| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard z-score calculations. Parts (i) and (ii) are routine textbook exercises requiring basic table lookups, while part (iii) adds mild complexity by requiring working backwards from a probability involving two boundaries, but the method is still standard for S1 level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(<570) = P\!\left(z < \frac{570-500}{91.5}\right) = P(z < 0.7650) = 0.7779\) | M1 | Standardising for either 570 or 390, no cc, no sq, no \(\sqrt{}\) |
| \(P(<390) = P\!\left(z < \frac{390-500}{91.5}\right) = P(z < -1.202) = 1 - 0.8853 = 0.1147\) | A1 | One correct \(z\) value |
| A1 | One correct \(\Phi\), final solution | |
| Large: \(0.222\) \((0.2221)\); Small: \(0.115\) \((0.1147)\) | A1 | Correct small and large |
| Medium: \(0.663\) \((0.6632)\) | A1FT | Correct medium rounding to \(0.66\) or ft \(1 -\) (their small + their large) |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.645 = \left(\frac{x-500}{91.5}\right)\) | B1 | \(\pm 1.645\) seen (critical value) |
| M1 | Standardising, accept cc, sq, sq rt | |
| \(x = 651\) | A1 | \(650 \leq \text{Ans} \leq 651\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(x > 610) = 0.1147\) (symmetry) | M1 | Attempt to find upper end prob \(x > 610\) or \(\Phi(x)\), ft their \(P(<390)\) from (i) |
| \(0.3 + 0.1147 = 0.4147 \Rightarrow \Phi(x) = 0.5853\) | M1 | Adding \(0.3\) to their \(P(x>610)\) or subt \(0.5\) from \(\Phi(x)\) or \(0.8853 - 0.3\) |
| \(z = 0.215\) or \(0.216\) | M1 | Finding \(z = \Phi^{-1}(0.5853)\) |
| \(0.215 = \frac{k-500}{91.5}\) | M1 | Standardising and solving, accept cc, sq, sq rt |
| \(k = 520\) | A1 | |
| 5 |
## Question 7:
### Part 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(<570) = P\!\left(z < \frac{570-500}{91.5}\right) = P(z < 0.7650) = 0.7779$ | M1 | Standardising for either 570 or 390, no cc, no sq, no $\sqrt{}$ |
| $P(<390) = P\!\left(z < \frac{390-500}{91.5}\right) = P(z < -1.202) = 1 - 0.8853 = 0.1147$ | A1 | One correct $z$ value |
| | A1 | One correct $\Phi$, final solution |
| Large: $0.222$ $(0.2221)$; Small: $0.115$ $(0.1147)$ | A1 | Correct small and large |
| Medium: $0.663$ $(0.6632)$ | A1FT | Correct medium rounding to $0.66$ or ft $1 -$ (their small + their large) |
| | **5** | |
### Part 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.645 = \left(\frac{x-500}{91.5}\right)$ | B1 | $\pm 1.645$ seen (critical value) |
| | M1 | Standardising, accept cc, sq, sq rt |
| $x = 651$ | A1 | $650 \leq \text{Ans} \leq 651$ |
| | **3** | |
### Part 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(x > 610) = 0.1147$ (symmetry) | M1 | Attempt to find upper end prob $x > 610$ or $\Phi(x)$, ft their $P(<390)$ from **(i)** |
| $0.3 + 0.1147 = 0.4147 \Rightarrow \Phi(x) = 0.5853$ | M1 | Adding $0.3$ to their $P(x>610)$ or subt $0.5$ from $\Phi(x)$ or $0.8853 - 0.3$ |
| $z = 0.215$ or $0.216$ | M1 | Finding $z = \Phi^{-1}(0.5853)$ |
| $0.215 = \frac{k-500}{91.5}$ | M1 | Standardising and solving, accept cc, sq, sq rt |
| $k = 520$ | A1 | |
| | **5** | |7 The weight, in grams, of pineapples is denoted by the random variable $X$ which has a normal distribution with mean 500 and standard deviation 91.5. Pineapples weighing over 570 grams are classified as 'large'. Those weighing under 390 grams are classified as 'small' and the rest are classified as 'medium'.\\
(i) Find the proportions of large, small and medium pineapples.\\
(ii) Find the weight exceeded by the heaviest $5 \%$ of pineapples.\\
(iii) Find the value of $k$ such that $\mathrm { P } ( k < X < 610 ) = 0.3$.\\
\hfill \mbox{\textit{CAIE S1 2017 Q7 [13]}}