| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Known variance confidence interval |
| Difficulty | Standard +0.3 This is a straightforward S3 confidence interval question with standard parts: calculating a CI from summary statistics, using normal distribution for percentages, and finding probability of a difference. Part (iii) requires recognizing that B₁-B₂ is normal with variance 2σ², which is a standard S3 technique. Part (iv) tests understanding of CLT. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(s^2 = (68636.41 - 2605^2/100)/99 (=7.84)\) | B1 | |
| \(\bar{x} = 26.05\) | B1 | |
| \(26.05\pm z\sqrt{10}\) | M1 | |
| \(z = 2.326\) or \(\phi^{-1}(0.99)\) | B1 | |
| ART \((25.4, 26.7)\) | A1 | 5 Allow \((t99)=2.365\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(N(26.05,7.84)\) | M1 | |
| \(P(c \geq 30) = 1-\Phi([30-26.05]/\sqrt{7.84})\) | M1 | |
| \(= 0.0792 = 7.92\%\) | A1 | 3 \(s^2\) from (i) M0 for 7.84/100; No "cc" allow either: ART 0.08 or 8% |
| Answer | Marks |
|---|---|
| Use \(B_1 - B_2 \sim N(0, 15.68)\) | M1 |
| A1 | |
| \(P(< 5) = \Phi(5/\sigma)\) | A1 |
| A1 | 4 With \(\mu = 0\); For variance \(\sigma^2\); Their \(\sigma\); \(\Phi(\pm 5) \Rightarrow\) M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) only since sample size of 100 is large enough (for CLT to hold) | B1 | 1(13) Must be clear which part and with correct reason. |
**6(i)**
$s^2 = (68636.41 - 2605^2/100)/99 (=7.84)$ | B1 |
$\bar{x} = 26.05$ | B1 |
$26.05\pm z\sqrt{10}$ | M1 |
$z = 2.326$ or $\phi^{-1}(0.99)$ | B1 |
ART $(25.4, 26.7)$ | A1 | 5 Allow $(t99)=2.365$
**6(ii)**
Use $N(26.05,7.84)$ | M1 |
$P(c \geq 30) = 1-\Phi([30-26.05]/\sqrt{7.84})$ | M1 |
$= 0.0792 = 7.92\%$ | A1 | 3 $s^2$ from (i) M0 for 7.84/100; No "cc" allow either: ART 0.08 or 8%
**6(iii)**
Use $B_1 - B_2 \sim N(0, 15.68)$ | M1 |
| A1 |
$P(< 5) = \Phi(5/\sigma)$ | A1 |
| A1 | 4 With $\mu = 0$; For variance $\sigma^2$; Their $\sigma$; $\Phi(\pm 5) \Rightarrow$ M1
**6(iv)**
(i) only since sample size of 100 is large enough (for CLT to hold) | B1 | 1(13) Must be clear which part and with correct reason.
---6 The Body Mass Index (BMI) of each of a random sample of 100 army recruits from a large intake in 2008 was measured. The results are summarised by
$$\Sigma x = 2605.0 , \quad \Sigma x ^ { 2 } = 68636.41 .$$
It may be assumed that BMI has a normal distribution.\\
(i) Find a 98\% confidence interval for the mean BMI of all recruits in 2008.\\
(ii) Estimate the percentage of the intake with a BMI greater than 30.0.\\
(iii) The BMIs of two randomly chosen recruits are denoted by $\boldsymbol { B } _ { 1 }$ and $\boldsymbol { B } _ { 2 }$. Estimate $\mathrm { P } \left( \boldsymbol { B } _ { 1 } - \boldsymbol { B } _ { 2 } < 5 \right)$.\\
(iv) State, giving a reason, for which of the above calculations the normality assumption is unnecessary.
\hfill \mbox{\textit{OCR S3 2011 Q6 [13]}}