| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | General probability threshold |
| Difficulty | Moderate -0.3 This is a straightforward three-part binomial distribution question covering standard techniques: direct binomial calculation for small n, solving an inequality for minimum sample size, and normal approximation for large n. All parts are routine applications of textbook methods with no novel problem-solving required, making it 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 |
| \(P(4,5,6) = {}^6C_4\, 0.35^4\, 0.65^2 + {}^6C_5\, 0.35^5\, 0.65^1 + 0.35^6\) | M1 | Binomial term of form \({}^6C_x p^x(1-p)^{6-x}\), \(0 < p < 1\), any \(p\), \(x \neq 6, 0\) |
| Correct unsimplified answer | A1 | |
| \(= 0.117\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.65^n > 0.95\); \(0.65^n < 0.05\) | M1 | Equation or inequality involving '\(0.65^n\) or \(0.35^n\)' and '0.95 or 0.05' |
| \(n > \frac{\log 0.05}{\log 0.65} = 6.95\) | M1 | Attempt to solve their exponential equation using logs or Trial and Error |
| \(n = 7\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 0.35 \times 100 = 35\); Variance \(= 0.35 \times 0.65 \times 100 = 22.75\) | B1 | Correct unsimplified \(np\) and \(npq\) |
| \(P\!\left(z > \frac{39.5 - 35}{\sqrt{22.75}}\right) = P(z > 0.943)\) | M1 | Substituting their \(\mu\) and \(\sigma\) (condone \(\sigma^2\)) into the \(\pm\) standardisation formula with a numerical value for '39.5' |
| M1 | Using continuity correction 39.5 or 40.5 | |
| \(= 1 - 0.8272\) | M1 | Appropriate area \(\Phi\) from standardisation formula \(P(z>\ldots)\) in final solution |
| \(= 0.173\) | A1 | Final answer |
## Question 6:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(4,5,6) = {}^6C_4\, 0.35^4\, 0.65^2 + {}^6C_5\, 0.35^5\, 0.65^1 + 0.35^6$ | M1 | Binomial term of form ${}^6C_x p^x(1-p)^{6-x}$, $0 < p < 1$, any $p$, $x \neq 6, 0$ |
| Correct unsimplified answer | A1 | |
| $= 0.117$ | A1 | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.65^n > 0.95$; $0.65^n < 0.05$ | M1 | Equation or inequality involving '$0.65^n$ or $0.35^n$' and '0.95 or 0.05' |
| $n > \frac{\log 0.05}{\log 0.65} = 6.95$ | M1 | Attempt to solve their exponential equation using logs or Trial and Error |
| $n = 7$ | A1 | CAO |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 0.35 \times 100 = 35$; Variance $= 0.35 \times 0.65 \times 100 = 22.75$ | B1 | Correct unsimplified $np$ and $npq$ |
| $P\!\left(z > \frac{39.5 - 35}{\sqrt{22.75}}\right) = P(z > 0.943)$ | M1 | Substituting their $\mu$ and $\sigma$ (condone $\sigma^2$) into the $\pm$ standardisation formula with a numerical value for '39.5' |
| | M1 | Using continuity correction 39.5 or 40.5 |
| $= 1 - 0.8272$ | M1 | Appropriate area $\Phi$ from standardisation formula $P(z>\ldots)$ in final solution |
| $= 0.173$ | A1 | Final answer |6 The results of a survey by a large supermarket show that $35 \%$ of its customers shop online.\\
(i) Six customers are chosen at random. Find the probability that more than three of them shop online.\\
(ii) For a random sample of $n$ customers, the probability that at least one of them shops online is greater than 0.95 . Find the least possible value of $n$.\\
(iii) For a random sample of 100 customers, use a suitable approximating distribution to find the probability that more than 39 shop online.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [11]}}