| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Normal approximation to binomial |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution and normal approximation. Part (i) requires calculating P(X ≥ 3) using binomial tables or complement rule. Part (ii) is a standard normal approximation with continuity correction where n=56, p=1/7 gives np=8 and npq≈6.86, both satisfying approximation conditions. The question follows a predictable template with no conceptual challenges beyond applying learned procedures. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\geq 3) = 1 - P(0, 1, 2)\) | M1 | For attempt at \(1 - P(0,1,2)\) or \(1 - P(0,1,2,3)\) or \(P(3\ldots15)\) or \(P(4\ldots15)\) |
| \(= 1 - (6/7)^{15} - {}_{15}C_1(1/7)(6/7)^{14} - {}_{15}C_2(1/7)^2(6/7)^{13}\) | M1 | For 1 or more terms with \(1/7\) and \(6/7\) to powers which sum to 15 and \({}_{15}C_{\text{something}}\) |
| \(= 1 - 0.0990 - 0.2476 - 0.2889\) | A1 | Completely correct unsimplified form |
| \(= 0.365\) (accept 0.364) | A1 4 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu = 56 \times 1/7\ (= 8)\); \(\sigma^2 = 56 \times 1/7 \times 6/7\ (= 6.857)\) | B1 | 8 and 6.857 or 6.86 or 2.618 seen or implied |
| \(P(\text{more than } 7) = 1 - \Phi\!\left(\dfrac{7.5 - 8}{\sqrt{6.857}}\right)\) | M1 | Standardising attempt with or without cc, must have square root |
| \(= \Phi\!\left(\dfrac{8 - 7.5}{\sqrt{6.857}}\right) = \Phi(0.1909)\) | M1 | Continuity correction either 7.5 or 6.5 |
| M1 | Final answer \(> 0.5\) (award this if the long way is used and the final answer is \(> 0.5\)) | |
| \(= 0.576\) | A1 5 | Correct final answer |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\geq 3) = 1 - P(0, 1, 2)$ | M1 | For attempt at $1 - P(0,1,2)$ or $1 - P(0,1,2,3)$ or $P(3\ldots15)$ or $P(4\ldots15)$ |
| $= 1 - (6/7)^{15} - {}_{15}C_1(1/7)(6/7)^{14} - {}_{15}C_2(1/7)^2(6/7)^{13}$ | M1 | For 1 or more terms with $1/7$ and $6/7$ to powers which sum to 15 and ${}_{15}C_{\text{something}}$ |
| $= 1 - 0.0990 - 0.2476 - 0.2889$ | A1 | Completely correct unsimplified form |
| $= 0.365$ (accept 0.364) | A1 **4** | Correct final answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu = 56 \times 1/7\ (= 8)$; $\sigma^2 = 56 \times 1/7 \times 6/7\ (= 6.857)$ | B1 | 8 and 6.857 or 6.86 or 2.618 seen or implied |
| $P(\text{more than } 7) = 1 - \Phi\!\left(\dfrac{7.5 - 8}{\sqrt{6.857}}\right)$ | M1 | Standardising attempt with or without cc, must have square root |
| $= \Phi\!\left(\dfrac{8 - 7.5}{\sqrt{6.857}}\right) = \Phi(0.1909)$ | M1 | Continuity correction either 7.5 or 6.5 |
| | M1 | Final answer $> 0.5$ (award this if the long way is used and the final answer is $> 0.5$) |
| $= 0.576$ | A1 **5** | Correct final answer |
---6 The probability that New Year's Day is on a Saturday in a randomly chosen year is $\frac { 1 } { 7 }$.\\
(i) 15 years are chosen randomly. Find the probability that at least 3 of these years have New Year's Day on a Saturday.\\
(ii) 56 years are chosen randomly. Use a suitable approximation to find the probability that more than 7 of these years have New Year's Day on a Saturday.
\hfill \mbox{\textit{CAIE S1 2007 Q6 [9]}}