| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Geometric distribution probability |
| Difficulty | Moderate -0.5 This is a straightforward application of normal approximation to binomial distribution with standard continuity correction, followed by a basic geometric distribution calculation. The question requires routine application of formulas rather than problem-solving insight, making it slightly easier than average for A-level statistics. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(p = \frac{1}{6}\): mean \(= np = 90 \times \frac{1}{6} = 15\) | B1 | Correct mean |
| Variance \(= npq = \frac{75}{6}\) | B1 | Correct variance |
| \(P(X < 18) = P\left(Z < \frac{17.5 - 15}{\sqrt{\frac{75}{6}}}\right) = P(Z < 0.7071)\) | M1 | Standardising equation, allow square root, continuity correction |
| \(= 0.760\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 15 > 5\) and \(nq = 75 > 5\), so normal justified | B1 | Both parts needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \left(\frac{5}{6}\right)^6\) | M1 | |
| \(= 0.665\) | A1 | |
| Total: 2 |
## Question 5:
**Part 5(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $p = \frac{1}{6}$: mean $= np = 90 \times \frac{1}{6} = 15$ | B1 | Correct mean |
| Variance $= npq = \frac{75}{6}$ | B1 | Correct variance |
| $P(X < 18) = P\left(Z < \frac{17.5 - 15}{\sqrt{\frac{75}{6}}}\right) = P(Z < 0.7071)$ | M1 | Standardising equation, allow square root, continuity correction |
| $= 0.760$ | A1 | |
| **Total: 4** | | |
**Part 5(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 15 > 5$ and $nq = 75 > 5$, so normal justified | B1 | Both parts needed |
**Part 5(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \left(\frac{5}{6}\right)^6$ | M1 | |
| $= 0.665$ | A1 | |
| **Total: 2** | | |
---5 A fair six-sided die, with faces marked 1, 2, 3, 4, 5, 6, is thrown 90 times.\\
(a) Use an approximation to find the probability that a 3 is obtained fewer than 18 times.\\
(b) Justify your use of the approximation in part (a).\\
On another occasion, the same die is thrown repeatedly until a 3 is obtained.\\
(c) Find the probability that obtaining a 3 requires fewer than 7 throws.\\
\hfill \mbox{\textit{CAIE S1 2020 Q5 [7]}}