| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sample size determination |
| Difficulty | Standard +0.3 This question tests standard application of sampling distribution of the mean with known formulas. Part (i) is routine calculation using σ/√n. Part (ii) requires working backwards from a probability to find n, involving one additional algebraic step (squaring), but remains a direct application of the Central Limit Theorem without requiring conceptual insight or problem-solving beyond textbook methods. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P\left(\overline{W} > 51\right) = P\left(z > \frac{51 - 48.5}{12.4/\sqrt{5}}\right)\) | M1 | Standardising with 51 and mean 48.5 |
| M1 | Standardising using \(\sqrt{5}\) | |
| \(= 1 - \Phi(0.451) = 1 - 0.674 = 0.326\) | A1 [3] | Correct answer |
| (ii) \(z = 1.5\) or \(1.499\) | B1 | 1.5 or 1.499 seen |
| \(\frac{51 - 48.5}{(12.4/\sqrt{n})} = 1.5\) | M1 | Standardising must have \(\sqrt{n}\) (no cc) |
| \(\sqrt{n} = 6\) | M1 | Attempt to solve equation with \(\sqrt{n}\), their \(z\) in correct answer |
| \(n = 36\) | A1 [4] | Correct answer |
**(i)** $P\left(\overline{W} > 51\right) = P\left(z > \frac{51 - 48.5}{12.4/\sqrt{5}}\right)$ | M1 | Standardising with 51 and mean 48.5
| M1 | Standardising using $\sqrt{5}$
$= 1 - \Phi(0.451) = 1 - 0.674 = 0.326$ | A1 [3] | Correct answer
**(ii)** $z = 1.5$ or $1.499$ | B1 | 1.5 or 1.499 seen
$\frac{51 - 48.5}{(12.4/\sqrt{n})} = 1.5$ | M1 | Standardising must have $\sqrt{n}$ (no cc)
$\sqrt{n} = 6$ | M1 | Attempt to solve equation with $\sqrt{n}$, their $z$ in correct answer
$n = 36$ | A1 [4] | Correct answer3 The weights of pebbles on a beach are normally distributed with mean 48.5 grams and standard deviation 12.4 grams.\\
(i) Find the probability that the mean weight of a random sample of 5 pebbles is greater than 51 grams.\\
(ii) The probability that the mean weight of a random sample of $n$ pebbles is less than 51.6 grams is 0.9332 . Find the value of $n$.
\hfill \mbox{\textit{CAIE S2 2009 Q3 [7]}}