independent $90 \%$ confidence intervals for $\mu$ are obtained. Calculate the probability that at least three of the intervals contain $\mu$.
\hfill \mbox{\textit{OCR S3 2010 Q4 [8]}}