| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | CI with known population variance |
| Difficulty | Challenging +1.2 This is a multi-part confidence interval question requiring knowledge of sampling distributions and normal theory. Part (i) is routine (finding critical value), part (ii) is standard (variance of difference of independent normals), but part (iii) requires careful probabilistic reasoning about non-overlapping intervals, which elevates it above typical textbook exercises while remaining within S3 scope. |
| Spec | 5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = z\sqrt{(1/4)}\) | M1 | From \(\sigma/\sqrt{n}\) |
| \(z = 2.326\) | B1 | |
| \(a = 1.163\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Normal with mean 0 | B1 | |
| \(\sigma^2 = \frac{1}{4} + \frac{9}{16} = 0.8125\) or \(\frac{13}{16}\) | B1 | Allow 0.813 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either \(\bar{x} + 1.163 < \bar{y} - 1.234\) or \(\bar{y} + 1.234 < \bar{x} - 1.163\) | M1 | Ft their \(a\) for M1M1. Can be implied. |
| giving \(\bar{x} - \bar{y} < -2.397\) or \(> 2.397\) | M1, A1 | |
| \(P(<-2.397) = P(Z < -2.659)\) oe \(= 0.0039(2)\) | M1, A1 | Correct use of N for their interval(s). 0.0039 from tables. |
| So required probability \(= 0.0078(4)\) | A1ft | ART 0.0078, ft \(2\times\) their 0.0039 |
| [6] |
# Question 7:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = z\sqrt{(1/4)}$ | M1 | From $\sigma/\sqrt{n}$ |
| $z = 2.326$ | B1 | |
| $a = 1.163$ | A1 | |
| | **[3]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normal with mean 0 | B1 | |
| $\sigma^2 = \frac{1}{4} + \frac{9}{16} = 0.8125$ or $\frac{13}{16}$ | B1 | Allow 0.813 |
| | **[2]** | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Either $\bar{x} + 1.163 < \bar{y} - 1.234$ or $\bar{y} + 1.234 < \bar{x} - 1.163$ | M1 | Ft their $a$ for M1M1. Can be implied. |
| giving $\bar{x} - \bar{y} < -2.397$ or $> 2.397$ | M1, A1 | |
| $P(<-2.397) = P(Z < -2.659)$ oe $= 0.0039(2)$ | M1, A1 | Correct use of N for their interval(s). 0.0039 from tables. |
| So required probability $= 0.0078(4)$ | A1ft | ART 0.0078, ft $2\times$ their 0.0039 |
| | **[6]** | |
---7 The random variable $X$ has distribution $\mathrm { N } ( \mu , 1 )$. A random sample of 4 observations of $X$ is taken. The sample mean is denoted by $\bar { X }$.\\
(i) Find the value of the constant $a$ for which ( $\bar { X } - a , \bar { X } + a$ ) is a $98 \%$ confidence interval for $\mu$.
The independent random variable $Y$ has distribution $\mathrm { N } ( \mu , 9 )$. A random sample of 16 observations of $Y$ is taken. The sample mean is denoted by $\bar { Y }$.\\
(ii) Write down the distribution of $\bar { X } - \bar { Y }$.\\
(iii) A $90 \%$ confidence interval for $\mu$ based on $\bar { Y }$ is given by ( $\bar { Y } - 1.234 , \bar { Y } + 1.234$ ). Find the probability that this interval does not overlap with the interval in part (i).
\hfill \mbox{\textit{OCR S3 2013 Q7 [11]}}