| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a straightforward two-part question applying standard binomial probability (part i) and normal approximation to binomial (part ii). Part (i) requires calculating P(X > 2) = 1 - P(X ≤ 2) for X ~ B(30, 0.04), which is routine. Part (ii) involves recognizing that the number of rejected boxes follows a binomial distribution, then applying the normal approximation with continuity correction. Both parts are textbook applications with no novel insight required, making this slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(P(>2) = 1 - P(0,1,2)\) | (M1 | Binomial term of form \({}^{30}C_x p^x (1-p)^{30-x}\), \(0 < p < 1\) any \(p\) |
| \(= 1 - (0.96)^{30} - {}^{30}C_1(0.04)(0.96)^{29} - {}^{30}C_2(0.04)^2(0.96)^{28}\) | A1 | Correct unsimplified answer |
| \(= 1 - 0.883103 = 0.117\) (0.116896) | A1) | |
| OR: \(P(>2) = P(3,4,5,\ldots,30)\) | (M1 | Binomial term of form \({}^{30}C_x p^x (1-p)^{30-x}\), \(0 < p < 1\) any \(p\) |
| \(= {}^{30}C_3(0.04)^3(0.96)^{27} + {}^{30}C_4(0.04)^4(0.96)^{26} + \ldots + (0.04)^{30}\) | A1 | Correct unsimplified answer |
| \(= 0.117\) | A1) | |
| 3 total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(np = 280 \times 0.1169 = 32.73\), \(npq = 280 \times 0.1169 \times 0.8831 = 28.9\) | M1 FT | Correct unsimplified \(np\) and \(npq\), FT their \(p\) from (i) |
| \(P(\geq 30) = P\left(z > \frac{29.5 - 32.73}{\sqrt{28.9}}\right) = P(z > -0.6008)\) | M1 | Substituting their \(\mu\) and \(\sigma\) (\(\sqrt{npq}\) only) into standardisation formula |
| M1 | Using continuity correction of 29.5 or 30.5 | |
| M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z > \ldots)\) in final solution | |
| \(= 0.726\) | A1 | |
| 5 total |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| EITHER: $P(>2) = 1 - P(0,1,2)$ | (M1 | Binomial term of form ${}^{30}C_x p^x (1-p)^{30-x}$, $0 < p < 1$ any $p$ |
| $= 1 - (0.96)^{30} - {}^{30}C_1(0.04)(0.96)^{29} - {}^{30}C_2(0.04)^2(0.96)^{28}$ | A1 | Correct unsimplified answer |
| $= 1 - 0.883103 = 0.117$ (0.116896) | A1) | |
| OR: $P(>2) = P(3,4,5,\ldots,30)$ | (M1 | Binomial term of form ${}^{30}C_x p^x (1-p)^{30-x}$, $0 < p < 1$ any $p$ |
| $= {}^{30}C_3(0.04)^3(0.96)^{27} + {}^{30}C_4(0.04)^4(0.96)^{26} + \ldots + (0.04)^{30}$ | A1 | Correct unsimplified answer |
| $= 0.117$ | A1) | |
| | **3 total** | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $np = 280 \times 0.1169 = 32.73$, $npq = 280 \times 0.1169 \times 0.8831 = 28.9$ | M1 FT | Correct unsimplified $np$ and $npq$, FT their $p$ from (i) |
| $P(\geq 30) = P\left(z > \frac{29.5 - 32.73}{\sqrt{28.9}}\right) = P(z > -0.6008)$ | M1 | Substituting their $\mu$ and $\sigma$ ($\sqrt{npq}$ only) into standardisation formula |
| | M1 | Using continuity correction of 29.5 or 30.5 |
| | M1 | Appropriate area $\Phi$ from standardisation formula $P(z > \ldots)$ in final solution |
| $= 0.726$ | A1 | |
| | **5 total** | |
---5 Blank CDs are packed in boxes of 30 . The probability that a blank CD is faulty is 0.04 . A box is rejected if more than 2 of the blank CDs are faulty.\\
(i) Find the probability that a box is rejected.\\
(ii) 280 boxes are chosen randomly. Use an approximation to find the probability that at least 30 of these boxes are rejected.\\
\hfill \mbox{\textit{CAIE S1 2017 Q5 [8]}}