| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum versus sum comparison |
| Difficulty | Standard +0.8 This question requires understanding of linear combinations of normal distributions, including the variance of sums and differences, and inverse normal calculations. Part (i) involves comparing two sample totals (requiring recognition that Var(X_A - X_B) = Var(X_A) + Var(X_B) with n=20 for each), while part (ii) requires working backwards from a probability to find sample size using the sampling distribution of the mean. These are non-trivial applications requiring careful setup and multiple conceptual steps beyond routine S2 exercises. |
| Spec | 5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(20 \text{ of } A \sim A^* \sim N(401, 20 \times 0.15^2)\) | \(B1\) | For correct mean for either |
| \(\sim N(401, 20 \times 0.15^2)\) | ||
| \(\sim N(401, 0.45)\) | For variance \(20 \times 0.15^2\) or \(20 \times 0.27^2\) | |
| \(20 \text{ of } B \sim B^* \sim N(401, 1.458)\) | \(B1\) | |
| \(A^* - B^* \sim N(0, 1.908)\) | \(M1\) | For adding their two variances |
| \(P(A^* - B^* > 2) = 1 - \Phi\left(\frac{2 - 0}{\sqrt{1.908}}\right)\) | \(M1\) | For consideration of their \(A^* - B^* > 2\) |
| \(= 1 - \Phi(1.4479)\) | \(M1\) | For standardising and finding correct area |
| \(= 0.0738\) | \(A1\) | For correct answer |
| OR \(\bar{A} \sim N(20.05, 0.15^2/20)\) | \(B1\) | For correct mean for either |
| \(B \sim N(20.05, 0.27^2/20)\) | \(B1\) | For variance \(0.15^2/20\) or \(0.27^2/20\) |
| \(\bar{A} - \bar{B} \sim N(0, 0.00477)\) | \(M1\) | For adding their variances |
| \(P(\bar{A} - \bar{B} > 0.1)\) | \(M1\) | For consideration of their \(\bar{A} - \bar{B} > 0.1\) |
| \(= 1 - \Phi\left(\frac{0.1 - 0}{\sqrt{0.00477}}\right)\) | \(M1\) | For standardising and finding correct area |
| \(= 0.0738\) | \(A1\) | For correct answer |
| (ii) \(1.96 = \frac{20.07 - 20.05}{(0.15/\sqrt{n})}\) | \(M1\) | For an equation of correct form on RHS involving \(\sqrt{n}\) |
| \(B1\) | For 1.96 used | |
| \(n = 216\) | \(M1\) | For solving an equation of correct form (any \(z\)) |
| \(A1\) | For correct answer |
**(i)** $20 \text{ of } A \sim A^* \sim N(401, 20 \times 0.15^2)$ | $B1$ | For correct mean for either
$\sim N(401, 20 \times 0.15^2)$ | |
$\sim N(401, 0.45)$ | | For variance $20 \times 0.15^2$ or $20 \times 0.27^2$
$20 \text{ of } B \sim B^* \sim N(401, 1.458)$ | $B1$ |
$A^* - B^* \sim N(0, 1.908)$ | $M1$ | For adding their two variances
$P(A^* - B^* > 2) = 1 - \Phi\left(\frac{2 - 0}{\sqrt{1.908}}\right)$ | $M1$ | For consideration of their $A^* - B^* > 2$
$= 1 - \Phi(1.4479)$ | $M1$ | For standardising and finding correct area
$= 0.0738$ | $A1$ | For correct answer
**OR** $\bar{A} \sim N(20.05, 0.15^2/20)$ | $B1$ | For correct mean for either
$B \sim N(20.05, 0.27^2/20)$ | $B1$ | For variance $0.15^2/20$ or $0.27^2/20$
$\bar{A} - \bar{B} \sim N(0, 0.00477)$ | $M1$ | For adding their variances
$P(\bar{A} - \bar{B} > 0.1)$ | $M1$ | For consideration of their $\bar{A} - \bar{B} > 0.1$
$= 1 - \Phi\left(\frac{0.1 - 0}{\sqrt{0.00477}}\right)$ | $M1$ | For standardising and finding correct area
$= 0.0738$ | $A1$ | For correct answer
**(ii)** $1.96 = \frac{20.07 - 20.05}{(0.15/\sqrt{n})}$ | $M1$ | For an equation of correct form on RHS involving $\sqrt{n}$
| $B1$ | For 1.96 used
$n = 216$ | $M1$ | For solving an equation of correct form (any $z$)
| $A1$ | For correct answer7 Machine $A$ fills bags of fertiliser so that their weights follow a normal distribution with mean 20.05 kg and standard deviation 0.15 kg . Machine $B$ fills bags of fertiliser so that their weights follow a normal distribution with mean 20.05 kg and standard deviation 0.27 kg .\\
(i) Find the probability that the total weight of a random sample of 20 bags filled by machine $A$ is at least 2 kg more than the total weight of a random sample of 20 bags filled by machine $B$. [6]\\
(ii) A random sample of $n$ bags filled by machine $A$ is taken. The probability that the sample mean weight of the bags is greater than 20.07 kg is denoted by $p$. Find the value of $n$, given that $p = 0.0250$ correct to 4 decimal places.
\hfill \mbox{\textit{CAIE S2 2003 Q7 [10]}}